Essentials Of Statistics For Business & Economics
Essentials Of Statistics For Business & Economics
9th Edition
ISBN: 9780357045435
Author: David R. Anderson, Dennis J. Sweeney, Thomas A. Williams, Jeffrey D. Camm, James J. Cochran
Publisher: South-Western College Pub
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Chapter 11.2, Problem 15E

a.

To determine

Conduct the required hypothesis test and state the conclusion using p-value approach.

a.

Expert Solution
Check Mark

Answer to Problem 15E

There is enough evidence to infer that the two variances are equal. That is, σ12=σ22.

Explanation of Solution

Calculation:

The given information is that the sample variance for first sample is s12=8.2, the sample variance for second sample is s22=4, the sample size for first sample is n1=21, and the sample size for second sample is n2=26. The significance level is α=0.05.

The hypotheses are given below:

Null hypothesis:

H0:σ12=σ22

Alternative hypothesis:

Ha:σ12σ22

Test statistic:

The formula for test statistic is given below:

F=s12s22 with n11 degrees of freedom for the numerator and n21 degrees of freedom for the denominator.

Where s12 is the sample variance for population 1, and s22 is the sample variance for population 2.

Substitute s12=8.2 and s22=4 in test statistic formula.

F=s12s22=8.24=2.05

Thus, the test statistic is 2.05.

Degrees of freedom for numerator:

n11=211=20

Degrees of freedom for denominator:

n21=261=25

p-value:

Software procedure:

Step-by-step procedure to obtain the probability value using Excel:

  • Open an EXCEL sheet and select the cell A1.
  • Enter the formula =F.DIST.RT(2.05,20,25) in the cell A1.
  • Press Enter.

Output obtained using EXCEL software is given below:

Essentials Of Statistics For Business & Economics, Chapter 11.2, Problem 15E

From the output, the p-value for the upper tail is 0.0452.

For a two-tailed test, the p-value is two times the upper-tail area.

From the output:

p-value=2(0.0452)=0.0904

Thus, the p-value is 0.0904.

p-value approach:

Rejection rule:

If p-valueα,, reject the null hypothesis H0.

If p-valueα, do not reject the null hypothesis H0.

Conclusion:

Here, the p-value is greater than the level of significance.

That is, p-value(=0.0904)>α(=0.05).

Therefore, the null hypothesis is not rejected.

There is enough evidence to infer that the two variances are equal. That is, σ12=σ22.

b.

To determine

Conduct the required hypothesis test and state the conclusions using critical value approach.

b.

Expert Solution
Check Mark

Answer to Problem 15E

There is enough evidence to infer that the two variances are equal. That is, σ12=σ22.

Explanation of Solution

Calculation:

In Part (a), the test statistic is 2.05.

Critical value:

Step-by-step procedure to obtain the value of F0.025 using Table 4 of Appendix B:

  • Locate the value 20 and α=0.025 in the left column of Table 4.
  • Go through the row corresponding to the value 20 and the column corresponding to the value 25 of the table.
  • The intersecting value that corresponds to the degrees of freedom 20 and 25 with level of significance 0.025 is 2.30.

The value of F0.025 with degrees of freedom 20 and 25 is 2.30.

Rejection rule:

If FFα2, reject the null hypothesis H0.

If F<Fα2, do not reject the null hypothesis H0.

Conclusion:

Here, the value of test statistic is lesser than the critical value.

That is, teststatistic(=2.05)<Fα2(=2.30).

Therefore, the null hypothesis is not rejected.

There is enough evidence to infer that the two variances are equal. That is, σ12=σ22.

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Essentials Of Statistics For Business & Economics

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