Welcome to Chapter 11 MBA 541

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Welcome to Chapter 11 MBA 541

BENEDICTINE UNIVERSITY

• Two-Sample Tests & Analysis of Variance

• Two-Sample Tests of Hypothesis

• Chapter 11

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Chapter 11

Please, Read Chapter 11 in

Lind before viewing this presentation.

StatisticalTechniques in

Business &Economics

Lind

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GoalsWhen you have completed this chapter, you will be

able to:

• ONE– Conduct a test of a hypothesis about the difference

between two independent populations means.• TWO

– Conduct a test of a hypothesis about the difference between two population proportions.

• THREE– Conduct a test of a hypothesis about the mean

difference between paired or dependent observations.• FOUR

– Understand the difference between dependent and independent samples.

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Comparing Two Populations• Does the distribution of the differences in sample

means have a mean of 0?

• If both samples contain at least 30 observations, use the z distribution as the test statistic.

• The samples are from independent populations.

• No assumptions about the shape of the populations are required.

• The formula for computing the value of z is:

1 2

2 21 2

1 2

X Xz

s sn n

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Example 1• Two cities, Bradford and Kane, are separated only

by the Conewango River. There is competition between the two cities.

• The local paper recently reported that the mean household income in Bradford is $38,000 with a standard deviation of $6,000 for a sample of 40 households.

• The same article reported the mean income in Kane is $35,000 with a standard deviation of $7,000 for a sample of 35 households.

• At the 0.01 significance level can we conclude that the mean income of Bradford is more?

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Example 1 (Continued)

• Step 1: State the null and alternate hypotheses.H0: μB≤ μK

H1: μB> μK

• Step 2: State the level of significance.The 0.01 significance is stated in the problem.

• Step 3: Find the appropriate test statistic.Because both samples are more than 30, use z as the test statistic.

• Step 4: State the decision rule.The null hypothesis is rejected if z is greater then 2.33 or if p is less than 0.01.

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Example 1 (Continued)

• Step 5: Compute the value of z and make a decision.

The p(z>1.98) is 0.0239 for a one-tailed test of significance.

Because the computed z of 1.98 is less than the critical z value of 2.33, and the p-value of 0.0239 is greater than the required level of significance of 0.01, the decision is to not reject the null hypothesis.

We cannot conclude that the mean household income in Bradford is larger.

2 2

$38,000 $35,000z 1.98

$6,000 $7,000

40 35

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Two-SampleTests of Proportions

• Two-Sample Tests of Proportions investigate whether the two samples came from populations with equal proportions of successes.

• The two samples are pooled using the following formula,

where X1 and X2 refer to the number of successes in the respective samples of n1 and n2.

• The value of the test statistic is computed from the followingformula,

1 2

c c c c

1 2

p pz .

p 1 p p 1 p

n n

1 2c

1 2

X Xp

n n

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Example 2

• Are unmarried workers more likely to be absent from work than married workers?

• A sample of 250 married workers showed that 22 missed more than 5 days last year.

• A sample of 300 unmarried workers showed that 35 missed more than 5 days last year.

• Use a 0.05 significance level.

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Example 2 (Continued)

• Step 1: State the null and alternate hypotheses.H0: πU≤ πM

H1: πU> πM

• Step 2: State the level of significance.The 0.05 significance is stated in the problem.

• Step 3: Find the appropriate test statistic.Use the two-sample test of proportions with z as the test statistic.

• Step 4: State the decision rule.The null hypothesis is rejected if z is greater then 1.65 or if p is less than 0.05.

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Example 2 (Continued)• Step 5: Compute the value

of z and make a decision.

The p(z>1.10) is 0.136 for a one-tailed test of significance.

Because the computed z of 1.10 is less than the critical z value of 1.96, and the p-value of 0.136 is greater than the required level of significance of 0.05, the decision is to not reject the null hypothesis.

We cannot conclude that a higher proportion of unmarried workers miss more days in a year than the married workers.

c

35 22p 0.1036

300 250

35 22300 250z 1.10

0.1036 1 .1036 0.1036 1 .1036

300 250

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Small Sample Tests of Means

• The t distribution is used as the test statistic if one or more of the samples have fewer than 30 observations.

• The required assumptions are:

1. Both populations must follow the normal distribution,

2. The populations must have equal standard deviations, and

3. The samples are from independent populations.

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Small Sample Tests of Means(Continued)

• Finding the value of the test statistic requires two steps.

• Step 1: Pool the sample variances.

• Step 2: Determine the value of t from the following formula.

2 21 1 2 22

p1 2

n 1 s n 1 ss

n n 2

1 2

2p

1 2

X Xt

1 1s

n n

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Example 3

• A recent EPA study compared the highway fuel economy of domestic and imported passenger cars.

• A sample of 15 domestic cars revealed a mean of 33.7 mpg with a standard deviation of 2.4 mpg.

• A sample of 12 imported cars revealed a mean of 35.7 mpg with a standard deviation of 3.9 mpg.

• At the 0.05 significance level, can the EPA conclude that the mpg is higher with the imported cars?

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Example 3 (Continued)• Step 1: State the null and alternate hypotheses.

H0: μI≤ μD

H1: μI> μD

• Step 2: State the level of significance.The 0.05 significance is stated in the problem.

• Step 3: Find the appropriate test statistic.Because both samples are less than 30, use t as the test statistic.

• Step 4: State the decision rule.The null hypothesis is rejected if t is greater then 1.708 or if p is less than 0.05. There are n1+ n2-2 or 25 degrees of freedom.

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Example 3 (Continued)

• Step 5: Compute the value of t and make a decision.Compute the pooled variance.

2 2

1 1 2 22p

1 2

n 1 s n 1 ss

n n 2

2 2

2p

12 1 3.9 15 1 2.4s 9.918

12 15 2

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Example 3 (Continued)

• Step 5 (Continued): Compute the value of t and make a decision.Compute the value of t.

1 2

2p

1 2

X Xt

1 1s

n n

35.7 33.7t 1.640

1 19.918

12 15

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Example 3 (Continued)

• Step 5 (Continued): Compute the value of t and make a decision.

Make a decision.

The p(t>1.640) is 0.0567 for a one-tailed t-test of significance.

Because the computed t of 1.640 is less than the critical t value of 1.708, and the p-value of 0.0567 is greater than the required level of significance of 0.05, the decision is to not reject the null hypothesis.

There is insufficient sample evidence to claim a higher mpg for the imported cars.

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Hypothesis Testing InvolvingPaired Observations

• Independent samples are samples that are not related in any way.

• Dependent samples are samples that are paired or related in some fashion. For example,

– If you wished to buy a car you would look at the same car at two (or more) different dealerships and compare the prices.

– If you wished to measure the effectiveness of a new diet you would weigh the dieters at the start and at the finish of the program.

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Hypothesis Testing InvolvingPaired Observations

• Use the following test, paired t test, when the samples are dependent.

• Where, is the mean of the differences,

sd is the standard deviation of the differences, and

n is the number of pairs of differences.

d

dt

s / n

d

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Example 4

• An independent testing agency is comparing the daily rental cost for renting a compact car from Hertz and Avis.

• A random sample of eight cities revealed the following information.

• At the 0.05 significance level, can the testing agency conclude that there is a difference in the rental charge?

City Hertz($)

Avis($)

Atlanta 42 40

Chicago 56 54

Cleveland 45 43

Denver 48 48

Honolulu 37 32

Kansas City 45 48

Miami 41 39

Seattle 46 50

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Example 4 (Continued)• Step 1: State the null and alternate hypotheses.

H0: μd=0

H1: μd≠0

• Step 2: State the level of significance.The 0.05 significance is stated in the problem.

• Step 3: Find the appropriate test statistic.The appropriate test statistic is the paired t-test.

• Step 4: State the decision rule.The null hypothesis is rejected if -2.36 > t > 2.365 or if p is less than 0.05. There are n-1 or 7 degrees of freedom.

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Example 4 (Continued)

• Step 5: Compute the value of t and make a decision.

City Hertz($)

Avis($)

d d²

Atlanta 42 40 2 4

Chicago 56 54 4 16

Cleveland 45 43 2 4

Denver 48 48 0 0

Honolulu 37 32 5 25

Kansas City 45 48 -3 9

Miami 41 39 2 4

Seattle 46 50 -4 16

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Example 4 (Continued)

• Step 5: Compute the value of t and make a decision.

d 8.0d 1.0

n 8

22

2

d

d 878d

8ns 3.1623n 1 8 1

c

d 1.0t 0.894

s / n 3.1623 / 8

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Example 4 (Continued)• Step 5 (Continued): Compute the value of t and

make a decision.Make a decision.

The p(t>0.894) is 0.20 for a one-tailed t-test at 7 degrees of freedom.

Because the computed t of 0.894 is less than the critical t value of 2.365, and the p-value of 0.20 is greater than the required level of significance of 0.05, do not reject the null hypothesis.

There is no significant difference in the mean amount charged by Hertz and Avis.

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Comparing Dependent and Independent Samples

• Two types of dependent samples:– The same subjects measured at two different

points in time.– Matched or paired observations.

• Disadvantage of dependent samples:– Degrees of freedom are halved.

• Advantage of dependent samples:– Reduction in variation in the sampling

distribution.