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Testing Hypotheses: Two-Sample Tests

Chapter 9

Statistics for Management

Levin and Rubin

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Basics

• Hypotheses based on one group• A two-sample case involves comparing a

parameter between two distinct groups, such as males versus females.

A two-sample hypothesis test might be to analyze the difference in the mean cholesterol level between men and women. Or, to determine if the percentage of job success is different between men and women.

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Different Parameters

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Testing Hypothesis about Difference between Two Population Means (Two Sample Tests of

Hypothesis)

•Null Hypothesis:

H0: (µ1 - µ2)=D

or

µ1 = µ2

•Alternate Hypothesis

H1:

(µ1 - µ2)>D

(µ1 - µ2)<D [µ1 > µ2

µ1 < µ2] ..one-tailed testAlternate Hypothesis

H1: (µ1 - µ2) ≠D

[µ1 ≠ µ2

]

..

two-tailed test•Test Statistics:

2

22

1

21

21

2

22

1

21

021 )()(

nn

xxzor

nn

Dxxz

(replace s when σ not available)

•Rejection Region

z > zα or z < -zα

(one tailed test) z > zα/2 or z < -zα/2

(two tailed test)

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Mary Jo Fitzpatric is the Vice President for Nursing Services at St. Luke’s Memorial Hospital. Recently she noticed in the job posting for nurses that those that are unionised seem to offer higher wages. She decided to investigate and gathered the following sample information.

Group Mean Wage

Sample Standard Deviation

Sample Size

Union $20.75 $2.25 40

Nonunion $19.80 $1.90 45

Would it be reasonable for her to conclude that there is significant difference in earning between union and non-union nurses? Use the .01 significance level.

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Example

• A manpower-development statistician is asked to determine whether the hourly wages of semiskilled workers are the same in two cities. The results of the survey are presented in the following table:

Hourly Wage Rate

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SolutionStep 1: This is a two-tailed test. The hypothesis is stated below. The significance level is 0.05 (given)

H0: μ1 = μ2 versus H1: μ1 ≠ μ2 Step 2: Since this is a test of the means and the degrees of freedom (n1 + n2 - 2) is in excess of 30, a z test is appropriate. The critical values are ± 1.96 (from a z table).

Step 3:

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Step 4: Sketch the distribution, locate the critical values and the test statistic. Step 5: Decide! Since the test statistic values lies within the rejection region then there is sufficient statistical evidence based on this sample to reject H0. The test for a difference between parameters does not have to be zero, it can be non-zero. For example: H0: μ1 - μ2 ≤ 0.10 versus H1: μ1 - μ2 > 0.10

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Problem: 9-3

• Two research laboratories have independently produced drugs that provide relief to arthritis sufferers. The first drug was tested on a group of 90 arthri tis sufferers and produced an average of 8.5 hours of relief, and a sample standard deviation of 1.8 hours. The second drug was tested on 80 arthritis sufferers, producing an average of 7.9 hours of relief, and a sample standard deviation of 2.1 hours. At the 0.05 level of significance, does the second drug provide a significantly shorter period of relief?

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Problem: 9-6

• Notwithstanding the Equal Pay Act of 1963, in 1993 it still appeared that men earned more than women in similar jobs. A random sample of 38 male machine tool operators found a mean hourly wage of $11.38, and the sample standard de viation was $1.84. A random sample of 45 female machine-tool operators found their mean wage to be $8.42, and the sample standard deviation was $1.31. On the basis of these samples, is it reasonable to conclude (at a = 0.01) that the male operators are earning over $2.00 more per hour than the female operators?

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Testing Hypothesis about Difference between Two Population Proportion

1.Null Hypothesis: H0: (p1 - p2)=D or p1 = p2

2.Alternate Hypothesis H1: (p1 - p2)>D (p1 - p2)<D [p1 > p2 p1 < p2] ..one-tailed testAlternate Hypothesis H1: (p1 - p2) ≠D[p1 ≠ p2 ] .. two-tailed test3. Test Statistics:

Rejection Region z > zα or z < -zα (one tailed test) z > zα/2 or z < -zα/2 (two tailed test)

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Testing Hypothesis about Difference between Two Population Proportion: Pooled

When population parameters ‘p’ and ‘q’ are unknown: 1. Null Hypothesis: H0: p1 = p2 [p1 -

p2 = 0] 2. Alternate Hypothesis H1: p1 > p2 p1 < p2 ..one-tailed test

Alternate Hypothesis H1: p1 ≠ p2 two-tailed test

3. Pooled proportion: 21

21ˆnn

xxp

4. Test Statistics: ]

11[ˆˆ

)ˆˆ(

21

21

nnqp

ppz

5. Rejection Region z > zα or z < -zα (one tailed test) z > zα/2 or z < -zα/2(two tailed test)

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Example:1. According to a report by the American Cancer Society, more men

than women smoke and twice as many smokers die prematurely than nonsmokers. In random samples of 200 males and 200 females, 62 of the males and 54 of the females were smokers. Is there sufficient evidence to conclude that the proportion of male smokers higher from the proportion of female smokers when α = .01?

2. A financial analyst wants to compare the turnover rates, in percent, for shares of oil related stocks versus other stocks. She selected 32 oil-related stocks and 49 other stocks. The mean turnover of oil related stocks is 31.4 percent and the standard deviation 5.1 percent. For the other stocks, the mean rate was computed to be 34.9 percent and the standard deviation 6.7 percent. Is there a significant difference in the turnover rates of the two types of stock?

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Problem: 9-22

• A coal-fired power plant is considering two different systems for pollution abatement. The first system has reduced the emission of pollutants to accept able levels 68 percent of the time, as determined from 200 air samples. The sec ond, more expensive system has reduced the emission of pollutants to ac ceptable levels 76 percent of the time, as determined from 250 air samples. If the expensive system is significantly more effective than the inexpensive sys tem in reducing pollutants to acceptable levels, then the management of the power plant will install the expensive system. Which system will be installed if management uses a significance level of 0.02 in making its decision?

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Problem: 9-23 • A group of clinical physicians is performing tests on patients to

determine the effectiveness of a new antihypertensive drug. Patients with high blood pressure were randomly chosen and then randomly assigned to either the control group (which received a well-established antihypertensive) or the treatment group (which received the new drug). The doctors noted the per centage of patients whose blood pressure was reduced to a normal level within 1 year. At the 0.01 level of significance, test appropriate hypotheses to determine whether the new drug is significantly more effective than the older drug in reducing high blood pressure.

Group Proportion That Improved Number of Patients

Treatment 0.45 120

Control 0.36 150

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Test for difference between Means: small sample size

• For small samples sizes, we must estimate a 'pooled' estimate (a.k.a. a weighted average) of the variances for the two groups. This estimate is:

• and then, the estimated standard error is:

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Tests for Difference Between Two Means: Small Sample Size

1. Null Hypothesis H0: µ1=µ2 2. Alternative Hypothesis H1: µ1 > µ2 or µ1 < µ2

(one tailed test) H1: µ1 ≠ µ 2 (two-tailed test)

3. Pooled Estimator: )1()1(

)1()1(

21

222

2112

nn

snsns p

4. Test Statistics: 21

ˆ21

xx

xxt

21

21

11

nns

xxt

p

)2( 21 nndf 5. Rejection Region t > tα or t < -tα(one tailed test)

t > tα/2 or t < -tα/2(two tailed test)

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Example:

• A company wishes to test when the sensitivity achieved by a new program is significantly higher than achieved under the legacy program. The following information is available from test results.

Sensitivity

Mean Standard Deviation

Sample Size

Proposed 92 15 12

P.M.O. 84 19 15

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SolutionStep 1: This is a one-tailed test. The hypothesis is stated below. The significance level is 0.05 (given) H0: μ1 ≤ μ2 versus H1: μ1 > μ2

Step 2: Since this is a test of the means and neither n1 or n2 is in excess of 30, a t test is appropriate. The critical value is 1.708 (from a t table with 25 degrees of freedom).

Step 3:

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Step 4: Sketch the distribution, locate the critical values and the test statistic. Step 5: Decide! Since the test statistic values lies within the retention region then there is no sufficient statistical evidence based on this sample to reject H0.

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Example:

Five vice presidents and four market research analysts of a large industry were asked to estimate what they consider to be the optimal market share for the company. Their responses are given in the following table. Do the data suggest that corporate vice presidents and market research nalysts tend to disagree when estimating their firm’s optimal market share? Test at a 5% level of significance.

Vice Presidents 22.5 25.0 30.0 27.5 20.0 Market Analysts 21.0 17.5 17.0 20.0

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Problem: 9-8

• A credit-insurance organization has developed a new high-tech method of training new sales personnel. The company sampled 16 employees who were trained the original way and found average daily sales to be $688 and the sample standard deviation was $32.63. They also sampled 11 employees who were trained using the new method and found aver age daily sales to be $706 and the sample standard deviation was $24.84. At a = 0.05, can the company conclude that average daily sales have in creased under the new plan?

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Problem: 9-9

• A large stock-brokerage firm wants to determine how successful its new ac count executives have been at recruiting clients. After completing their training, new account execs spend several weeks calling prospective clients, trying to get the prospects to open accounts with the firm. The following data give the numbers of new accounts opened in their first 2 weeks by 10 randomly chosen female account execs and by 8 randomly chosen male ac count execs. At a = 0.05, does it appear that the women are more effective at generating new accounts than the men are?

Number of New Accounts Female account execs 12 11 14 13 13 14 13 12 14 12 Male account execs 13 10 11 12 13 12 10 12

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Paired Difference Test: Dependent Paired Difference Test: Dependent SamplesSamples

Paired Difference Test: Dependent Samples

1. Null Hypothesis H0:µ1 - µ2 =µd=D0=0 2. Alternative Hypothesis H1: µd > 0 or µd < 0

(one tailed test) H1: µd ≠ 0 (two-tailed test)

3. Test Statistics ns

xt 0

4. Rejection Region t > tα or t < -tα (one tailed test) t > tα/2 or t < -tα/2 (two tailed test)

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Problem: 9-15 • Jeff Richardson, the receiving clerk for a chemical-products

distributor, is faced with the continuing problem of broken glassware, including test tubes, petri dishes, and flasks. Jeff has determined some additional ship ping precautions that can be undertaken to prevent breakage, and he has asked the Purchasing Director to inform the suppliers of the new measures. Data for 8 suppliers are given below in terms of average number of broken items per shipment. Do the data indicate, at a = 0.05, that the new measures have lowered the average number of broken items?

Supplier 1 2 3 4 5 6 7 8

Before 16 12 18 7 14 19 6 17 After 14 13 12 6 9 15 8 15

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Problem: 9-17 • Aquarius Health Club has been advertising a rigorous

program for body conditioning. The club claims that after 1 month in the program, the aver age participant should be able to do eight more push-ups in 2 minutes than he or she could do at the start. Does the random sample of 10 program par ticipants given below support the club's claim? Use the 0.025 level of sig nificance.

Participant 1 2 3 4 5 6 7 8 9 10

Before 38 11 34 25 17 38 12 27 32 29 After 45 24 41 39 30 44 30 39 40 41

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Another way to look athypothesis testing (more common)

• In typical hypothesis testing, the analyst does not go through the procedures manually. Instead, they use a computer and a statistical analysis software package (such as SAS, SPSS, Minitab, etc.)

• Instead of drawing the distribution and determining the location of the test statistic value (within the rejection or retention region), the computer calculates the area under the normal curve in excess of the test statistic value. This area is called the probability value or p-value

• So, you do not have to determine the location of the test statistic to make your decision on H0. You can simply look at the p-value and decide.

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Benefits

• Saves time since all computer statistics packages compute the p-value, and

• Not all people share your personal level of risk assessment. That is, you might use an alpha value of 0.05 but someone else might use a higher/lower value. You might reject H0 using α = 0.05 but someone else might retain H0 at a different level of risk.

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Decision rule using p-values