Author Archives: Bherulal

Formation of Urine

Posted on by
Reply

The nephrons of the kidneys process blood and create urine through a process of filtration, reabsorption, and secretion. Urine is about 95% water and 5% waste products. Nitrogenous wastes excreted in urine include urea, creatinine, ammonia, and uric acid.

Urine formation depends on three functions:

  • Filtration is accomplished by the movement of fluids from the blood into the Bowman’s Capsule
  • Reabsorption involves the transfer of essential solutes and water from the nephron back into the blood
  • Secretion involves the movement of materials from the blood back into the nephron
  • For a detailed diagram and summary, see Fig. 1 on p. 350 and Table 2 on p. 351

Filtration

  • Blood running through the afferent arteriole into the glomerulus is under high pressure (65 mmHg compared to 25 mmHg normally found in capillary beds)
  • Most dissolved solutes (see Table 1, p. 349 for a list) pass through the walls of the glomerulus into the Bowman’s capsule

Reabsorption

  • On average, 600 mL of fluid flows through the kidneys every minute
  • About 20% (or 120 mL) is filtered into the nephron
  • If all of that fluid left in urine, dehydration would be a constant danger
  • Fortunately, only 1 mL of urine is formed for every 120 mL, meaning that 119 mL of fluids and solutes are reabsorbed
  • Selective reabsorption occurs by both active and passive transport
  • Carrier molecules move Na+ ions across the cell membranes of the cells that line the nephron
  • Negative ions (Cl- and HCO3-) follow the positive Na+ ions by charge attraction
  • Many mitochondria supply energy needed for active transport
  • Reabsorption occurs until the threshold level of a substance is reached
  • Excess (like NaCl) remains in the nephron and is excreted with urine
  • Other molecules are actively transported from the proximal tubule
  • Glucose and amino acids attach to specific carrier molecules, which shuttle them out of the nephron and into the blood
  • The amount of solute that can be reabsorbed is limited
  • Ex – individuals with high blood glucose will excrete some in their urine
  • The solutes that are actively transported out of the nephron create an osmotic gradient that draws water in from the nephron
  • A second osmotic force, created by the proteins not filtered into the nephron, also help reabsorption
  • The proteins remain in the blood stream and draw water from the interstitial fluid into the blood
  • As water is reabsorbed from the nephron, the remaining solutes become more concentrated
  • Molecules like urea and uric acid will diffuse from the nephron back into the blood

Secretion

  • Secretion is the movement of wastes from the blood into the nephron
  • Nitrogen containing wastes, excess H+ ions, and minerals like K+ ions are examples of substances secreted
  • Drugs (like penicillin) can also be secreted
  • Cells loaded with mitochondria line the distal tubule, providing energy for active transport

What is vital capacity and its importance?

Posted on by
Reply

What is vital capacity and its importance?

Vital capacity (VC) is the maximum amount of air a person can expel from the lungs after a maximum inspiration. It is equal to the sum of inspiratory reserve volume, tidal volume, and expiratory reserve volume i.e, VC = IRV + TV + ERV.

The breathing capacity of the lungs expressed as the number of cubic inches or cubic centimeters of air that can be forcibly exhaled after a full inspiration. It is about 3.5 – 4.5 liters in the human body.

It promotes the act of supplying fresh air and getting rid of foul air, thereby increasing the gaseous exchange between the tissues and the environment. Thus, the greater the VC, the more is the energy available to the body. VC of a person gives important clues for diagnosing a lung problem.

Its measurement helps the doctor to decide about the possible causes of the diseases and about the line of treatment.

It determines the stamina of sportsperson and mountain climbers. Sportsperson and mountain dwellers have a higher vital capacity. Young persons have more vital capacity than aged.

Homeostasis

Posted on by
Reply

The term homeostasis is used by physiologists to mean maintenance of nearly constant conditions in the internal environment. Essentially all organs and tissues of the body perform functions that help maintain these relatively constant conditions. For instance, the lungs provide oxygen to the extracellular fluid to replenish the oxygen used by the cells, the kidneys maintain constant ion concentrations, and the gastrointestinal system provides nutrients.

The term homeostasis comes from the Ancient Greek ὅμοιος (hómoios (homeo), meaning “similar”), from στημι (hístēmi (stasis), “standing still”) and stasis, from στάσις (stásis, meaning “standing”). The concept of homeostasis was first described in 1865 by Claude Bernard, a French physiologist. However, the term was coined later in 1962 by the American physiologist Walter Bradford Cannon.

Concept of Extracellular and Intracellular fluid

  • Intracellular fluid—fluid contained within all of the cells of the body
  • Extracellular fluid—fluid outside the cells of the body and is the internal environment in which the cells live. It is composed of plasma and interstitial fluid.

 

 

 

 

 

 

Interdependent relationship of cells, systems and homeostasis

Factors homeostatically regulated

  • Concentration of nutrient molecules
  • Concentration of CO2 andO2
  • Concentration of waste products
  • pH
  • Concentration of water, salt and other electrolytes
  • Temperature
  • Volume and pressure

Contribution of body systems to Homeostasis

  • The Circulatory System
    • Extracellular fluid is transported through all parts of the body in two stages. The first stage is movement of blood through the body in the blood vessels, and the second is movement of fluid between the blood capillaries and the intercellular spaces between the tissue cells.
    • All the blood in the circulation traverses the entire circulatory circuit an average of once each minute when the body is at rest and as many as six times each minute when a person is extremely active.
    • As blood passes through the blood capillaries, continual exchange of extracellular fluid also occurs between the plasma portion of the blood and the interstitial fluid that fills the intercellular spaces.
  • The Digestive System
    • A large portion of the blood pumped by the heart also passes through the walls of the gastrointestinal tract. Here different dissolved nutrients, including carbohydrates, fatty acids, and amino acids, are absorbed from the ingested food into the extracellular fluid of the blood.
    • Undigested material that enters the gastrointestinal tract and some waste products of metabolism are eliminated in the feces.
  • The Respiratory System
    • The blood picks up oxygen in the alveoli, thus acquiring the oxygen needed by the cells. The membrane between the alveoli and the lumen of the pulmonary capillaries, the alveolar membrane, is only 0.4 to 2.0 micrometers thick, and oxygen rapidly diffuses by molecular motion through this membrane into the blood.
    • Removal of Carbon Dioxide by the Lungs. At the same time that blood picks up oxygen in the lungs, carbon dioxide is released from the blood into the lung alveoli; the respiratory movement of air into and out of the lungs carries the carbon dioxide to the atmosphere. Carbon dioxide is the most abundant of all the end products of metabolism.
  • The Urinary System
    • Passage of the blood through the kidneys removes from the plasma most of the other substances besides carbon dioxide that are not needed by the cells.
    • These substances include different end products of cellular metabolism, such as urea and uric acid; they also include excesses of ions and water from the food that might have accumulated in the extracellular fluid.
    • The kidneys perform their function by first filtering large quantities of plasma through the glomeruli into the tubules and then reabsorbing into the blood those substances needed by the body, such as glucose, amino acids, appropriate amounts of water, and many of the ions. Most of the other substances that are not needed by the body, especially the metabolic end products such as urea, are reabsorbed poorly and pass through the renal tubules into the urine.
  • The Musculoskeletal system
    • How does the musculoskeletal system contribute to homeostasis? The answer is obvious and simple: Were it not for the muscles, the body could not move to the appropriate place at the appropriate time to obtain the foods required for nutrition. The musculoskeletal system also provides motility for protection against adverse surroundings, without which the entire body, along with its homeostatic mechanisms, could be destroyed instantaneously.
  • The Integumentary System
    • The skin and its various appendages, including the hair, nails, glands, and other structures, cover, cushion, and protect the deeper tissues and organs of the body and generally provide a boundary between the body’s internal environment and the outside world. The integumentary system is also important for temperature regulation and excretion of wastes and it provides a sensory interface between the body and the external environment. The skin generally comprises about 12 to 15 percent of body weight.
  • The Immune System
    • The immune system consists of the white blood cells, tissue cells derived from white blood cells, the thymus, lymph nodes, and lymph vessels that protect the body from pathogens such as bacteria, viruses, parasites, and fungi. The immune system provides a mechanism for the body to (1) distinguish its own cells from foreign cells and substances and (2) destroy the invader by phagocytosis or by producing sensitized lymphocytes or specialized proteins (e.g., antibodies) that either destroy or neutralize the invader.
  • The Nervous System
    • The nervous system is composed of three major parts: the sensory input portion, the central nervous system (or integrative portion), and the motor output portion. Sensory receptors detect the state of the body or the state of the surroundings. For instance, receptors in the skin apprise one whenever an object touches the skin at any point. The eyes are sensory organs that give one a visual image of the surrounding area. The ears are also sensory organs. The central nervous system is composed of the brain and spinal cord. The brain can store information, generate thoughts, create ambition, and determine reactions that the body performs in response to the sensations. Appropriate signals are then transmitted through the motor output portion of the nervous system to carry out one’s desires.
  • The Endocrine System
    • Located in the body are eight major endocrine glands that secrete chemical substances called hormones. Hormones are transported in the extracellular fluid to all parts of the body to help regulate cellular function. For instance, thyroid hormone increases the rates of most chemical reactions in all cells, thus helping to set the tempo of bodily activity. Insulin controls glucose metabolism; adrenocortical hormones control sodium ion, potassium ion, and protein metabolism; and parathyroid hormone controls bone calcium and phosphate. Thus, the hormones provide a system for regulation that complements the nervous system.
  • Reproductive system
    • Sometimes reproduction is not considered a homeostatic function. It does, however, help maintain homeostasis by generating new beings to take the place of those that are dying. This may sound like a permissive usage of the term homeostasis, but it illustrates that, in the final analysis, essentially all body structures are organized such that they help maintain the automaticity and continuity of life.

Homeostatic Control System

It is a functionally interconnected network of body components that operate to maintain a given physical or chemical factor in the internal environment relatively constant around an optimal level.

The human body has thousands of control systems. The most intricate of these are the genetic control systems that operate in all cells to help control intracellular function and extracellular functions.

Many other control systems operate within the organs to control functions of the individual parts of the organs; others operate throughout the entire body to control the interrelations between the organs. For instance, the respiratory system, operating in association with the nervous system, regulates the concentration of carbon dioxide in the extracellular fluid. The liver and pancreas regulate the concentration of glucose in the extracellular fluid, and the kidneys regulate concentrations of hydrogen, sodium, potassium, phosphate, and other ions in the extracellular fluid. Can be classified as:

  • Intrinsic (local controls) are inherent compensatory responses of an organ to a change
  • Extrinsic controls are responses of an organ that are triggered by factors external to the organ, namely, by the nervous and endocrine systems

Both intrinsic and extrinsic control systems generally operate on the principle of

  • Negative feedback mechanism

In addition

  • Positive feedback mechanism
  • Feedforward mechanism

Negative feedback is a mechanism that reverses a deviation from the set point. Therefore, negative feedback maintains body parameters within their normal range. The maintenance of homeostasis by negative feedback goes on throughout the body at all times, and an understanding of negative feedback is thus fundamental to an understanding of human physiology.

Change in a homeostatically control factor triggers a response that seeks to restore the factor to normal by moving the factor in the opposite direction of its initial change or it is a pathway where the response opposes or removes the signal.

If the results suppress or stops the original signal, then it is said to be negative feedback mechanism.

Most of the mechanisms of the body belongs to this category

Feedback loop steps:

 

C:\Users\Bherulal\OneDrive\Pictures\Screenshots\Screenshot (785).png

  • Stimulus
  • Sensor
  • Integrator
  • Effector
  • Response
  • Result

 

 

 

Examples of Negative feedback:

Blood Pressure:

C:\Users\Bherulal\OneDrive\Pictures\Screenshots\Screenshot (789).png

CO2

C:\Users\Bherulal\OneDrive\Pictures\Screenshots\Screenshot (790).png

Thermoregulation:

C:\Users\Bherulal\OneDrive\Pictures\Screenshots\Screenshot (786).png

Positive feedback

Positive feedback intensifies a change in the body’s physiological condition rather than reversing it. A deviation from the normal range results in more change, and the system moves farther away from the normal range. Positive feedback in the body is normal only when there is a definite end point. Childbirth and the body’s response to blood loss are two examples of positive feedback loops that are normal but are activated only when needed.

Childbirth at full term is an example of a situation in which the maintenance of the existing body state is not desired. Enormous changes in the mother’s body are required to expel the baby at the end of pregnancy. And the events of childbirth, once begun, must progress rapidly to a conclusion or the life of the mother and the baby are at risk. The extreme muscular work of labor and delivery are the result of a positive feedback system.

  • The output is continually enhanced or amplified so that the controlled variable continues to be moved in the direction of the initial change or a pathway in which the response reinforces the stimulus.
  • It is also precisely defined as “if the response enhances the original stimulus it is called positive feedback mechanism.
  • It is also called vicious cycle in some cases.

Examples:

Childbirth:

C:\Users\Bherulal\OneDrive\Pictures\Screenshots\Screenshot (788).png

Blood clotting:

C:\Users\Bherulal\OneDrive\Pictures\Screenshots\Screenshot (787).png

Breastfeeding:

C:\Users\Bherulal\OneDrive\Pictures\Screenshots\Screenshot (784).png

Feedforward Mechanism

It brings about a compensatory response in anticipation of a change in a regulated variable.

In this mechanism our body gives response prior to signals sent by brain.

Examples

  • Production of saliva in the mouth when we see the food.
  • Secretion of HCL in stomach when we engulf the food.
  • Drawback of our hand instantly when we touch a hot pot.

Key Points

  • Homeostatic control mechanisms have at least three interdependent components: a receptor, integrating center, and effector.
  • The receptor senses environmental stimuli, sending the information to the integrating center.
  • The integrating center, generally a region of the brain called the hypothalamus, signals an effector (e.g. muscles or an organ) to respond to the stimuli.
  • Positive feedback enhances or accelerates output created by an activated stimulus. Platelet aggregation and accumulation in response to injury is an example of positive feedback.
  • Negative feedback brings a system back to its level of normal functioning. Adjustments of blood pressure, metabolism, and body temperature are all negative feedback.
  • Many diseases are a result of homeostatic imbalance, an inability of the body to restore a functional, stable internal environment.
  • Aging is a source of homeostatic imbalance as the control mechanisms of the feedback loops lose their efficiency, which can cause heart failure.
  • Diseases that result from a homeostatic imbalance include heart failure and diabetes, but many more examples exist.
  • Diabetes occurs when the control mechanism for insulin becomes imbalanced, either because there is a deficiency of insulin or because cells have become resistant to insulin.
  • Homeostasis is the ability of a system to regulate its internal environment through maintaining a stable, relatively constant set of properties such as temperature and pH.

Inferential Statistics

Posted on by
Reply

Statistical inference is the procedure by which we reach a conclusion about a population on the basis of the information contained in a sample drawn from that population. It consists of two techniques:

  • Estimation of parameters
  • Hypothesis testing

ESTIMATION OF PARAMETERS

The process of estimation entails calculating, from the data of a sample, some statistic that is offered as an approximation of the corresponding parameter of the population from which the sample was drawn.

Parameter estimation is used to estimate a single parameter, like a mean.

There are two types of estimates

  • Point Estimates
  • Interval Estimates (Confidence Interval).

POINT ESTIMATES

A point estimate is a single numerical value used to estimate the corresponding population parameter.

For example: the sample mean ‘x’ is a point estimate of the population mean μ. the sample variance S2 is a point estimate of the population variance σ2. These are point estimates — a single–valued guess of the parametric value.

A good estimator must satisfy three conditions:

  1. Unbiased: The expected value of the estimator must be equal to the mean of the parameter
  2. Consistent: The value of the estimator approaches the value of the parameter as the sample size increases
  3. Relatively Efficient: The estimator has the smallest variance of all estimators which could be used

CONFIDENCE INTERVAL (Interval Estimates)

An interval estimate consists of two numerical values defining a range of values that, with a specified degree of confidence, most likely includes the parameter being estimated.

Interval estimation of a parameter is more useful because it indicates a range of values within which the parameter has a specified probability of lying. With interval estimation, researchers construct a confidence interval around estimate; the upper and lower limits are called confidence limits.

Interval estimates provide a range of values for a parameter value, within which we have a stated degree of confidence that the parameter lies. A numeric range, based on a statistic and its sampling distribution that contains the population parameter of interest with a specified probability.

A confidence interval gives an estimated range of values which is likely to include an unknown population parameter, the estimated range being calculated from a given set of sample data

Calculating confidence interval when n ≥ 30 (Single Population Mean)

Example: A random sample of size 64 with mean 25 & Standard Deviation 4 is taken from a normal population. Construct 95 % confidence interval

We use following formula to solve Confidence Interval when n ≥ 30

Data

  • = 25

= 4

n = 64

25 4/ . x 1.96

25 4/8 x 1.96

25 0.5 x 1.96

25 0.98

25 – 0.98 ≤ µ ≤ 25 + 0.98

24.02≤ µ ≤ 25.98

We are 95% confident that population mean (µ) will have value between 24.02 & 25.98

Calculating confidence interval when n < 30 (Single Population Mean)

Example: A random sample of size 9 with mean 25 & Standard Deviation 4 is taken from a normal population. Construct 95 % confidence interval

We use following formula to solve Confidence Interval when n < 30

(OR)

Data

  • = 25

S = 4

n = 9

α/2 = 0.025

df = n – 1 (9 -1 = 8)

tα/2,df = 2.306

25 ± 4/√9 x 2.306

25 ± 4/3 x 2.306

25 ± 1.33 x 2.306

25 ± 3.07

25 – 3.07 ≤ µ ≤ 25 + 3.07

21.93 ≤ µ ≤ 28.07

We are 95% confident that population mean (µ) will have value between 21.93 & 28.07

Hypothesis:

A hypothesis may be defined simply as a statement about one or more populations. It is frequently concerned with the parameters of the populations about which the statement is made.

Types of Hypotheses

Researchers are concerned with two types of hypotheses

  1. Research hypotheses

The research hypothesis is the conjecture or supposition that motivates the research. It may be the result of years of observation on the part of the researcher.

  1. Statistical hypotheses

Statistical hypotheses are hypotheses that are stated in such a way that they may be evaluated by appropriate statistical techniques.

Types of statistical Hypothesis

There are two statistical hypotheses involved in hypothesis testing, and these should be stated explicitly.

  1. Null Hypothesis:

The null hypothesis is the hypothesis to be tested. It is designated by the symbol Ho. The null hypothesis is sometimes referred to as a hypothesis of no difference, since it is a statement of agreement with (or no difference from) conditions presumed to be true in the population of interest.

In general, the null hypothesis is set up for the express purpose of being discredited. Consequently, the complement of the conclusion that the researcher is seeking to reach becomes the statement of the null hypothesis. In the testing process the null hypothesis either is rejected or is not rejected. If the null hypothesis is not rejected, we will say that the data on which the test is based do not provide sufficient evidence to cause rejection. If the testing procedure leads to rejection, we will say that the data at hand are not compatible with the null hypothesis, but are supportive of some other hypothesis.

 

  1. Alternative Hypothesis

The alternative hypothesis is a statement of what we will believe is true if our sample data cause us to reject the null hypothesis. Usually the alternative hypothesis and the research hypothesis are the same, and in fact the two terms are used interchangeably. We shall designate the alternative hypothesis by the symbol HA orH1.

LEVEL OF SIGNIFICANCE

The level of significance is a probability and, in fact, is the probability of rejecting a true null hypothesis. The level of significance specifies the area under the curve of the distribution of the test statistic that is above the values on the horizontal axis constituting the rejection region. It is denoted by ‘α’.

Types of Error

In the context of testing of hypotheses, there are basically two types of errors:

  • TYPE I Error
  • TYPE II Error

Type I Error

  • A type I error, also known as an error of the first kind, occurs when the null hypothesis (H0) is true, but is rejected.
  • A type I error may be compared with a so called false positive.
  • The rate of the type I error is called the size of the test and denoted by the Greek letter α (alpha).
  • It usually equals the significance level of a test.
  • If type I error is fixed at 5 %, it means that there are about 5 chances in 100 that we will reject H0 when H0 is true.

Type II Error

  • Type II error, also known as an error of the second kind, occurs when the null hypothesis is false, but erroneously fails to be rejected.
  • Type II error means accepting the hypothesis which should have been rejected.
  • A Type II error is committed when we fail to believe a truth.
  • A type II error occurs when one rejects the alternative hypothesis (fails to reject the null hypothesis) when the alternative hypothesis is true.
  • The rate of the type II error is denoted by the Greek letter β (beta) and related to the power of a test (which equals 1-β ).

In the tabular form two errors can be presented as follows:

Null hypothesis (H0) is true Null hypothesis (H0) is false
Reject null hypothesis Type I error
False positive		 Correct outcome
True positive
Fail to reject null hypothesis Correct outcome
True negative		 Type II error
False negative

D:\Type-I-and-II-Errors.topicArticleId-267532,articleId-267497_files\267185.pngGraphical depiction of the relation between Type I and Type II errors

What are the differences between Type 1 errors and Type 2 errors?

Type 1 Error Type 2 Error
  • A type 1 error is when a statistic calls for the rejection of a null hypothesis which is factually true.
  • We may reject H0 when H0 is true is known as Type I error .
  • A type 1 error is called a false positive.
  • It denoted by the Greek letter α (alpha).
  • Null hypothesis and type I error
  • A type 2 error is when a statistic does not give enough evidence to reject a null hypothesis even when the null hypothesis should factually be rejected.
  • We may accept H0 when infect H0 is not true is known as Type II Error.
  • A type 2 error is a false negative.
  • It denoted by “β” (beta)
  • Alternative hypothesis and type II error.

Reducing Type I Errors

  • Prescriptive testing is used to increase the level of confidence, which in turn reduces Type I errors. The chances of making a Type I error are reduced by increasing the level of confidence.

Reducing Type II Errors

  • Descriptive testing is used to better describe the test condition and acceptance criteria, which in turn reduces type ii errors. This increases the number of times we reject the null hypothesis – with a resulting increase in the number of type I errors (rejecting H0 when it was really true and should not have been rejected).
  • Therefore, reducing one type of error comes at the expense of increasing the other type of error! The same means cannot reduce both types of errors simultaneously.

Power of Test:

Statistical power is defined as the probability of rejecting the null hypothesis while the alternative hypothesis is true.

Power = P(reject H0 | H1 is true)

= 1 – P(type II error)

= 1 – β

That is, the power of a hypothesis test is the probability that it will reject when it’s supposed to.

Distribution under H0

Distribution under H1

 Power

Factors that affect statistical power include

  • The sample size
  • The specification of the parameter(s) in the null and alternative hypothesis, i.e. how far they are from each other, the precision or uncertainty the researcher allows for the study (generally the confidence or significance level)
  • The distribution of the parameter to be estimated. For example, if a researcher knows that the statistics in the study follow a Z or standard normal distribution, there are two parameters that he/she needs to estimate, the population mean (μ) and the population variance (σ2). Most of the time, the researcher know one of the parameters and need to estimate the other. If that is not the case, some other distribution may be used, for example, if the researcher does not know the population variance, he/she can estimate it using the sample variance and that ends up with using a T distribution.

Application:

In research, statistical power is generally calculated for two purposes.

  1. It can be calculated before data collection based on information from previous research to decide the sample size needed for the study.
  2. It can also be calculated after data analysis. It usually happens when the result turns out to be non-significant. In this case, statistical power is calculated to verify whether the non-significant result is due to really no relation in the sample or due to a lack of statistical power.

Relation with sample size:

Statistical power is positively correlated with the sample size, which means that given the level of the other factors, a larger sample size gives greater power. However, researchers are also faced with the decision to make a difference between statistical difference and scientific difference. Although a larger sample size enables researchers to find smaller difference statistically significant, that difference may not be large enough be scientifically meaningful. Therefore, this would be recommended that researcher have an idea of what they would expect to be a scientifically meaningful difference before doing a power analysis to determine the actual sample size needed.

HYPOTHESIS TESTING

Statistical hypothesis testing provides objective criteria for deciding whether hypotheses are supported by empirical evidence.

The purpose of hypothesis testing is to aid the clinician, researcher, or administrator in reaching a conclusion concerning a population by examining a sample from that population.

STEPS IN STATISTICAL HYPOTHESIS TESTING

Step # 01: State the Null hypothesis and Alternative hypothesis.

The alternative hypothesis represents what the researcher is trying to prove. The null hypothesis represents the negation of what the researcher is trying to prove.

Step # 02: State the significance level, α (0.01, 0.05, or 0.1), for the test

The significance level is the probability of making a Type I error. A Type I Error is a decision in favor of the alternative hypothesis when, in fact, the null hypothesis is true.

Type II Error is a decision to fail to reject the null hypothesis when, in fact, the null hypothesis is false.

Step # 03: State the test statistic that will be used to conduct the hypothesis test

The appropriate test statistic for different kinds of hypothesis tests (i.e. t-test, z-test, ANOVA, Chi-square etc.) are stated in this step

Step # 04: Computation/ calculation of test statistic

Different kinds of hypothesis tests (i.e. t-test, z-test, ANOVA, Chi-square etc.) are computed in this step.

Step # 05: Find Critical Value or Rejection (critical) Region of the test

Use the value of α (0.01, 0.05, or 0.1) from Step # 02 and the distribution of the test statistics from Step # 03.

Step # 06: Conclusion (Making statistical decision and interpretation of results)

If calculated value of test statistics falls in the rejection (critical) region, the null hypothesis is rejected, while, if calculated value of test statistics falls in the acceptance (noncritical) region, the null hypothesis is not rejected i.e. it is accepted.

Note: In case if we conclude on the basis of p-value then we compare calculated p-value to the chosen level of significance. If p-value is less than α, then the null hypothesis will be rejected and alternative will be affirmed. If p-value is greater than α, then the null hypothesis will not be rejected

If the decision is to reject, the statement of the conclusion should read as follows: “we reject at the _______ level of significance. There is sufficient evidence to conclude that (statement of alternative hypothesis.)”

If the decision is to fail to reject, the statement of the conclusion should read as follows: “we fail to reject at the _______ level of significance. There is no sufficient evidence to conclude that (statement of alternative hypothesis.)”

Rules for Stating Statistical Hypotheses

When hypotheses are stated, an indication of equality (either = ,≤ or ≥ ) must appear in the null hypothesis.

Example:

We want to answer the question: Can we conclude that a certain population mean is not 50? The null hypothesis is

Ho : µ = 50

And the alternative is

HA : µ ≠ 50

Suppose we want to know if we can conclude that the population mean is greater than

50. Our hypotheses are

Ho: µ ≤ 50

HA: µ >

If we want to know if we can conclude that the population mean is less than 50, the hypotheses are

Ho : µ ≥ 50

HA: µ < 50

We may state the following rules of thumb for deciding what statement goes in the null hypothesis and what statement goes in the alternative hypothesis:

  • What you hope or expect to be able to conclude as a result of the test usually should be placed in the alternative hypothesis.
  • The null hypothesis should contain a statement of equality, either = ,≤ or ≥.
  • The null hypothesis is the hypothesis that is tested.
  • The null and alternative hypotheses are complementary. That is, the two together exhaust all possibilities regarding the value that the hypothesized parameter can assume.

T- TEST

T-test is used to test hypotheses about μ when the population standard deviation is unknown and Sample size can be small (n<30).

The distribution is symmetrical, bell-shaped, and similar to the normal but more spread out.

Calculating one sample t-test

Example: A random sample of size 16 with mean 25 and Standard Deviation 5 is taken from a normal population Test at 5% LOS that; : µ= 22

: µ≠22

SOLUTION

Step # 01: State the Null hypothesis and Alternative hypothesis.

: µ= 22

: µ≠22

Step # 02: State the significance level

α = 0.05 or 5% Level of Significance

http://onlinepubs.trb.org/onlinepubs/nchrp/cd-22/v2appendixa_files/image040.gifStep # 03: State the test statistic (n<30)

t-test statistic

Step # 04: Computation/ calculation of test statistic

Data

  • = 25

µ = 22

S = 5

n = 16

t calculated = 2.4

Step # 05: Find Critical Value or Rejection (critical) Region

For critical value we find and on the basis of its answer we see critical value from t-distribution table.

Critical value = α/2(v = 16-1)

= 0.05/2(v = 15)

= (0.025, 15)

t tabulated = ± 2.131

t calculated = 2.4

Step # 06: Conclusion: Since t calculated = 2.4 falls in the region of rejection therefore we reject at the 5% level of significance. There is sufficient evidence to conclude that Population mean is not equal to 22.

Z- TEST

  1. Z-test is applied when the distribution is normal and the population standard deviation σ is known or when the sample size n is large (n ≥ 30) and with unknown σ (by taking S as estimator of σ).
  2. Z-test is used to test hypotheses about μ when the population standard deviation is known and population distribution is normal or sample size is large (n ≥ 30)

Calculating one sample z-test

Example: A random sample of size 49 with mean 32 is taken from a normal population whose standard deviation is 4. Test at 5% LOS that : µ= 25

: µ≠25

SOLUTION

Step # 01: : µ= 25

: µ≠25

Step # 02: α = 0.05

Step # 03:Since (n<30), we apply z-test statistic

http://onlinepubs.trb.org/onlinepubs/nchrp/cd-22/v2appendixa_files/image008.gif

Step # 04: Calculation of test statistic

Data

  • = 32

µ = 25

= 4

n = 49

Zcalculated = 12.28

Step # 05: Find Critical Value or Rejection (critical) Region

Critical Value (5%) (2-tail) = ±1.96

Zcalculated = 12.28

Step # 06: Conclusion: Since Zcalculated = 12.28 falls in the region of rejection therefore we reject at the 5% level of significance. There is sufficient evidence to conclude that Population mean is not equal to 25.

CHI-SQUARE

A statistic which measures the discrepancy (difference) between KObserved Frequencies fo1, fo2… fok and the corresponding ExpectedFrequencies fe1, fe2……. fek

The chi-square is useful in making statistical inferences about categorical data in whichthe categories are two and above.

Characteristics

  1. Every χ2 distribution extends indefinitely to the right from 0.
  2. Every χ2 distribution has only one (right sided) tail.
  3. As df increases, the χ2 curves get more bell shaped and approach the normal curve in appearance (but remember that a chi square curvestarts at 0, not at – ∞ )

Calculating Chi-Square

Example 1: census of U.S. determine four categories of doctors practiced in different areas as

Specialty % Probability
General Practice 18% 0.18
Medical 33.9 % 0.339
Surgical 27 % 0.27
Others 21.1 % 0.211
Total 100 % 1.000

A searcher conduct a test after 5 years to check this data for changes and select 500 doctors and asked their speciality. The result were:

Specialty frequency
General Practice 80
Medical 162
Surgical 156
Others 102
Total 500

Hypothesis testing:

Step 01”

Null Hypothesis (Ho):

There is no difference in specialty distribution (or) the current specialty distribution of US physician is same as declared in the census.

Alternative Hypothesis (HA):

There is difference in specialty distribution of US doctors. (or) the current specialty distribution of US physician is different as declared in the census.

Step 02: Level of Significance

α = 0.05

Step # 03:Chi-squire Test Statistic

Step # 04:

Statistical Calculation

fe (80) = 18 % x 500 = 90

fe (162) = 33.9 % x 500 = 169.5

fe (156) = 27 % x 500 = 135

fe (102) = 21.1 % x 500 = 105.5

S # (n) Specialty fo fe (fo – fe) (fo – fe) 2 (fo – fe) 2 / fe
1 General Practice 80 90 -10 100 1.11
2 Medical 162 169.5 -7.5 56.25 0.33
3 Surgical 156 135 21 441 3.26
4 Others 102 105.5 -3.5 12.25 0.116
  4.816

χ2cal= = 4.816

Step # 05:

Find critical region using X2– chi-squire distribution table

χ2 = χ2 = χ2 = 7.815

tab (α,d.f) (0.05,3)

(d.f = n – 1)

Step # 06:

Conclusion: Since χ2cal value lies in the region of acceptance therefore we accept the HO and reject HA. There is no difference in specialty distribution among U.S. doctors.

Example2: A sample of 150 chronic Carriers of certain antigen and a sample of 500 Non-carriers revealed the following blood group distributions. Can one conclude from these data that the two population from which samples were drawn differ with respect to blood group distribution? Let α = 0.05.

Blood Group Carriers Non-carriers Total
O 72 230 302
A 54 192 246
B 16 63 79
AB 8 15 23
Total 150 500 650

Hypothesis Testing

Step # 01: HO: There is no association b/w Antigen and Blood Group

HA: There is some association b/w Antigen and Blood Group

Step # 02:α = 0.05

Step # 03:Chi-squire Test Statistic

Step # 04:

Calculation

fe (72) = 302*150/650 = 70

fe (230) = 302*500/ 650 = 232

fe (54) = 246*150/650 = 57

fe (192) = 246*500/650 = 189

fe (16) = 79*150/650 = 18

fe (63) = 79*500/650 = 61

fe (8) = 23*150/650 = 05

fe (15) = 23*500/650 = 18

fo fe (fo – fe) (fo – fe) 2 (fo – fe) 2 / fe
72 70 2 4 0.0571
230 232 -2 4 0.0172
54 57 -3 9 0.1578
192 189 3 9 0.0476
16 18 -2 4 0.2222
63 61 2 4 0.0655
8 5 3 9 1.8
15 18 -3 9 0.5
2.8674

X2 = = 2.8674

X2cal = 2.8674

Step # 05:

Find critical region using X2– chi-squire distribution table

X2 = (α, d.f) = (0.05, 3) = 7.815

Step # 06:

Conclusion: Since X2cal value lies in the region of acceptance therefore we accept the HO andreject HA. Means There is no association b/w Antigen and Blood Group

WHAT IS TEST OF SIGNIFICANCE? WHY IT IS NECESSARY? MENTION NAMES OF IMPORTANT TESTS.

1. Test of significance

A procedure used to establish the validity of a claim by determining whether or not the test statistic falls in the critical region. If it does, the results are referred to as significant. This test is sometimes called the hypothesis test.

The methods of inference used to support or reject claims based on sample data are known as tests of significance.

Why it is necessary

A significance test is performed;

  • To determine if an observed value of a statistic differs enough from a hypothesized value of a parameter
  • To draw the inference that the hypothesized value of the parameter is not the true value. The hypothesized value of the parameter is called the “null hypothesis.”

Types of test of significance

  1. Parametric
  2. t-test (one sample & two sample)
  3. z-test (one sample & two Sample)
  4. F-test.
  5. Non-parametric
  6. Chi-squire test
  7. Mann-Whitney U test
  8. Coefficient of concordance (W)
  9. Median test
  10. Kruskal-Wallis test
  11. Friedman test
  12. Rank difference methods (Spearman rho and Kendal’s tau)

P –Value:

A p-value is the probability that the computed value of a test statistic is at least as extreme as a specified value of the test statistic when the null hypothesis is true. Thus, the p value is the smallest value of for which we can reject a null hypothesis.

Simply the p value for a test may be defined also as the smallest value of α for which the null hypothesis can be rejected.

The p value is a number that tells us how unusual our sample results are, given that the null hypothesis is true. A p value indicating that the sample results are not likely to have occurred, if the null hypothesis is true, provides justification for doubting the truth of the null hypothesis.

Test Decisions with p-value

The decision about whether there is enough evidence to reject the null hypothesis can be made by comparing the p-values to the value of α, the level of significance of the test.

A general rule worth remembering is:

  • If the p value is less than or equal to, we reject the null hypothesis
  • If the p value is greater than, we do not reject the null hypothesis.
If p-value ≤ α reject the null hypothesis
If p-value ≥ α fail to reject the null hypothesis

Observational Study:

An observational study is a scientific investigation in which neither the subjects under study nor any of the variables of interest are manipulated in any way.

An observational study, in other words, may be defined simply as an investigation that is not an experiment. The simplest form of observational study is one in which there are only two variables of interest. One of the variables is called the risk factor, or independent variable, and the other variable is referred to as the outcome, or dependent variable.

Risk Factor:

The term risk factor is used to designate a variable that is thought to be related to some outcome variable. The risk factor may be a suspected cause of some specific state of the outcome variable.

Types of Observational Studies

There are two basic types of observational studies, prospective studies and retrospective studies.

Prospective Study:

A prospective study is an observational study in which two random samples of subjects are selected. One sample consists of subjects who possess the risk factor, and the other sample consists of subjects who do not possess the risk factor. The subjects are followed into the future (that is, they are followed prospectively), and a record is kept on the number of subjects in each sample who, at some point in time, are classifiable into each of the categories of the outcome variable.

The data resulting from a prospective study involving two dichotomous variables can be displayed in a 2 x 2 contingency table that usually provides information regarding the number of subjects with and without the risk factor and the number who did and did not

Retrospective Study:

A retrospective study is the reverse of a prospective study. The samples are selected from those falling into the categories of the outcome variable. The investigator then looks back (that is, takes a retrospective look) at the subjects and determines which ones have (or had) and which ones do not have (or did not have) the risk factor.

From the data of a retrospective study we may construct a contingency table

Relative Risk:

Relative risk is the ratio of the risk of developing a disease among subjects with the risk factor to the risk of developing the disease among subjects without the risk factor.

We represent the relative risk from a prospective study symbolically as

We may construct a confidence interval for RR

100 (1 – α)%CI=

Where zα is the two-sided z value corresponding to the chosen confidence coefficient and X2is computed by Equation

Interpretation of RR

  • The value of RR may range anywhere between zero and infinity.
  • A value of 1 indicates that there is no association between the status of the risk factor and the status of the dependent variable.
  • A value of RR greater than 1 indicates that the risk of acquiring the disease is greater among subjects with the risk factor than among subjects without the risk factor.
  • An RR value that is less than 1 indicates less risk of acquiring the disease among subjects with the risk factor than among subjects without the risk factor.

EXAMPLE

In a prospective study of pregnant women, Magann et al. (A-16) collected extensive information on exercise level of low-risk pregnant working women. A group of 217 women did no voluntary or mandatory exercise during the pregnancy, while a group of

238 women exercised extensively. One outcome variable of interest was experiencing preterm labor. The results are summarized in Table

Estimate the relative risk of preterm labor when pregnant women exercise extensively.

Solution:

By Equation

These data indicate that the risk of experiencing preterm labor when a woman exercises heavily is 1.1 times as great as it is among women who do not exercise at all.

Confidence Interval for RR

We compute the 95 percent confidence interval for RR as follows.

The lower and upper confidence limits are, respectively

= 0.65 and = 1.86

Conclusion:

Since the interval includes 1, we conclude, at the .05 level of significance, that the population risk may be 1. In other words, we conclude that, in the population, there may not be an increased risk of experiencing preterm labor when a pregnant woman exercises extensively.

Odds Ratio

An odds ratio (OR) is a measure of association between an exposure and an outcome. The OR represents the odds that an outcome will occur given a particular exposure, compared to the odds of the outcome occurring in the absence of that exposure.

It is the appropriate measure for comparing cases and controls in a retrospective study.

Odds:

The odds for success are the ratio of the probability of success to the probability of failure.

Two odds that we can calculate from data displayed as in contingency Table of retrospective study

  • The odds of being a case (having the disease) to being a control (not having the disease) among subjects with the risk factor is [a/ (a +b)] / [b/ (a + b)] = a/b
  • The odds of being a case (having the disease) to being a control (not having the disease) among subjects without the risk factor is [c/(c +d)] / [d/(c + d)] = c/d

The estimate of the population odds ratio is

We may construct a confidence interval for OR by the following method:

100 (1 – α) %CI=

Where is the two-sided z value corresponding to the chosen confidence coefficient and X2 is computed by Equation

Interpretation of the Odds Ratio:

In the case of a rare disease, the population odds ratio provides a good approximation to the population relative risk. Consequently, the sample odds ratio, being an estimate of the population odds ratio, provides an indirect estimate of the population relative risk in the case of a rare disease.

  • The odds ratio can assume values between zero and ∞.
  • A value of 1 indicates no association between the risk factor and disease status.
  • A value less than 1 indicates reduced odds of the disease among subjects with the risk factor.
  • A value greater than 1 indicates increased odds of having the disease among subjects in whom the risk factor is present.

EXAMPLE

Toschke et al. (A-17) collected data on obesity status of children ages 5–6 years and the smoking status of the mother during the pregnancy. Table below shows 3970 subjects classified as cases or noncases of obesity and also classified according to smoking status of the mother during pregnancy (the risk factor).

We wish to compare the odds of obesity at ages 5–6 among those whose mother smoked throughout the pregnancy with the odds of obesity at age 5–6 among those whose mother did not smoke during pregnancy.

Solution

By formula:

We see that obese children (cases) are 9.62 times as likely as nonobese children (noncases) to have had a mother who smoked throughout the pregnancy.

We compute the 95 percent confidence interval for OR as follows.

The lower and upper confidence limits for the population OR, respectively, are

= 7.12 and = = 13.00

We conclude with 95 percent confidence that the population OR is somewhere between

7.12 And 13.00. Because the interval does not include 1, we conclude that, in the population, obese children (cases) are more likely than nonobese children (noncases) to have had a mother who smoked throughout the pregnancy.

 

Measures of Dispersion

Posted on by
Reply

This term is used commonly to mean scatter, Deviation, Fluctuation, Spread or variability of data.

The degree to which the individual values of the variate scatter away from the average or the central value, is called a dispersion.

Types of Measures of Dispersions:

  • Absolute Measures of Dispersion: The measures of dispersion which are expressed in terms of original units of a data are termed as Absolute Measures.
  • Relative Measures of Dispersion: Relative measures of dispersion, are also known as coefficients of dispersion, are obtained as ratios or percentages. These are pure numbers independent of the units of measurement and used to compare two or more sets of data values.

http://www.freewebs.com/maeconomics/Statistics/measures_of_dispersion_files/measur3.gif

Absolute Measures

  • Range
  • Quartile Deviation
  • Mean Deviation
  • Standard Deviation

Relative Measure

  • Co-efficient of Range
  • Co-efficient of Quartile Deviation
  • Co-efficient of mean Deviation
  • Co-efficient of Variation.

The Range:

1.      The range is the simplest measure of dispersion.  It is defined as the difference between the largest value and the smallest value in the data:

http://www.freewebs.com/maeconomics/Statistics/measures_of_dispersion_files/measur4.gif

2. For grouped data, the range is defined as the difference between the upper class boundary (UCB) of the highest class and the lower class boundary (LCB) of the lowest class.

MERITS OF RANGE:-

  • Easiest to calculate and simplest to understand.
  • Gives a quick answer.

DEMERITS OF RANGE:-

  • It gives a rough answer.
  • It is not based on all observations.
  • It changes from one sample to the next in a population.
  • It can’t be calculated in open-end distributions.
  • It is affected by sampling fluctuations.
  • It gives no indication how the values within the two extremes are distributed

Quartile Deviation (QD):

1.      It is also known as the Semi-Interquartile Range.  The range is a poor measure of dispersion where extremely large values are present.  The quartile deviation is defined half of the difference between the third and the first quartiles:

QD = Q3 – Q1/2

Inter-Quartile Range

The difference between third and first quartiles is called the ‘Inter-Quartile Range’.

IQR = Q3 – Q1

Mean Deviation (MD):

1.      The MD is defined as the average of the deviations of the values from an average:

http://www.freewebs.com/maeconomics/Statistics/measures_of_dispersion_files/measur6.gif

It is also known as Mean Absolute Deviation.

2.      MD from median is expressed as follows:

http://www.freewebs.com/maeconomics/Statistics/measures_of_dispersion_files/measur7.gif

3.      for grouped data:

http://www.freewebs.com/maeconomics/Statistics/measures_of_dispersion_files/measur8.gif

http://www.freewebs.com/maeconomics/Statistics/measures_of_dispersion_files/measur9.gif

Mean Deviation:

  1. The MD is simple to understand and to interpret.
  2. It is affected by the value of every observation.
  3. It is less affected by absolute deviations than the standard deviation.
  4. It is not suited to further mathematical treatment.  It is, therefore, not as logical as convenient measure of dispersion as the SD.

The Variance:

  • Mean of all squared deviations from the mean is called as variance
  • (Sample variance=S2; population variance= σ2sigma squared (standard deviation squared). A high variance means most scores are far away from the mean, a low variance indicates most scores cluster tightly about the mean.

Variance2JPGFormula

OR S2 =

Calculating variance: Heart rate of certain patient is 80, 84, 80, 72, 76, 88, 84, 80, 78, & 78. Calculate variance for this data.

Solution:

Step 1:

Find mean of this data

measures-of-central-tendency-4

= 800/10 Mean = 80

Step 2:

Draw two Columns respectively ‘X’ and deviation about mean (X- ). In column ‘X’ put all values of X and in (X- ) subtract each ‘X’ value with .

Step 3:

Draw another Column of (X- ) 2, in which put square of deviation about mean.

X (X- )

Deviation about mean

 (X- )2

Square of Deviation about mean
80

84

80

72

76

88

84

80

78

78

 80 – 80 = 0

84 – 80 = 4

80 – 80 = 0

72 – 80 = -8

76 – 80 = -4

88 – 80 = 8

84 – 80 = 4

80 – 80 = 0

78 – 80 = -2

78 – 80 = -2

 0 x 0 = 00

4 x 4 = 16

0 x 0 = 00

-8 x -8 = 64

-4 x -4 = 16

8 x 8 = 64

4 x 4 = 16

0 x 0 = 00

-2 x -2 = 04

-2 x -2 = 04
∑X = 800

= 80

 ∑(X- ) = 0

Summation of Deviation about mean is always zero

 ∑(X- )2 = 184

Summation of Square of Deviation about mean

Step 4

Apply formula and put following values

∑(X- ) 2= 184

n = 10

Variance2JPG

Variance = 184/ 10-1 = 184/9

Variance = 20.44

Standard Deviation

  • The SD is defined as the positive Square root of the mean of the squared deviations of the values from their mean.
  • The square root of the variance.
  • It measures the spread of data around the mean. One standard deviation includes 68% of the values in a sample population and two standard deviations include 95% of the values & 3 standard deviations include 99.7% of the values
  • The SD is affected by the value of every observation.
  • In general, it is less affected by fluctuations of sampling than the other measures of dispersion.
  • It has a definite mathematical meaning and is perfectly adaptable to algebraic treatment.

http://www.bmj.com/statsbk/2-20.gifFormula:

OR S =

Calculating Standard Deviation (we use same example): Heart rate of certain patient is 80, 84, 80, 72, 76, 88, 84, 80, 78, & 78. Calculate standard deviation for this data.

SOLUTION:

Step 1: Find mean of this data

measures-of-central-tendency-4

= 800/10 Mean = 80

Step 2:

Draw two Columns respectively ‘X’ and deviation about mean (X-). In column ‘X’ put all values of X and in (X-) subtract each ‘X’ value with.

Step 3:

Draw another Column of (X- ) 2, in which put square of deviation about mean.

X (X- )

Deviation about mean

 (X- )2

Square of Deviation about mean
80

84

80

72

76

88

84

80

78

78

 80 – 80 = 0

84 – 80 = 4

80 – 80 = 0

72 – 80 = -8

76 – 80 = -4

88 – 80 = 8

84 – 80 = 4

80 – 80 = 0

78 – 80 = -2

78 – 80 = -2

 0 x 0 = 00

4 x 4 = 16

0 x 0 = 00

-8 x -8 = 64

-4 x -4 = 16

8 x 8 = 64

4 x 4 = 16

0 x 0 = 00

-2 x -2 = 04

-2 x -2 = 04
∑X = 800

= 80

 ∑(X- ) = 0

Summation of Deviation about mean is always zero

 ∑(X- )2 = 184

Summation of Square of Deviation about mean

Step 4

Apply formula and put following values

∑(X- )2 = 184

n = 10

sd calculation

MERITS AND DEMERITS OF STD. DEVIATION

  • Std. Dev. summarizes the deviation of a large distribution from mean in one figure used as a unit of variation.
  • It indicates whether the variation of difference of a individual from the mean is real or by chance.
  • Std. Dev. helps in finding the suitable size of sample for valid conclusions.
  • It helps in calculating the Standard error.

DEMERITS-

  • It gives weightage to only extreme values. The process of squaring deviations and then taking square root involves lengthy calculations.

 Relative measure of dispersion:

(a)    Coefficient of Variation,

(b)   Coefficient of Dispersion,

(c)    Quartile Coefficient of Dispersion, and

(d)   Mean Coefficient of Dispersion.

Coefficient of Variation (CV):

1.      Coefficient of variation was introduced by Karl Pearson.  The CV expresses the SD as a percentage in terms of AM:

http://www.freewebs.com/maeconomics/Statistics/measures_of_dispersion_files/measur50.gif   —————- For sample data

http://www.freewebs.com/maeconomics/Statistics/measures_of_dispersion_files/measur51.gif   ————— For population data

  • It is frequently used in comparing dispersion of two or more series.  It is also used as a criterion of consistent performance, the smaller the CV the more consistent is the performance.
  • The disadvantage of CV is that it fails to be useful when  http://www.freewebs.com/maeconomics/Statistics/measures_of_dispersion_files/measur52.gif   is close to zero.
  • It is sometimes also referred to as ‘coefficient of standard deviation’.
  • It is used to determine the stability or consistency of a data.
  • The higher the CV, the higher is instability or variability in data, and vice versa.

Coefficient of Dispersion (CD):

If Xm and Xn are respectively the maximum and the minimum values in a set of data, then the coefficient of dispersion is defined as:

http://www.freewebs.com/maeconomics/Statistics/measures_of_dispersion_files/measur53.gif

Coefficient of Quartile Deviation (CQD):

1.      If Q1 and Q3 are given for a set of data, then (Q1 + Q3)/2 is a measure of central tendency or average of data.  Then the measure of relative dispersion for quartile deviation is expressed as follows:

http://www.freewebs.com/maeconomics/Statistics/measures_of_dispersion_files/measur54.gif

CQD may also be expressed in percentage.

Mean Coefficient of Dispersion (CMD):

The relative measure for mean deviation is ‘mean coefficient of dispersion’ or ‘coefficient of mean deviation’:

http://www.freewebs.com/maeconomics/Statistics/measures_of_dispersion_files/measur55.gif   ——————– for arithmetic mean

http://www.freewebs.com/maeconomics/Statistics/measures_of_dispersion_files/measur56.gif   ——————– for median

Percentiles and Quartiles

The mean and median are special cases of a family of parameters known as location parameters. These descriptive measures are called location parameters because they can be used to designate certain positions on the horizontal axis when the distribution of a variable is graphed.

Percentile:

  1. Percentiles are numerical values that divide an ordered data set into 100 groups of values with at the most 1% of the data values in each group. There can be maximum 99 percentile in a data set.
  2. A percentile is a measure that tells us what percent of the total frequency scored at or below that measure.

Percentiles corresponding to a given data value: The percentile in a set corresponding to a specific data value is obtained by using the following formula

Number of values below X + 0.5

Percentile = ——————————————–

Number of total values in data set

Example: Calculate percentile for value 12 from the following data

13 11 10 13 11 10 8 12 9 9 8 9

Solution:

Step # 01: Arrange data values in ascending order from smallest to largest

S. No 1 2 3 4 5 6 7 8 9 10 11 12
Observations or values 8 8 9 9 9 10 10 11 11 12 13 13

Step # 02: The number of values below 12 is 9 and total number in the data set is 12

Step # 03: Use percentile formula

9 + 0.5

Percentile for 12 = ——— x 100 = 79.17%

12

It means the value of 12 corresponds to 79th percentile

Example2: Find out 25th percentile for the following data

6 12 18 12 13 8 13 11

10 16 13 11 10 10 2 14

SOLUTION

Step # 01: Arrange data values in ascending order from smallest to largest

S. No 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Observations or values 2 6 8 10 10 10 11 11 12 12 13 13 13 14 16 18

Step # 2 Calculate the position of percentile (n x k/ 100). Here n = No: of observation = 16 and k (percentile) = 25

16 x 25 16 x 1

Therefore Percentile = ———- = ——— = 4

100 4

Therefore, 25th percentile will be the average of values located at the 4th and 5th position in the ordered set. Here values for 4th and 5th correspond to the value of 10 each.

(10 + 10)

Thus, P25 (=Pk) = ————– = 10

2

Quartiles

These are measures of position which divide the data into four equal parts when the data is arranged in ascending or descending order. The quartiles are denoted by Q.

quartile

Quartiles Formula for Ungrouped Data Formula for Grouped Data
Q1 = First Quartile below which first 25% of the observations are present Q1_ungroup
Q2 = Second Quartile below which first 50% of the observations are present.

It can easily be located as the median value.

 Q2_ungroup Q2
Q3 = Third Quartile below which first 75% of the observations are present Q3_ungroup Q3

Symbol Key:

 

Probability

Posted on by
Reply

Probability is used to measure the ‘likelihood’ or ‘chances’ of certain events (pre-specified outcomes) of an experiment.

If an event can occur in N mutually exclusive and equally likely ways, and if m of these possess a trait E, the probability of the occurrence of E expressed as:

Number of favourable cases

=

Total number of outcome (sample Space)

Characteristics of probability:

  • It is usually expressed by the symbol ‘P’
  • It ranges from 0 to 1
  • When P = 0, it means there is no chance of happening or impossible.
  • If P = 1, it means the chances of an event happening is 100%.
  • The total sum of probabilities of all the possible outcomes in a sample space is always equal to one (1).
  • If the probability of occurrence is p(o)= A, then the probability of non-occurrence is 1-A.

Terminology

Random Experiment:

Any natural phenomenon, yielding some result will be termed as random experiment when it is not possible to predict a particular result to turn out.

An Outcome:

The result of an experiment in all possible form are said to be event of that experiment. e.g. When you toss a coin once, you either get head or tail.

A trial:

This refers to an activity of carrying out an experiment like tossing a coin or rolling a die or dices.

Sample Space:

A set of All possible outcomes of a probability experiment.

Example 1: In tossing a coin, the outcomes are either Head (H) or tail (T) i.e. there are only two possible outcomes in tossing a coin. The chances of obtaining a head or a tail are equal. It can be solved as follow;

n(s) = 2 ways

S = {H, T}

Example 2: what is sample space when single dice is rolled?

n(s) = 6 ways

S = {1, 2, 3, 4, 5, 6}

A Simple Event

In an experimental probability, an event with only one outcome is called a simple event.

Compound Events

When two or more events occur in connection with each other, then their simultaneous occurrence is called a compound event.

Mutually exhaustive:

If in an experiment the occurrence of one event prevents or rules out the happening of all other events in the same experiment then these event are said to be mutually exhaustive events.

Mutually exclusive:

Two events are said to be mutually exclusive if they cannot occur simultaneously.

Example: tossing a coin, the events head and tail are mutually exclusive because if the outcome is head then the possibilities of getting a tail in the same trial is ruled out.

Equally likely events:

Events are said to be equally likely if there is no reason to expect any one in preference to other.

Example: in a single cast of a fair die each of the events 1, 2, 3, 4, 5, 6 is equally likely to occur.

Favourable case:

The cases which ensure the occurrence of an event are said to be favourable to the events.

Independent event:

When the experiments are conducted in such a way that the occurrence of an event in one trial does not have any effect on the occurrence of the other events at a subsequent experiment, then the events are said to be independent.

Example:

If we draw a card from a pack of cards and again draw a second a card from the pack by replacing the first card drawn, the second draw is known as independent f the first.

Dependent Event:

When the experiments are conducted in such a way that the occurrence of an event in one trial does have some effect on the occurrence of the other events at a subsequent experiment, then the event are said to be dependent event.

Example:

If we draw a card from a pack and again draw a card from the rest of pack of cards (containing 51 cards) then the second draw is dependent on the first.

Conditional Probability:

The probability of happening of an event A, when it is known that B has already happened, is called conditional probability of A and is denoted by P (A/B) i.e.

  • P(A/B) = conditional probability of A given that B has already occurred.
  • P (A/B) = conditional Probability of B given that A has already occurred.

Types of Probability:

The Classical or mathematical:

Probability is the ratio of the number of favorable cases as compared to the total likely cases.

The probability of non-occurrence of the same event is given by {1-p (occurrence)}.

The probability of occurrence plus non-occurrence is equal to one.

If probability occurrence; p (O) and probability of non-occurrence (O’), then p(O)+p(O’)=1.

Statistical or Empirical

Empirical probability arises when frequency distributions are used. For example:

Observation ( X) 0 1 2 3 4
Frequency ( f) 3 7 10 16 11

The probability of observation (X) occurring 2 times is given by the formulae

RULES OF PROBABILITY

Addition Rule

  1. Rule 1: When two events A and B are mutually exclusive, then probability of any one of them is equal to the sum of the probabilities of the happening of the separate events;

Mathematically:

P (A or B) =P (A) +P (B)

Example: When a die or dice is rolled, find the probability of getting a 3 or 5.

Solution: P (3) =1/6 and P (5) =1/6.

Therefore P (3 or 5) = P (3) + P (5) = 1/6+1/6 =2/6=1/3.

2) Rule 2: If A and B are two events that are NOT mutually exclusive, then

P (A or B) = P(A) + P(B) – P(A and B), where A and B means the number of outcomes that event A and B have in common.

Given two events A and B, the probability that event A, or event B, or both occur is equal to the probability that event A occurs, plus the probability that event B occurs, minus the probability that the events occur simultaneously.

Example: When a card is drawn from a pack of 52 cards, find the probability that the card is a 10 or a heart.

Solution: P (10) = 4/52 and P (heart) =13/52

P (10 that is Heart) = 1/52

P (A or B) = P (A) +P (B)-P (A and B) = 4/52 _ 13/52 – 1/52 = 16/52.

Multiplication Rule

  1. Rule 1: For two independent events A and B, then

P (A and B) = P (A) x P (B).

Example: Determine the probability of obtaining a 5 on a die and a tail on a coin in one throw.

Solution: P (5) =1/6 and P (T) =1/2.

P (5 and T) = P (5) x P (T) = 1/6 x ½= 1/12.

  1. Rule 2: When two events are dependent, the probability of both events occurring is P (A and B) =P (A) x P (B|A), where P (B|A) is the probability that event B occurs given that event A has already occurred.

Example: Find the probability of obtaining two Aces from a pack of 52 cards without replacement.

Solution: P (Ace) =2/52 and P (second Ace if NO replacement) = 3/51

Therefore P (Ace and Ace) = P (Ace) x P (Second Ace) = 4/52 x 3/51 = 1/221

Construct sample space, when two dice are rolled

n(s) = n1 x n2 = 6 x 6 = 36

(1,1) (2,1) (3,1) (4,1) (5,1) (6,1)
(1,2) (2, 2) (3, 2) (4, 2) (5, 2) (6, 2)
(1, 3) (2, 3) (3, 3) (4, 3) (5, 3) (6, 3)
(1, 4) (2, 4) (3, 4) (4, 4) (5, 4) (6, 4)
(1, 5) (2, 5) (3, 5) (4, 5) (5, 5) (6, 5)
(1, 6) (2, 6) (3, 6) (4, 6) (5, 6) (6, 6)

EXAMPLE OF FINDING PROBABILITY OF AN EVENT

If 3 coins are tossed together, construct a tree diagram & find the followings;

a) Event showing No head b) Event showing 01 head,

c) Event showing 02 heads d) Event showing 03 heads

n (s) = n1 x n2 x n3

= 2 x 2 x2 = 8

tree diagram

    1. Event showing no head = P(X = 0)

Answer: TTT, 1/8 = 0.125

    1. Event showing 01 head = P(X = 1)

Answer: HTT, THT, TTH 3/8 = 0.375

    1. Event showing 02 heads = P(X = 2)

Answer: HHT, HTH, THH 3/8 = 0.375

    1. Event showing 03 heads = P(X = 3)

Answer: HHH 1/8 = 0.125

Complementary Events

Complementary events happen when there are only two outcomes, like getting a job, or not getting a job. In other words, the complement of an event happening is the exact opposite: the probability of it not happening.

The probability of not occurrence of an event.

The probability of an event A is equal to 1 minus the probability of its complement, which is written as Ā and

P (Ā) = 1 – P (A)

CONDITIONAL PROBABILITY &SCREENING TESTS

Sensitivity, Specificity, and Predictive Value Positive and Negative

In the health sciences field a widely used application of probability laws and concepts is found in the evaluation of screening tests and diagnostic criteria. Of interest to clinicians is an enhanced ability to correctly predict the presence or absence of a particular disease from knowledge of test results (positive or negative) and/or the status of presenting symptoms (present or absent). Also of interest is information regarding the likelihood of positive and negative test results and the likelihood of the presence or absence of a particular symptom in patients with and without a particular disease.

In consideration of screening tests, one must be aware of the fact that they are not always infallible. That is, a testing procedure may yield a false positive or a false negative.

False Positive:

A false positive results when a test indicates a positive status when the true status is negative.

False Negative:

A false negative results when a test indicates a negative status when the true status is positive.

Sensitivity:

The sensitivity of a test (or symptom) is the probability of a positive test result (or presence of the symptom) given the presence of the disease.

Specificity:

The specificity of a test (or symptom) is the probability of a negative test result (or absence of the symptom) given the absence of the disease.

Predictive value positive:

The predictive value positive of a screening test (or symptom) is the probability that a subject has the disease given that the subject has a positive screening test result (or has the symptom).

Predictive value negative:

The predictive value negative of a screening test (or symptom) is the probability that a subject does not have the disease, given that the subject has a negative screening test result (or does not have the symptom).

Summary of formulae:

Symbols

COUNTING RULES

1) FACTORIALS (number of ways)

The result of multiplying a sequence of descending natural numbers down to a number. It is denoted by “!”

Examples:
4! = 4 × 3 × 2 × 1×0! = 24
7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040

Remember : 0! = 1

General Method:

n! = n (n -1) (n -2) (n -3)……….. (n – n)!

2) PERMUTATION RULES

All possible arrangements of a collection of things, where the order is important in a subset.

Repetition of same items with different arrangement is allowed.

Examples

  1. COMBINATIONS

The order of the objects in a subset is immaterial.

Repetition of same objects in not allowed with different arrangement

Examples:

Binomial distribution:

Binomial distribution is a probability distribution which is obtained when the probability ‘P’ of the happening of an event is same in all the trials and there are only two event in each trial.

Conditions:

  • Each trial results in one of two possible, mutually exclusive, outcomes. One of the possible outcomes is denoted (arbitrarily) as a success, and the other is denoted a failure.
  • The probability of a success, denoted by p, remains constant from trial to trial. The probability of a failure (1 – p) is denoted by q.
  • The trials are independent; that is, the outcome of any particular trial is not affected by the outcome of any other trial.
  • Parameter should be available; (n & p) are parameters.

Formula:

b (X: n, p) = nCx px qn – x (OR) f (x) = nCx px qn – x

Where

X = Random variable

n = Number of Trials

p = Probability of Success

q = Probability of Failure

Microbiology

Posted on by
Reply

Microbiology MCQs/BCQs

1. The media used for culturing salmonella is
a. VL-broth
b. Sabouruad agar
c. Slanetz bartley
d. Meat broth
e. Selenite Broth

2. Structural and functional as well as basic unit of life which was discovered by Robert hook.
a. Cell
b. Tissue
c. Organ
d. System
e. Human
3. Area of the cytoplasm that contains the single bacterial DNA molecule
a. Plasma
b. Nucleoid region
c. Cytoplasm
d. Ribosomes
e. Vacuole

4. Cells have “little organs” in it are collectively known as
a. Cell parts
b. Inter cell
c. Cell organelles
d. Cytoplasm
e. Golgi bodies

5. It is called the power house of cell.
a. Golgi bodies
b. Plasma membrane
c. Mitochondria
d. Cell membrane
e. Food vacuole

6. The bodies ability to fight against pathogens is known as
a. Microbiology
b. Histology
c. Pathology
d. Immunity
e. Biology

7. A patient skin-tested with purified protein derivative (PPD) to determine previous exposure to Mycobacterium tuberculosis develops induration at the skin test site 48 hours later. Histologically, the reaction site would MOST probably show:
a. Neutrophils
b. Helper T cells and macrophages
c. B cells
d. Eosinophils
e. A cells

8. It is highly complex jelly like material in which other parts are embedded.
a. Apoptosis
b. Cytoplasm
c. Ribosomes
d. Endoplasmic reticulum
e. Nucleus

9. Part of microscope that is hollow cylindrical tube & support the ocular lens is named as
a. Objective
b. Lens
c. Eye piece
d. Body tube
e. Nose piece

10. It acts as a barrier to invasion.
a. Skin
b. Bones
c. Organs
d. Cell
e. Brain

11. According to mitosis cell division has
a. 10 phases
b. 4 phases
c. 2 phases
d. 5 phases
e. 6 phases

12. It is a chemical substance derivable from a mold or bacterium that kills microorganisms and cures infections.
a. Antibiotic
b. Antigen
c. Protein
d. Virus
e. Bacteria

13. A child stung by a bee experiences respiratory distress within minutes and lapses into unconsciousness. This reaction is probably mediated by:
a. IgM antibody
b. Sensitized T cells
c. IgE antibody
d. IgF antibody
e. Sensitized T cells

14. organism made up of one of single cell are known as
a. multi cellular
b. muti nucleoid
c. nucleus
d. unicellular
e. animal

15. It is a milky body fluid that contains a type of white blood cells.
a. Body
b. Cell
c. Lymph
d. Plasma
e. Membrane

16. An injection of a weakened form of the actual antigen that causes the disease is called
a. Injection
b. Syringe
c. Vaccine
d. Inmmune system
e. Lymphocytes

17. A patient with a central nervous system disorder is maintained on the drug methyldopa. Hemolytic anemia develops, which resolves shortly after the drug is withdrawn. This is MOST probably an example of:
a. Cell-mediated hypersensitivity
b. Immune-complex hypersensitivity
c. Atopic hypersensitivity
d. Cytotoxic hypersensitivity
e. Non hypersensitivity

18. Are chemical substance as protein, carbohydrates, lipids (fats) or nutic acid which stimulates specific immune response.
a. Antibiotic
b. Antigen
c. Protein
d. Virus
e. Bacteria

19. The 3rd phase of mitosis (cell division) is known as
a. Anaphase
b. Metaphase
c. Telophase
d. Prophase
e. Cytokinesis

20. A small dense(thick) body in nucleus that contain ribo nucleic acid is known as
a. Metacarpals
b. Meta tarsals
c. Nucleoli
d. Nucleus
e. Ribosomes

21. An immunoglobulin is a
a. Carbohydrate
b. Glycoprotein
c. Protein
d. Minerals
e. Amino acid

22. It is useful to stimulate antibody production
a. An adjuvant
b. A hapten
c. A protein
d. Red blood cells
e. Purified antigen

23. The resistance power of body against infectious agents like bacteria, virus etc.
a. Microbiology
b. Histology
c. Pathology
d. Immunity
e. Biology

24. In this type of cell division four haploid gametes are produced.
a. Mitosis
b. Meiosis
c. Decomposition
d. Photosynthesis
e. Sterilization

25. A part of microscope that is circular shutter which regulates the size of opening through which light processes in the condenser.
a. Condenser
b. Mirror
c. Arm
d. Diaphragm
e. Inclinations joint

26. A process of A-sexual reproduction or simple method of cell division that occurs in unicellular organisms.
a. Sterilization
b. Fertilization
c. Meiosis
d. Mitosis
e. Sanitization

27. The first person to identify microbes as causing disease was
a. Louis Pasteur
b. Robert Koch
c. Robert hook
d. Ivan lewin hook
e. Edward jenner
28. A minute animal or vegetable which can’t be seen by naked eye but just can be seen by micro-scope.
a. Macro organism
b. Micro organism
c. Human being
d. Ant
e. Elephant
29. A disease in which minute organisms, invisible to the naked eye, invade and multiply within the body.
a. Infectious disease
b. Healthy disease
c. Disease
d. Infection
e. Microscopic disease

30. Those diseases that are found normally in a population are named as
a. Epidemic
b. Endemic
c. Pendemic
d. Epizootic
e. Incidence

 

31. Respiratory acidosis and alkalosis are due to a problem with the
a. Heart
b. Lungs
c. Hands
d. Foot
e. Brain

32. Metabolic acidosis and alkalosis are due to a problem with the
a. Kidney
b. Pancreas
c. Gall bladder
d. Urinary bladder
e. Bones

33. The movement of molecules across the cell membrane and does not requires energy is known as
a. Passive transport
b. Active transport
c. Cellulose
d. Movement
e. Motion

34. A term for the language deficit accompanying cerebral stroke is
a. Agraphia
b. Anosognosia
c. Dysphagia
d. Aphasia
e. Euphoria

35. The inability to carry out a motor task on command given adequate strength, sensation, coordination and comprehension is called:
a. Aphasia
b. Apraxia
c. Alexia
d. Aprosodia
e. Abulia

36. A 64 year-old right handed male presents with right upper limb plegia, right lower limb paresis, a hemi sensory deficit, a decreased ability to comprehend verbal or written commands and poor language output. His lesion is most likely in the:
a. Basal ganglia
b. Middle cerebral artery distribution
c. Posterior cerebral artery distribution
d. Brain stem
e. Anterior cerebral artery distribution

37. A stroke affecting the back part of a brain could affect
a. Lungs
b. Heart
c. Sight
d. Taste
e. Smell

38. Hemorrhagic Stroke is associated with
a. Hypertension
b. Dialysis
c. Resting
d. Fatigue
e. Hepatitis

39. The carrier-mediated transport of large molecules through the cell membrane using transport proteins embedded within the cell membrane is known as
a. Passive transport
b. Active transport
c. Facilitated diffusion
d. Movement
e. Motion
40. A semi permeable membrane separating the blood from the cerebrospinal fluid, and constituting a barrier to the passage of cells, particles, and large molecules.
a. Permeable membrane
b. Cell membrane
c. Cell wall
d. Blood brain barrier
e. Kidney blood barrier

 

41. “The movement of solute molecules and water across a membrane by normal cardiovascular pressure” refers to
a. Permeability
b. Semi permeability
c. Filtration
d. Vibration
e. Movement

42. The sodium-potassium pump carries out a form of
a. Active transport
b. Passive transport
c. Sterilization
d. Filtration
e. Disinfection

43. The diffusion of a solute across a selectively permeable membrane. In this case the solute molecules always move from the stronger concentration (hypertonic) to the weaker (hypotonic).
a. Diffusion
b. Osmosis
c. Dialysis
d. Filtration
e. De colorization

44. The most important carbohydrate is
a. Fructose
b. Lactose
c. Glucose (monosaccharide)
d. Sucrose
e. Glycoprotein

45. Cholesterol is a type of
a. Protein
b. Carbohydrate
c. Amino acids
d. Lipids
e. Fats
46. Anti lipolytic hormone is
a. Insulin
b. Epinephrine
c. Nor epinephrine
d. Thyroid
e. Serotonin

47. The maintenance of steady levels of glucose in the body is known as
a. Lipid regulation
b. Fat regulation
c. Serotonin regulation
d. Gluco regulation
e. Thyroid regulation

48. The entire spectrum of chemical reactions, occurring in the living system is termed as
a. Metabolism
b. Anabolism
c. Catabolism
d. Ion regulation
e. Insulin regulation

49. Synthesis of glucose from non carbohydrate compounds is known as
a. Glycoprotein
b. Glycolipids
c. Glucose
d. Lactogenesis
e. Glycogenesis

50. Non essential amino acids are named as
a. Dispensable amino acids
b. Non dispensable amino acids
c. Regulatory amino acids
d. Retained amino acids
e. Fatty amino acids

Subjective
Note: attempt any five questions. All questions carry equal marks
Qno1. Write a short note on blood brain barrier?
Qno2. Discuss briefly active and passive transport in general with examples and explain sodium potassium pump?
Qno3. Differentiate between trauma and cellular injury?
Qno4. Define microbiology and briefly discuss its importance in nursing?
Qno5. Define immunity and explain its types with at least two examples for each type?
Qno6 write short note any one of the following
a. Mitosis
b. meiosis
Qno7. Differentiate between unicellular and multi cellular organisms with examples and differentiate between eukaryotes and prokaryotes with examples?

Answer key
01 E
02 A
03 B
04 C
05 C
06 D
07 B
08 B
09 D
10 A
11 B
12 A
13 E
14 D
15 C
16 C
17 D
18 B
19 A
20 C
21 B
22 A
23 D
24 B
25 D

 

26 D
27 B
28 B
29 A
30 B
31 B
32 A
33 A
34 D
35 B
36 B
37 C
38 A
39 C
40 D
41 C
42 A
43 C
44 C
45 D
46 A
47 D
48 A
49 E
50 A

Health Assessment MCQs/BCQs

Posted on by
Reply

Health Assessment

B.Sc Nursing (Post RN) 1st year 1st semester Session

Q.No.1: Choose the correct answer

Objective Paper
Q.No.1: Choose the correct answer
1. A plan of care that identifies the specific needs of the client and that needs will be addressed by the healthcare system or skilled nursing facility is
a. Health identification
b. Health assessment
c. Health examination
d. Disease identification
e. Patients assessment
2. The process in which diseases detect early in people that may look and feel well is called
a. Medical assessment
b. Disease assessment
c. Investigation of disease
d. Health assessment
e. Health care
3. Nurses use physical assessment skills to
a. To identify and manage a variety of patient problems
b. To discharge the patient from hospital
c. To collect the health history
d. To realize the patient importance to relatives
e. To enhance the quality of care
4. When a client have a complain of sever headache a nurse assess that it is
a. Objective data
b. Subjective data
c. Client history
d. Chief complain
e. Present complain

5. A patient admit in general ward and have a complain of vertigo a nurse check blood pressure and inform to doctor it is called
a. Subjective data
b. Take vital sign of client
c. Health history
d. Objective data
e. duty of nurse

6. A seated position back unsupported and legs hanging freely is
a. Dorsal recumbent
b. Supine
c. Sims
d. Lithotomy
e. Sitting
7. Lies on abdomen with head torn to the side, wit or without a small pillow this is
a. Supine position
b. Lithotomy position
c. Horizontal recumbent position
d. Prone position
e. Sims position
8. A assessment technique in which critical observation of client done without touching by nurse or health care provider is
a. Inspection
b. Palpation
c. Percussion
d. Auscultation
e. Objective data
9. During assessment a sounds produced by striking body surface of individual this step of technique is called
a. Subjective data
b. Objective data
c. Inspection
d. Percussion
e. Diagnostic procedure

 

10. A Stethoscope is used to listening the sounds produced by the body of patient or individual this technique is called
a. Inspection
b. Palpation
c. Percussion
d. Auscultation
e. Physical examination
11. During the physical examination a lubricant like xylocain jell or liquid paraffin is used to
a. Ease the insertion of instrument
b. Visualize the body part
c. Heal the injury
d. Enhance the client’s complain
e. Document the main complain of patient
12. A physical examination in which tongue blades ( depressor) is used
a. To depress the tongue during assessment of nose and throat
b. To depress the tongue during assessment of mouth and larynx
c. To depress the tongue during assessment of mouth and pharynx
d. To depress the tongue during assessment of mouth and esophagus
e. To elevate the tongue during assessment of mouth and pharynx
13. Vaginal speculum is used to assess the
a. ovary
b. fallopian tube
c. Uterus
d. Cervix & vagina
e. Urethra
14. During assessment a lighted instrument is used to visualize the anterior of eye is called
a. Otoscope
b. Stethoscope
c. Laryngoscope
d. Nasal speculum
e. Ophthalmoscope

 

15. When client have a complain of congested chest and sounds are audible without stethoscope it is
a. Direct auscultation
b. Indirect auscultation
c. Inspection
d. Percussion
e. Palpation

16. Acknowledging the patient’s verbal and nonverbal communication conveys true interest and encourages further communication by
a. History taking
b. Interview
c. Data collection
d. Subjective data
e. Objective data
17. Otitis media is an
a. Inflammation of external ear
b. Inflammation of middle ear
c. Inflammation of inner ear
d. Inflammation of nasal cavity
e. Inflammation of oral cavity
18. The interviews require less time and are very effective for obtaining factual data with specific questions and are controlled by the nurse
a. Interview
b. Directive interview
c. Nondirective interview
d. History taking step
e. Open-ended question
19. In interview elicit a “yes” or “no” response, to client this type of question are
a. Open question
b. Closed question
c. Direct question
d. Indirect question
e. Simple question

 

20. The time during which a female is menstruating
a. Menopause
b. Menstrual period
c. Last menstrual period
d. Expected date of menstruation
e. Irregular cycle

21. X-ray of breast
a. Mammogram
b. Digital x-ray
c. Ct-scan
d. MRI
e. Barium scan
22. The process of identification of the condition, needs, abilities and preferences of a patient is
a. Nursing assessment
b. Patient assessment
c. Medical assessment
d. Professional assessment
e. Physical assessment
23. The process gathering of information about a patient’s physiological, psychological, sociological, and spiritual status in
a. Nursing assessment
b. Patient assessment
c. Medical assessment
d. Professional assessment
e. Physical assessment
24. When Blanch Test is performed and nails pressed between the fingers the nails return to usual color in less than
a. 4 seconds
b. 6 seconds
c. 8 second
d. 2 second
e. 3 second
25. The thyroid gland is not visible during the
a. Inspection
b. Palpation
c. Percussion
d. Auscultation
e. Surgery
26. Patient was able to read the newsprint at a distance of
a. 8 inches
b. 10 inches
c. 12 inches
d. 20 inches
e. 25 inches

27. Able to hear ticking on right ear at a distance of one inch and was able to hear the ticking on the left ear at the same distance this assessment test is called
a. Hearing Acuity Test
b. Watch Tick Test
c. Blanch Test
d. Weber test
e. Assessment test
28. An instrument used to measure the B.P of client is called
a. Stethoscope
b. Otoscope
c. Ophthalmoscope
d. Sphygmomanometer
e. Laryngoscope
29. The sweat to reduce the body temperature is eliminated by
a. Sweats gland
b. Apocrine gland
c. Eccrine gland
d. Thyroid gland
e. Hypothalamus gland
30. For the detection of hearing loss an instrument in physical examination is called
a. Otoscope
b. Ophthalmoscope
c. Hammer
d. Tuning fork
e. Speculum

 

31. Cleft palate is a congenital defect where the maxillary process fails to fuse. This causes a gap in the
a. hard palate and possibly the lower lip
b. soft palate and possibly the upper lip
c. hard palate and possibly the upper lip
d. hard palate and possibly the corner of lip
e. hard palate and possibly the mucous part of lip
32. A 70-year-old woman complains of dry mouth. The most frequent cause of this problem is:
a. The aging process
b. Related to medications she may be taking
c. The use of dentures
d. Related to a diminished sense of smell
e. Atrophy of esophagus
33. 72-year-old client is considered a normal process or aging the most common complain
a. My tongue feels swollen.”
b. “My tonsils are large and sore.”
c. “I have white and black spots under my tongue.”
d. “Food does not taste the same as it used to.”
e. Insomnia
34. A technique in which the hands and fingers are used to gather information by touch or it may be either superficial or deep
a. Inspection
b. Palpation
c. Percussion
d. Auscultation
e. Physical examination
35. During physical examination when using the stethoscope its exact position between
a. index and little fingers
b. thumb and all four fingers
c. index and ring fingers
d. thumb and index fingers
e. index and middle fingers
36. Occipital lobe of brain is said to be
a. memory storage center
b. Emotions control center
c. Visual center
d. Interpretation of sensory center
e. Auditory center
37. Name, Date of Birth, Age, Parents & siblings information of client are gather in
a. Present history
b. Past medical history
c. Bio-graphic data
d. Health history
e. Interview
38. When a nurse performed the physical examination of abdomen the sequence of examination should be
a. Inspection, auscultation, Percussion, palpation
b. Inspection, palpation ,Percussion, , auscultation,
c. auscultation , Inspection, , Percussion, palpation
d. Percussion, Inspection, auscultation, palpation
e. Palpation, Inspection, auscultation, Percussion,
39. The appropriate time to collect a urine specimen from a patient Is
a. before the physical examination
b. any time the patient feels he can provide a specimen
c. during the examination
d. after the examination
e. after follow up
40. The best examination position for the physician to evaluate the patient’s ability to fully expand the lungs is the
a. Sitting position
b. Prone position
c. Lithotomy position
d. knee-chest position
e. Fowler’s position
41. A patient who has low blood pressure or is in shock would be placed in a
a. Sitting position
b. Prone position
c. Lithotomy position
d. knee-chest position
e. Trendelenburg position
42. The normal range for body temperature is
a. 96°F to 98°F
b. 97°F to 99°F
c. 98°F to 99°F
d. 97°F to 100.4°F
e. 96°F to 97 °F
43. A temperature of 103°F is classified as
a. Normal
b. Hypo pyrexia
c. Hyper pyrexia
d. Low-grade fever
e. Pyrexia
44. One respiration consists of
a. One inhalation
b. One exhalation
c. One inhalation and one exhalation
d. The opening and closing of the valves of the heart
e. The opening and closing of the pulmonary valves of the lungs
45. The normal respiratory rate of an adult ranges from:
a. 8 to 16 respirations per minute
b. 10 to 18 respirations per minute
c. 12 to 20 respirations per minute
d. 16 to 22 respirations per minute
e. 14 to 20 respirations per minute
46. The abbreviation used to record oxygen saturation as measured by a pulse oximeter is:
a. SaO2
b. PCO2
c. PO2
d. SpO2
e. SpO4
47. Blood pressure is measured in:
a. Units
b. Degrees
c. Beats/min
d. Millimeters of mercury
e. Nanometer
48. Over which artery is the stethoscope placed when taking blood pressure:
a. Radial
b. Brachial
c. Apical
d. Carotid
e. Femoral
49. When measuring blood pressure, the patient’s arm should be positioned
a. Above heart level
b. At heart level
c. Across the chest
d. With the palm facing downward
e. With the palm facing upward
50. The term used to describe the point of lesser pressure on the arterial walls when assessing blood pressure:
a. Systolic pressure
b. Diastolic pressure
c. Diastole
d. Hypotension
e. Pulse pressure

Subjective paper
Q: 01. Define health assessment? Enlist the step of history taking.
Q: 02. What is interview? Explain the interview phases.
Q: 03. Describe the physical assessment skill & give any one example of each skill.
Q: 04. What is vital sign? Differentiate the value of infant, adult and older with example of normal range.
Q: 05. Define exercise? Enlist the type of exercise.
Q: 06. Define the following terms:
a. Tachycardia
b. Bradypnea
c. Otitis media
d. Percussion
e. Subjective & objective data
f. Temperature

Answer key

1 B
2 D
3 A
4 B
5 D
6 E
7 D
8 A
9 D
10 D
11 A
12 C
13 D
14 E
15 A
16 B
17 B
18 B
19 B
20 B

 

21 A
22 B
23 A
24 A
25 A
26 A
27 B
28 D
29 C
30 D
31 C
32 B
33 D
34 B
35 E
36 C
37 C
38 A
39 A

40 A
41 E
42 C
43 C
44 C
45 C
46 D
47 D
48 B
49 B
50 B

Multiple Sclerosis

Posted on by
Reply

Multiple sclerosis (MS) is a chronic demyelinating disease that affects the myelin sheath of neurons in the CNS. Multiple-Sclerosis-Infographic Multiple sclerosis is a disease that causes vision problems, numbness and tingling, muscle weakness, and other problems. It happens when the body’s infection-fighting system attacks and damages nerve cells and their connections in the brain and spinal cord. When the body’s infection-fighting system, called the “immune system,” attacks the body’s own cells, it is called an “autoimmune response.” It causes damage to myelin, the protective coating around the nerves. When myelin is damaged, messages can no longer be clearly transmitted from the brain and spinal cord to other parts of the body. Many people refer to multiple sclerosis as “MS.” INCIDENCE

  • Onset occurs between 20-40 years of age.
  • Women are more affected than men. (AANN, 2011).
  • Whites are more affected than Hispanics, blacks, or Asians.
  • Most prevalent in colder climates of North America & Europe.
  • Migration.

 

ETIOLOGY & RISK FACTORS

  • Exact cause is not known yet.
  • Most theories suggest that MS is an immunogenetic viral disease (with Epstein Barr virus).

Risk factors are:

  • Age (most of the time between 20-40 yrs).
  • Sex (women have more chance).
  • Family history (genetic susceptibility).
  • Certain infections (like Epsteinbarr virus).
  • Climate (more in cold climate areas).
  • Certain auto-immune diseases (higher risks with thyroid disease, type-1 DM or IBD).
  • Smoking.
  • Stress, fatigue.
  • Physical injury.
  • Pregnancy (may relating to stress to labour, or puerperium).

 

PATHOPHYSIOLOGY Due to etiological factors Activated T-cells (which recognise self Ag) expressed in CNS, & Macrophages (B-cells) enters the brain from peripherral circulation Production of inflammatory cytokines & reactive O2 species Inflammation Then activated T-cells & B-cells cause demyelination and destruction of oligodendrocytes Formation of plaque Causes scarring & destruction of sheath Compensatory system starts causing subsidation of edema & inflammation After that some remyelination process occurs which is often incomplete Multiple sclerosis. CLINICAL MANIFESTATIONS The course of illness varies from person to person.

  1. Fatigue is the lack of physical and mental energy that impacts daily tasks. Fatigue can be physical or mental and is not correlated to how much rest or sleep a person gets. It is one of the most common symptoms and impacts about 80 percent of people living with MS. It can be http://nursingfile.com/wp-content/uploads/2013/12/multiple-sclerosis-249x300.jpg the most debilitating factor, even for those who have minimal physical restrictions, and is one of the leading causes for people leaving the workforce.
  2. Heat intolerance in MS is a temporary worsening of symptoms with elevated body temperatures including hot and humid weather, exercising, sunbathing, or fevers. A small rise in body temperature (a quarter to a half a degree) can cause increased fatigue, tingling, blurry vision, or even the inability to walk. Most people living with MS have to avoid outdoor activity and/or use cooling garments to complete simple, daily activities due to this intolerance.
  3. Cognitive dysfunction affects high-level brain functions such as memory, attention/concentration, the ability to solve daily problems, understand and use language, and process information from different senses. Impaired cognition affects 50-65 percent of those living with MS and is another major reason for leaving the workforce early.
  4. Pain/abnormal sensation is a common symptom with MS and can be directly related to neuropathic pain (the disease process itself) or from musculoskeletal pain (changes to the body and immobility). The pain experience is unique to each person and can greatly limit his or her ability to participate in and enjoy socialization and activities. Those living with MS can also experience various abnormal sensations such as numbness and tingling, prickling, sharp/stabbing pains, hot/cold sensations, and burning pains which can also impact movement and daily function.
  5. Depression comes in various forms and can be one of the most common symptoms in MS, more common in people with MS than the general population. Depression can happen to anyone at any time during the disease course and does not correlate to disease severity, however it can greatly impact someone’s quality of life and ability to participate in daily activities.

 

COMPLICATIONS People with multiple sclerosis may also develop:

  • Muscle stiffness or spasms
  • Paralysis, typically in the legs
  • Problems with bladder, bowel or sexual function
  • Mental changes, such as forgetfulness or mood swings
  • Depression
  • Epilepsy

DIAGNOSTIC EVALUATION

  • There is no definitive test for MS.
  • Detailed history of episodes of neurologic dysfunction
  • Physical examination.

Other tests include:-

  • CSF evaluation (for presence of IgG antibody or oligoclonal bonding)
  • Evoked potentials of optic pathways & auditory system to assess presence of slowed nerve conduction.
  • MRI of brain and spinal cord (to determine the presence of MS plaques)
  • CT scan (to detect areas of demyelination, but with less detail as by MRI).

 

MEDICAL MANAGEMENT

• No exact cure.
• Aim is to prevent or postpone the long term disability (often evolves slowly over many years).
• The treatment falls into 3 categories:-
1. Treatment of acute relapses.
2. Treatment aimed at disease management.
3. Symptomatic treatment.

1. Treatment of acute relapse:-

  • Corticosteroid therapy ( anti-inflammatory & immunosuppressive property )
  • For example:
  • Methyl-prednisolone , (given I.V. or orally)
  • Azathioprine & cyclophosphamide (in severe cases)

2. Treat exacerbations:- (treatment aimed at disease management)

  • Interferon-Beta 1b
    • Betaseron, given subcutaneously. (antiviral & immuno-regulatory) (For ambulatory clients with relapsing –remitting).
  • Interferon Beta 1a
    • Avonex, (for treating replasing form of MS).
  • Glatiramer acetate
    • Copaxane, (for relapsing re-emitting MS).

3. Symptomatic treatment:-

  • For bladder dysfunction:
    • oxybutynin, propantheline.
  • For constipation:
    • psyllium hydrophilic mucilloid, suppositories.
  • For fatigue:
    • amantadine, modafinil .
  • For spasticity:
    • baclofen, diazefen, dantrolone.
  • For Tremor :
    • propanolol, phenobarbital, clonazepam.
  • For dysesthesias & trigeminal neurolgia:
    • carbamazepine, phenytoin, amitriptyline.
  • For dysesthesias:
    • Transcutaneous electrical nerve stimulation (TENS) is also helpful.

4. Nutritional therapy:-

  • Megavitamin therapy (cobalamin/vit. B12 and vit. C)
  • Low fat diet.
  • high roughage diet (to relieve constipation)

5. Other therapies:- (to improve neurological functioning)

  • Physical and speech therapies.
  • Exercise.
  • Water exercise.

 

SURGICAL MANAGEMENT

  • Deep brain stimulation:-
    • if other options have failed then a device is implanted that stimulates an area of brain. (in case of severe tremor in limbs).
  • Implantation of a drug catheter or pump:
    • a catheter is placed in lower spinal area to deliver a constant flow of drug like baclofen. (in case of severe pain or spasticity).

NURSING DIAGNOSIS:

    • Impaired physical mobility related to fatigue & weakness
    • Activity intolerance r/t weakness, dizziness, and unsteady gait
    • Self-esteem disturbance r/t loss of health & lifestyle changes

 

NURSING MANAGEMENT

    1. Promotes physical mobility – activity and rest
  • no vigorous physical exercise
  • frequent rest periods
  • walking and gait exercises
  • minimize spasticity and contractures – warm packs, daily muscle stretching
  • activities: swimming, stationary bike, progressive wt bearing
  • Minimize effects of immobility; skin integrity; cough and deep breathing exercises.
    1. Prevent injury – walk with feet wide apart, environment awareness and modification, gait training. Use of assistive devices – walker, cane etc.
    2. Promote bladder & bowel control – Urinal/bedpan readily available, po fluids intake schedule/voiding schedule, increase fiber in diet, intermittent self-catheterization
    3. Improve sensory and cognitive function:
  • Vision – eye patch for diplopia; prism glasses for reading; talking books
  • Speech – slurred, low volume, problems with phonation – speech therapist
  • Cognitive & emotional responses – forgetfulness, easily distracted, emotionally labile, social activities; hobbies.
    1. Development of coping strengths – education about diseases process; stress relief; network of services – social, speech, PT, psychological, homemaker/meal on wheels
    2. Improve self-care – assistive devices, raised toilet seat, shower bench, reached tongs, decrease physical and emotional stress, decrease exposure of extreme temperatures
    3. Adapting to sexual dysfunction – counseling, plan sexual activity, willingness to experiment.

Answer Key: Pre-Interview Written test for the post of Staff Nurse (BPS-16)- SPSC-2019

Posted on by
Reply

Sindh public service commission (SPSC)
Pre-Interview Written test for the post of Staff Nurse (BPS-16)
Question Paper: 24th August,2019

Answer key
1. C
2. B
3. B
4. A
5. C
6. C
7. A
8. C
9. A
10. A

11. C
12. B
13. B
14. D
15. C
16. B
17. A
18. C
19. D
20. B

21. A
22. D
23. A
24. C
25. D
26. B
27. B
28. B
29. C
30. C

31. C
32. D
33. B
34. B
35. A
36. B
37. C
38. B
39. B
40. B

41. B
42. C
43. C
44. D
45. A
46. D
47. D
48. B
49. B
50. A

 

For QUESTION PAPER, please click on the below link:

Pre-Interview Written test for the post of Staff Nurse (BPS-16)- SPSC-2019