block iii unit i testing of hypothesis

Unit I: Testing of Hypothesis

Block III/ Testing of Hypothesis

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Contents

1.1 Objectives ............................................................................................................................ 4

1.2 Introduction.......................................................................................................................... 4

1.3 Hypothesis Testing Methodology........................................................................................ 5

1.4 Z -test of Hypothesis for the Mean .................................................................................... 12

1.5 One-tailed Testing.............................................................................................................. 12

1.6 Two-tail Test ...................................................................................................................... 14

1.7 T-Test Of Hypothesis For The Mean................................................................................. 15

1.8 Z-Test Of Hypothesis For The Proportion......................................................................... 17

1.9 Test For The Difference Between Two Means And The Proportions ............................... 25

1.10 Self Evaluating Questions .............................................................................................. 28

1.11 References.......................................................................................................................... 31


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BLOCK III- Statistical Decision Theory

Unit I Testing Of Hypothesis

1.1 Objectives

After studying this unit, you will be able to:

explain the Hypothesis testing methodology

explain the one and two nail tests

describe the T test hypothesis for the mean and proportion

1.2 Introduction

A hypothesis is a judgment about a situation, outcome, or population parameter based simply on an assumption or intuition with no concrete backup information or analysis. Hypothesis testing is to take sample data and make on objective decision based on the results of the test within an appropriate significance level.

Hypothesis Tests

Statisticians follow a formal process to determine whether to reject a null hypothesis, based on sample data. This process, called hypothesis testing, consists of four steps.

State the hypotheses. This involves stating the null and alternative hypotheses. The hypotheses are stated in such a way that they are mutually exclusive. That is, if one is true, the other must be false.


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Formulate an analysis plan. The analysis plan describes how to use sample data to evaluate the null hypothesis. The evaluation often focuses around a single test statistic.

Analyze sample data. Find the value of the test statistic (mean score, proportion, t-score, z-score, etc.) described in the analysis plan.

Interpret results. Apply the decision rule described in the analysis plan. If the value of the test statistic is unlikely, based on the null hypothesis, reject the null hypothesis.

1.3 Hypothesis Testing Methodology

Suppose I have two coins in my pocket. One is a fair coini.e., p(Head) = p(Tail) = 0.5. The other coin is biased toward Tails: p(Head) = .15, p(Tail) = .85. I then place the two coins on a table, and allow you to choose one of them. You take the selected coin, and flip it 11 times, noting each time whether it showed Heads or Tails.

Let X = the number of Heads observed in 11 flips.

Let hypothesis A be that you selected and flipped the fair coin.

Let hypothesis B be that you selected and flipped the biased coin.

Under what circumstances would you decide that hypothesis A is true?

Under what circumstances would you decide that hypothesis B is true?

FAIR COIN

p = p(Head) = 0.5

q = p(Tail) = 0.5

BIASED COIN


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p = p(Head) = 0.15

q = p(Tail) = 0.85

A good way to start is to think about what kinds of outcomes you would expect for each

hypothesis. For example, if hypothesis A is true (i.e., the coin is fair), you expect the number of Heads to be somewhere in the middle of the 0-11 range. But if hypothesis B is true (i.e., the coin is biased towards tails), you probably expect the number of Heads to be quite small.

Note as well that a very large number of Heads is improbable in either case, but is less probable if the coin is biased towards tails.

In general terms, therefore, we will decide that the coin is biased towards tails (hypothesis B) if the number of Heads is quite low; otherwise, we will decide that the coin is fair (hypothesis A).

IF the number of heads is LOW

THEN decide that the coin is biased

ELSE decide that the coin is fair.

But an obvious problem now confronts us: That is, how low is low? Where do we draw the line between low and middling?

The answer is really quite simple. The key is to recognize that the variable X (the number of Heads) has a binomial distribution. Furthermore, if hypothesis A is true, X will have a binomial distribution with N = 11, p = .5, and q = .5. But if hypothesis B is true, then X will have a binomial distribution with N = 11, p = .15, and q = .85.

We can generate these two distributions using the formula you learned earlier.


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Two Binomial Distributions with N = 11 and X = # of Heads

X ( | ) A p X H ( | ) B p X H

0 .0005 .1673

1 .0054 .3248

2 .0269 .2866

3 .0806 .1517

4 .1611 .0536

5 .2256 .0132

6 .2256 .0023

7 .1611 .0003

8 .0806 .0000

9 .0269 .0000

10 .0054 .0000

11 .0005 .0000

1.0000 1.0000

We are now in a position to compare conditional probabilities for particular experimental outcomes. For example, if we actually did carry out the coin tossing experiment and obtained 3Heads (X=3), we would know that the probability of getting exactly 3 Heads is lower if hypothesis A is true (.0806) than it is if hypothesis B is true (.1517). Therefore, we might decide that hypothesis B is true if the outcome was X = 3 Heads.


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But what if we had obtained 4 Heads (X=4) rather than 3? In this case the probability of exactly 4 Heads is higher if hypothesis A is true (.1611) than it is if hypothesis B is true (.0536). So in this case, we would probably decide that hypothesis A is true (i.e., the coin is fair).

A decision rule to minimize the overall probability of error

In more general terms, we have been comparing the (conditional) probability of a particular outcome if hypothesis A is true to the (conditional) probability of that same outcome if hypothesis B is true. And we have gone with whichever hypothesis yields the higher conditional probability for the outcome. We could represent this decision rule symbolically as follows:

if p(X | A) > p(X | B) then choose A

if p(X | A) < p(X | B) then choose B

Note that even if the coin is biased towards tails, it is possible for the number of Heads to be very large; and if the coin is fair, it is possible to observe very few Heads. No matter which hypothesis we choose, therefore, there is always the possibility of making an error.3 However, the use of the decision rule described here will minimize the overall probability of error.

9 .0269 .0000

10 .0054 .0000

11 .0005 .0000

1.0 1.0000

Null and Alternative Hypotheses

In any experiment, there are two hypotheses that attempt to explain the results. They are the

alternative hypothesis and the null hypothesis.


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Alternative Hypothesis (H1 or H A ). In experiments that entail manipulation of an independent variable, the alternative hypothesis states that the results of the experiment are due to the effect of the independent variable. In the coin tossing example above, H 1 would state that the biased coin had been selected, and that p(Head) = 0.15.

Null Hypothesis (H 0 ). The null hypothesis is the complement of the alternative hypothesis. In other words, if H1 is not true, then H0 must be true, and vice versa. In the foregoing coin tossing situation, H0 asserts that the fair coin was selected, and that p(Head) = 0.50.

Thus, the decision rule to minimize the overall p(error) can be restated as follows:

if p(X | H0 ) > p(X |H 1 ) then do not reject H0

if p(X | H0 ) < p(X | H 1 ) then reject H0

Rejecting H0 is essentially the same thing as choosing H1 , but note that many statisticians are very careful about the terminology surrounding this topic. According to statistical purists, it is only proper to reject the null hypothesis or fail to reject the null hypothesis. Acceptance of either hypothesis is strictly forbidden.

Rejection Region

As implied by the foregoing, the rejection region is a range containing outcomes that lead to rejection of H0 . In the coin tossing example above, the rejection region consists of 0, 1, 2, and 3. For X > 3, we would fail to reject H0 .

Type I and Type II Errors

When one is making a decision about 0H0 (i.e., either to reject or fail to reject it), it is possible to make two different types of errors.

Type I Error. A Type I Error occurs when H0 is TRUE, but the experimenter decides to reject 0H0 . In other words, the experimenter attributes the results of the experiment to the effect of the


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independent variable, when in fact the independent variable had no effect. The probability of Type I error is symbolized by the Greek letter alpha:

p(Type I Error) =

Type II Error. A Type II Error occurs when H0 is FALSE (or 1 H is TRUE), and the experimenter fails to reject H0 . In this case, the experimenter concludes that the independent variable has no effect when in fact it does. The probability of Type II error is symbolized by the Greek letter beta:

p(Type II Error) =

The two types of errors are often illustrated using a 2x2 table as shown below.

The True State Of Nature

Your Decision H0 Is True H0 Is False

Reject H0 Type I error ( ) Correct rejection of H0

Fail to reject H0 Correct failure to reject H0 Type II error ( )

EXAMPLE: To illustrate how to calculate and , let us return to the coin tossing experiment Decision rule to minimize the overall probability of error

X p(X | H0 ) 1 p(X | H )

0 .0005 .1673

1 .0054 .3248

2 .0269 .2866 Reject 0 H

3 .0806 .1517


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4 .1611 .0536

5 .2256 .0132

6 .2256 .0023

7 .1611 .0003 Fail to reject 0 H

8 .0806 .0000

9 .0269 .0000

10 .0054 .0000

11 .0005 .0000

1.0000 1.0000

= .0005 + .0054 + .0269 + .0806 = .1134

= .0536 + .0132 + .0023 + .0003 + ... = .0694

Let us begin with , the probability of Type I error. According to the definitions given above, Type I error can only occur if the null hypothesis is TRUE. And if the null hypothesis is true, then we know that we should be taking our probability values from the distribution that is conditional on 0 H being true. In this case, that means we want the distribution on the left (binomial with N = 11, p = .5).

We now know which distribution to use, but still have to decide which region to take values from, above the line or below the line. We know that H 0 is true, and we know that an error has been made. In other words, we have rejected H 0 . We could reject 0 H for any value of X equal to or less than 3. And so, the probability of Type I error is the sum of the probabilities in the rejection region (from the distribution that is conditional on H0 being true):


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= .0005 + .0054 + .0269 + .0806 = .1134

1.4 Z -test of Hypothesis for the Mean

After deciding what level of significance to use, our next task in hypothesis testing is to determine the appropriate probability distribution. We have a choice between the normal distribution, and the t distribution. The rules for choosing the appropriate dis t r ibut ion are similar to those we encountered in the unit on estimation. The Table below summarizes when to use the normal and t distributions in making tests of means. Later in this unit, we shall examine the distributions appropriate for testing hypotheses about proportions.

Remember one more rule when testing the hypothesized values of a mean. As in estimation, use the finite population multiplier whenever the population is finite in size, sampling is

done without replacement, and the sample is more than 5 percent of the population

Conditions for using the Normal and t distributions in Testing Hypothesis about

When the Population

Standard Deviation is

When the Population

Standard Deviation is Sample size n is larger than Normal distribution, z -table

Normal distribution, z -table Sample size n is 30 or less

and we assume the

Normal distribution, z -table

t Distribution, t - table

1.5 One-tailed Testing

A test of a statistical hypothesis, where the region of rejection is on only one side of the sampling distribution, is called a one-tailed test.


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For example, suppose the null hypothesis states that the mean is less than or equal to 10. The alternative hypothesis would be that the mean is greater than 10. The region of rejection would consist of a range of numbers located on the right side of sampling distribution; that is, a set of numbers greater than 10.

A positive one-tailed test predicts that there will be an increase in the population mean as a result of the manipulation.

H0

H1 >

A negative one-tailed test predicts that there will be a decrease in the population mean as a result of the manipulation.

H0

H1 <


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1.6 Two-tail Test

A test of a statistical hypothesis, where the region of rejection is on both sides of the sampling distribution, is called a two-tailed test.

For example, suppose the null hypothesis states that the mean is equal to 10. The alternative hypothesis would be that the mean is less than 10 or greater than 10. The region of rejection would consist of a range of numbers located on both sides of sampling distribution; that is, the region of rejection would consist partly of numbers that were less than 10 and partly of numbers that were greater than 10.

A two tailed test is appropriate when the null hypothesis is = Ho (where Ho is some specified value) and the alternative hypothesis is Ho.

Example 1: Assume that a manufacturer of light bulbs wants to produce bulbs with a mean life of = Ho = 1,000 hours. If the lifetime is shorter, he will lose customers to his

competitions; if the lifetime is longer, he will have a very high production cost because the filaments will be excessively thick. In order to see whether his production


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process is working properly, he takes a sample of the output to test the hypothesis Ho; =

1,000. Because he does not want to deviate significantly from 1,000 hours in either direction, the appropriate alternative hypothesis is H1: 1,000, and he uses a

two-tailed test. That is, he rejects the null hypothesis if the mean life of bulbs in the sample is either too far above 1,000 hours or too far below 1,000 hours. However, there are situations in which a two-tailed test is not appropriate, and we must u se a one-tail

test.

1.7 T-Test Of Hypothesis For The Mean

T-test concerns a number of procedures concerned with comparing two averages. It can be used to compare the difference in weight between two groups on a different diet, or to compare the proportion of patients suffering from complications after two different types of operations, or the number of traffic accidents on two busy junctions. You can compare continuous averages, they can be above or below one, examples are the difference in mean length or weight between two groups of people. The certainty with which these averages are measured are expressed in the


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standard deviation. Also, you can compare proportion averages, basically a number divided by a larger number. Examples are the proportion of people suffering from complications comparing two different types of operation (number of complications on the number of operations), the proportion of a manufactured product damaged comparing two different methods of production (number damaged on the number manufactured). The certainty of these averages is directly related to the number of cases observed.

The t-test gives the probability that the difference between the two means is caused by chance. It is customary to say that if this probability is less than 0.05, that the difference is significant, the difference is not caused by chance.

The t-test is basically not valid for testing the difference between two proportions. However, the t-test in proportions has been extensively studied, has been found to be robust, and is widely and successfully used in proportional data. With one exception: if one of the proportions is very close to zero, one or minus one, you will do better with Fishers exact test.

If the calculated t value is equal or larger than the table value, then the null hypothesis is rejected.

Independent and non-independent samples

Independent samples are samples that are randomly formed, that is formed without any type of matching.

Non independent samples are samples formed by some type of matching.

t test for independent samples

The t test for independent samples is used to determine whether there is probably a significant difference between the means of two independent samples.

_ _


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t = X1 - X2 / square root of (SS1 + SS2) / (n1 + n2 -2) * (1 / n1 + 1 / n2)

t test for nonindependent samples

The t test for non-independent samples is used to determine whether there is probably a significant difference between the means of two matched or non-independent samples or between the means for one sample at two different times.

When samples are non-independent, the error term of the t test tends to be smaller and therefore there is a higher probability that the null hypothesis will be rejected.

t = D / square root of (sum of D square - sum square of D / N) / N(N-1)

1.8 Z-Test Of Hypothesis For The Proportion

In hypothesis testing for the proportion we test the assumption about the value of the population proportion. In the same way for the mean, we take a sample from this

+

+

+

=

2121

21 112

21

nnnn

SSSS

XXt

( ))1(

22

=

NNND

D

Dt


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population, d e t e r mine the sample proportion, and measure the difference between this proportion and the hypothesized population value. If the difference between the sample proportion and the hypothesized population proportion is small, then the higher is the probability that our hypothesized population proportion value is correct. If the difference is large then the probability that our hypothesized value is correct is low.

The sampling method for each population is simple random sampling.

The samples are independent.

Each sample includes at least 10 successes and 10 failures. (Some texts say that 5 successes and 5 failures are enough.)

Each population is at least 10 times as big as its sample.

This approach consists of four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results.

State the Hypotheses

Every hypothesis test requires the analyst to state a null hypothesis and an alternative hypothesis. The table below shows three sets of hypotheses. Each makes a statement about the difference d between two population proportions, P1 and P2. (In the table, the symbol means " not equal to ".)

Set Null hypothesis Alternative hypothesis Number of tails

1 P1 - P2 = 0 P1 - P2 0 2

2 P1 - P2 > 0 P1 - P2 < 0 1

3 P1 - P2 < 0 P1 - P2 > 0 1

The first set of hypotheses (Set 1) is an example of a two-tailed test, since an extreme value on either side of the sampling distribution would cause a researcher to reject the null hypothesis. The other two sets of hypotheses (Sets 2 and 3) are one-tailed tests, since an extreme value on only one side of the sampling distribution would cause a researcher to reject the null hypothesis.


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When the null hypothesis states that there is no difference between the two population proportions (i.e., d = 0), the null and alternative hypothesis for a two-tailed test are often stated in the following form.

H0: P1 = P2

Ha: P1 P2

Formulate an Analysis Plan

The analysis plan describes how to use sample data to accept or reject the null hypothesis. It should specify the following elements.

Significance level. Often, researchers choose significance levels equal to 0.01, 0.05, or 0.10; but any value between 0 and 1 can be used.

Test method. Use the two-proportion z-test (described in the next section) to determine whether the hypothesized difference between population proportions differs significantly from the observed sample difference.

Analyze Sample Data

Using sample data, complete the following computations to find the test statistic and its associated P-Value.

Pooled sample proportion. Since the null hypothesis states that P1=P2, we use a pooled sample proportion (p) to compute the standard error of the sampling distribution.

p = (p1 * n1 + p2 * n2) / (n1 + n2)

where p1 is the sample proportion from population 1, p2 is the sample proportion from

population 2, n1 is the size of sample 1, and n2 is the size of sample 2.


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Standard error. Compute the standard error (SE) of the sampling distribution difference between two proportions.

SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }

where p is the pooled sample proportion, n1 is the size of sample 1, and n2 is the size of sample 2.

Test statistic. The test statistic is a z-score (z) defined by the following equation.

z = (p1 - p2) / SE

where p1 is the proportion from sample 1, p2 is the proportion from sample 2, and SE is the standard error of the sampling distribution.

P-value. The P-value is the probability of observing a sample statistic as extreme as the test statistic. Since the test statistic is a z-score, use the Normal Distribution Calculator to assess the probability associated with the z-score. (See sample problems at the end of this lesson for examples of how this is done.)

The analysis described above is a two-proportion z-test.

Interpret Results

If the sample findings are unlikely, given the null hypothesis, the researcher rejects the null hypothesis. Typically, this involves comparing the P-value to the significance level, and rejecting the null hypothesis when the P-value is less than the significance level.

Test Your Understanding of This Lesson

In this section, two sample problems illustrate how to conduct a hypothesis test for the difference between two proportions. The first problem involves a a two-tailed test; the second problem, a one-tailed test.


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Problem 1: Two-Tailed Test

Suppose the Acme Drug Company develops a new drug, designed to prevent colds. The company states that the drug is equally effective for men and women. To test this claim, they choose a a simple random sample of 100 women and 200 men from a population of 100,000 volunteers.

At the end of the study, 38% of the women caught a cold; and 51% of the men caught a cold. Based on these findings, can we reject the company's claim that the drug is equally effective for men and women? Use a 0.05 level of significance.

Solution: The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P1 = P2

Alternative hypothesis: P1 P2

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the proportion from population 1 is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a two-proportion z-test.

Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).

p = (p1 * n1 + p2 * n2) / (n1 + n2) = [(0.38 * 100) + (0.51 * 200)] / (100 + 200) = 140/300 = 0.467

SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }


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SE = sqrt [ 0.467 * 0.533 * ( 1/100 + 1/200 ) ] = sqrt [0.003733] = 0.061

z = (p1 - p2) / SE = (0.51 - 0.38)/0.061 = 2.13

where p1 is the sample proportion in sample 1, where p2 is the sample proportion in sample 2, n1

is the size of sample 2, and n2 is the size of sample 2.

Since we have a two-tailed test, the P-value is the probability that the z-score is less than -2.13 or greater than 2.13.

We use the Normal Distribution Calculator to find P(z < -2.13) = 0.017, and P(z > 2.13) = 0.017. Thus, the P-value = 0.017 + 0.017 = 0.034.

Interpret results. Since the P-value (0.034) is less than the significance level (0.05), we cannot accept the null hypothesis.

Note: If you use this approach on an exam, you may also want to mention why this approach is

appropriate. Specifically, the approach is appropriate because the sampling method was simple random sampling, the samples were independent, each population was at least 10 times larger than its sample, and each sample included at least 10 successes and 10 failures.

Problem 2: One-Tailed Test

Suppose the previous example is stated a little bit differently. Suppose the Acme Drug Company develops a new drug, designed to prevent colds. The company states that the drug is more effective for women than for men. To test this claim, they choose a simple random sample of 100 women and 200 men from a population of 100,000 volunteers.

At the end of the study, 38% of the women caught a cold; and 51% of the men caught a cold. Based on these findings, can we conclude that the drug is more effective for women than for men? Use a 0.01 level of significance.


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Null hypothesis: P1 >= P2

Alternative hypothesis: P1 < P2

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the proportion of women catching cold (p1) is sufficiently smaller than the proportion of men catching cold (p2).

Formulate an analysis plan. For this analysis, the significance level is 0.01. The test method is a two-proportion z-test.

Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).

p = (p1 * n1 + p2 * n2) / (n1 + n2) = [(0.38 * 100) + (0.51 * 200)] / (100 + 200) = 40/300 = 0.467

SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] } SE = sqrt [ 0.467 * 0.533 * ( 1/100 + 1/200 ) ] = sqrt [0.003733] = 0.061

z = (p1 - p2) / SE = (0.38 - 0.51)/0.061 = -2.13

where p1 is the sample proportion in sample 1, where p2 is the sample proportion in sample 2, n1

is the size of sample 2, and n2 is the size of sample 2.


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Since we have a one-tailed test, the P-value is the probability that the z-score is less than -2.13. We use the Normal Distribution Calculator to find P(z < -2.13) = 0.017. Thus, the P-value = 0.017.

Interpret results. Since the P-value (0.017) is greater than the significance level (0.01), we cannot reject the null hypothesis.

Classification of Test Statistics

Statistics used for Testing of Hypothesis can be classified as follows

A Large Samples (n > 30) Attributes (proportions)

Test No. Description of Test Test Statistics Notes 1 Test for specified

P - Ps

P Population proportion

2 Test for

specified proportion

Finite

P - Ps

Z = PQ 1/2 N n

P = Population proportion

Ps = Sample 3 Test between proportions

different Population

P1 P2

Z = P1Q1 P2 Q2 1/2

P1 -first sample

proportion P2 -second

sample proportion

Q1 = 1 P, Q2 = 1-P2

4 Test between

proportion

same population

P1 P2

Z = PQ (1/n1 + 1/n2) 1/2

P1 -first sample proportion

P2 -second

sample


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1.9 Test For The Difference Between Two Means And The Proportions

Suppose a researcher wishes to determine whether there is a difference in the average age of nursing students who enroll in a nursing program at a community college and those who enroll in a nursing program at a university. In this case, the researcher is not interested in the average age of all beginning nursing students; instead, he is interested in comparing the means of the two groups. His research question is: Does the mean age of nursing students who enroll at a community college differ from the mean age of nursing students who enroll at a university? Here, the hypotheses are

H0: m1 _ m2

H1: m1 _ m2

where

m1 _ mean age of all beginning nursing students at the community college

m2 _ mean age of all beginning nursing students at the university

Another way of stating the hypotheses for this situation is

H0: m1 _ m2 _ 0

H1: m1 _ m2 _ 0

If there is no difference in population means, subtracting them will give a difference of

Zero. If they are different, subtracting will give a number other than zero. 1

1. The samples must be independent of each other. That is, there can be no relationship between the subjects in each sample.


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2. The populations from which the samples were obtained must be normally distributed, and the standard deviations of the variable must be known, or the sample sizes must be greater than or equal to 30.

The theory behind testing the difference between two means is based on selecting pairs of samples and comparing the means of the pairs. The population means need not be known.

All possible pairs of samples are taken from populations. The means for each pair of samples are computed and then subtracted, and the differences are plotted. If both populations have the same mean, then most of the differences will be zero or close to zero.

Occasionally, there will be a few large differences due to chance alone, some positive and others negative. If the differences are plotted, the curve will be shaped like a normal distribution and have a mean of zero.

If both independent sample are large or both populations from which the sample is selected are drawn from a normal distributions with known standard deviations, use the normal distribution to test 1 2.

If both independent sample are large or both populations from which the sample is selected are drawn from a normal distributions with unknown standard deviations, use the t-distribution with = n1 + n2 2 degrees of freedom to test 1 2 .

If the sample size is large or the population of paired differences is normally distributed with unknown standard deviations, then use the t-distribution with = n 1 degrees of freedom to test d .

Testing the Difference Between Two Means: Small Independent Samples

Excel has a two-sample t test in its Data Analysis tools. To perform the t test for the difference between means, see Example XL93.

Example XL93

Test the hypothesis that there is no difference between population means based on these sample


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data. Assume the population variances are not equal. Use = 0.05

A 32 38 37 36 36 34 39 36 37 4Set B 30 36 35 36 31 34 37 33 32

1. Enter the 10-number data set A in column A.

2. Enter the 9-number data set B in column B.

3. Select Tools>Data Analysis and choose t-Test: Two-Sample Assuming Unequal

Variances.

4. Enter the data ranges, hypothesized mean difference (here, 0), and a.

5. Select a location for output and click [OK].


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The output reports both one- and two-tailed P-values.

If the variances are equal, use the two-sample t test, assuming equal-variances procedure.

1.10 Self Evaluating Questions

Problem 1: Two-Tailed Test

An inventor has developed a new, energy-efficient lawn mower engine. He claims that the engine will run continuously for 5 hours (300 minutes) on a single gallon of regular gasoline. Suppose a simple random sample of 50 engines is tested. The engines run for an average of 295 minutes, with a standard deviation of 20 minutes. Test the null hypothesis that the mean run time is 300 minutes against the alternative hypothesis that the mean run time is not 300 minutes. Use a 0.05 level of significance. (Assume that run times for the population of engines are normally distributed.)




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Null hypothesis: = 300 Alternative hypothesis: 300

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample mean is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t-score test statistic (t).

SE = s / sqrt(n) = 20 / sqrt(50) = 20/7.07 = 2.83 DF = n - 1 = 50 - 1 = 49 t = (x - ) / SE = (295 - 300)/2.83 = 1.77

where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.

Since we have a two-tailed test, the P-value is the probability that the t-score having 49 degrees of freedom is less than -1.77 or greater than 1.77.

We use the t Distribution Calculator to find P(t < -1.77) = 0.04, and P(t > 1.75) = 0.04. Thus, the P-value = 0.04 + 0.04 = 0.08.


Note: If you use this approach on an exam, you may also want to mention why this approach is

appropriate. Specifically, the approach is appropriate because the sampling method was simple random sampling, and the population was normally distributed.


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Problem 2: One-Tailed Test

Bon Air Elementary School has 300 students. The principal of the school thinks that the average IQ of students at Bon Air is at least 110. To prove her point, she administers an IQ test to 20 randomly selected students. Among the sampled students, the average IQ is 108 with a standard deviation of 10. Based on these results, should the principal accept or reject her original hypothesis? Assume a significance level of 0.01.



Null hypothesis: >= 110 Alternative hypothesis: < 110

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.01. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t-score test statistic (t).

SE = s / sqrt(n) = 10 / sqrt(20) = 10/4.472 = 2.236 DF = n - 1 = 20 - 1 = 19 t = (x - ) / SE = (108 - 110)/2.236 = -0.894

where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.


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Since we have a one-tailed test, the P-value is the probability that the t-score having 19 degrees of freedom is less than -0.894.

We use the t Distribution Calculator to find P(t < -0.894) = 0.19. Thus, the P-value is 0.19.


1.11 References

Koontz and ODonnell. Essentials of Management. E-McGraw Hill, New Delhi

Fred Luthan S. Introduction to Management. McGraw Hill, New Delhi, 2008

Peter.F.Drucker. The Practice of Management. Allied Publishers, 2008

Stoner, Freemen and Gilbert. Management. Pearson (6th Edition), 1995

Griffin. Management. South Western Educational Publishing, 2006

Peter. F. Drucker. Management- Tasks and Responsibilities. Harper Business 1993

Theo Haimann. Professional Management. Houghton Miller, 1998

Richard L.Draft. Organization Theory and Design. Thomson Learning, 2004

Peter F.Drucker. People and Performance. Harvard Business School Press, 2007

block iii unit i testing of hypothesis

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