Beginning Statistics, 2nd Edition
Beginning Statistics, 2nd Edition
2nd Edition
ISBN: 9781932628678
Author: Carolyn Warren; Kimberly Denley; Emily Atchley
Publisher: Hawkes Learning Systems
Question
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Chapter 11.5, Problem 25E
To determine

a.

To State:

Null hypotheses and alternative hypotheses.

Expert Solution
Check Mark

Answer to Problem 25E

Solution:

Null Hypotheses –

H0: σ12 = σ22

Alternative Hypotheses –

Ha : σ12  σ22

Explanation of Solution

Given:

One study claims that the variance in the resting heart rates of smokers is different than the variance in the resting heart rates of nonsmokers.

A medical student decides to test this claim.

The sample variance of resting heart rates, measured in beats per minute, for a random sample of 5 smokers is 545.1.

The sample variance for a random sample of 5 nonsmokers is 103.7.

Test the study’s claim using a 0.01 level of significance.

Calculation:

Let population variances in the resting heart rates of smokers be represented by σ12 and population variances in the resting heart rates of non-smokers be represented by σ22. One study claims that the variance in the resting heart rates of smokers is different than the variance in the resting heart rates of nonsmokers that is σ12  σ22.

The mathematical opposite of this claim is σ12 = σ22.

The null hypothesis and alternative hypothesis are stated as follows-

H0 : σ12 = σ22

Ha : σ12  σ22

To determine

b.

The type of distribution to use for the test statistics and state the level of significance.

Expert Solution
Check Mark

Answer to Problem 25E

Solution:

The F- test statistic is appropriate and the level of significance for this test is α = 0.01.

Explanation of Solution

Given:

One study claims that the variance in the resting heart rates of smokers is different than the variance in the resting heart rates of nonsmokers.

A medical student decides to test this claim.

The sample variance of resting heart rates, measured in beats per minute, for a random sample of 5 smokers is 545.1.

The sample variance for a random sample of 5 nonsmokers is 103.7.

Test the study’s claim using a 0.01 level of significance.

Calculation:

For comparing the variances of two normally distributed populations using independent, simple random samples, F- test statistic is used.

The level of significance for this test is α = 0.01.

Therefore, the F-test statistic is appropriate and the level of significance for this test is α = 0.01

To determine

c.

To Calculate:

The test statistic.

Expert Solution
Check Mark

Answer to Problem 25E

Solution:

The test statistic 5.2565.

Explanation of Solution

Given:

One study claims that the variance in the resting heart rates of smokers is different than the variance in the resting heart rates of nonsmokers.

A medical student decides to test this claim.

The sample variance of resting heart rates, measured in beats per minute, for a random sample of 5 smokers is 545.1.

The sample variance for a random sample of 5 nonsmokers is 103.7.

Test the study’s claim using a 0.01 level of significance.

Formula used:

When the samples are given to be independent, the given population distribution are approximately normal, then the test statistics for the hypothesis test for two population variances is given by,

F=s12s22

Where s12 and s22 are the sample variances.

n1 is the total number of data of population 1.

n2 is the total number of data of population 2.

The degree of freedom for the numerator is df1=n11

The degree of freedom for the denominator is df2=n21

Given n1= 5, s12= 545.1, n2= 5, s22 = 103.7, α = 0.01

Null Hypotheses –

H0: σ12 = σ22

Alternative Hypotheses –

Ha : σ12  σ22

The test statistic value is given by,

F = s12s22

F = 545.1103.7= 5.2565

Therefore, the test statistic is 5.2565.

To determine

d.

To Draw:

The conclusion and interpret the decision.

Expert Solution
Check Mark

Answer to Problem 25E

Solution:

The null hypothesis is accepted and it is concluded that there is no sufficient evidence at the 0.01 level of significance to support the claim that variances in the resting heart rates of smokers is different than the variance in the resting heart rates of nonsmokers.

Explanation of Solution

Given:

One study claims that the variance in the resting heart rates of smokers is different than the variance in the resting heart rates of nonsmokers.

A medical student decides to test this claim.

The sample variance of resting heart rates, measured in beats per minute, for a random sample of 5 smokers is 545.1.

The sample variance for a random sample of 5 nonsmokers is 103.7.

Test the study’s claim using a 0.01 level of significance.

Approach:

The null hypothesis is rejected if,

FF(1α) for the left tailed test.

FFα for the right tailed test.

FF(1α2) or FFα/2 for the two tail test.

Calculation:

The level of significance, α=0.01.

The degree of freedom for numerator is,

df1=n11

Substitute 4 for n1 in the above equation.

df1= n1-1= 5-1= 4

The degree of freedom for denominator is,

df2=n21

Substitute 4 for n2 in the above equation.

df2= n2-1= 5-1= 4

From the F-table, the critical values for the required degrees is,

Fα/2= F0.005=23.1545

Another critical value for the same degrees of freedom is,

F1α2= F0.995=0.0432

By comparing the test statistic value and the critical value, the F value is less than the critical value, so by the two tailed test the null hypothesis is rejected.

That is,

RejectH0ifF0.0432orF23.1545

The critical value is, p-value0.1369

Conclusion:

Thus, there is no sufficient evidence at 0.01 level of significance to support the claim that the variances in the resting heart rates of smokers is different than the variance in the resting heart rates of nonsmokers

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Chapter 11 Solutions

Beginning Statistics, 2nd Edition

Ch. 11.1 - Prob. 11ECh. 11.1 - Prob. 12ECh. 11.1 - Prob. 13ECh. 11.1 - Prob. 14ECh. 11.1 - Prob. 15ECh. 11.1 - Prob. 16ECh. 11.1 - Prob. 17ECh. 11.1 - Prob. 18ECh. 11.1 - Prob. 19ECh. 11.1 - Prob. 20ECh. 11.1 - Prob. 21ECh. 11.2 - Prob. 1ECh. 11.2 - Prob. 2ECh. 11.2 - Prob. 3ECh. 11.2 - Prob. 4ECh. 11.2 - Prob. 5ECh. 11.2 - Prob. 6ECh. 11.2 - Prob. 7ECh. 11.2 - Prob. 8ECh. 11.2 - Prob. 9ECh. 11.2 - Prob. 10ECh. 11.2 - Prob. 11ECh. 11.2 - Prob. 12ECh. 11.2 - Prob. 13ECh. 11.2 - Prob. 14ECh. 11.2 - Prob. 15ECh. 11.2 - Prob. 16ECh. 11.2 - Prob. 17ECh. 11.3 - Prob. 1ECh. 11.3 - Prob. 2ECh. 11.3 - Prob. 3ECh. 11.3 - Prob. 4ECh. 11.3 - Prob. 5ECh. 11.3 - Prob. 6ECh. 11.3 - Prob. 7ECh. 11.3 - Prob. 8ECh. 11.3 - Prob. 9ECh. 11.3 - Prob. 10ECh. 11.4 - Prob. 1ECh. 11.4 - Prob. 2ECh. 11.4 - Prob. 3ECh. 11.4 - Prob. 4ECh. 11.4 - Prob. 5ECh. 11.4 - Prob. 6ECh. 11.4 - Prob. 7ECh. 11.4 - Prob. 8ECh. 11.4 - Prob. 9ECh. 11.4 - Prob. 10ECh. 11.4 - Prob. 11ECh. 11.5 - Prob. 1ECh. 11.5 - Prob. 2ECh. 11.5 - Prob. 3ECh. 11.5 - Prob. 4ECh. 11.5 - Prob. 5ECh. 11.5 - Prob. 6ECh. 11.5 - Prob. 7ECh. 11.5 - Prob. 8ECh. 11.5 - Prob. 9ECh. 11.5 - Prob. 10ECh. 11.5 - Prob. 11ECh. 11.5 - Prob. 12ECh. 11.5 - Prob. 13ECh. 11.5 - Prob. 14ECh. 11.5 - Prob. 15ECh. 11.5 - Prob. 16ECh. 11.5 - Prob. 17ECh. 11.5 - Prob. 18ECh. 11.5 - Prob. 19ECh. 11.5 - Prob. 20ECh. 11.5 - Prob. 21ECh. 11.5 - Prob. 22ECh. 11.5 - Prob. 23ECh. 11.5 - Prob. 24ECh. 11.5 - Prob. 25ECh. 11.5 - Prob. 26ECh. 11.6 - Prob. 1ECh. 11.6 - Prob. 2ECh. 11.6 - Prob. 3ECh. 11.6 - Prob. 4ECh. 11.6 - Prob. 5ECh. 11.6 - Prob. 6ECh. 11.6 - Prob. 7ECh. 11.6 - Prob. 8ECh. 11.6 - Prob. 9ECh. 11.6 - Prob. 10ECh. 11.6 - Prob. 11ECh. 11.6 - Prob. 12ECh. 11.6 - Prob. 13ECh. 11.6 - Prob. 14ECh. 11.6 - Prob. 15ECh. 11.6 - Prob. 16ECh. 11.CR - Prob. 1CRCh. 11.CR - Prob. 2CRCh. 11.CR - Prob. 3CRCh. 11.CR - Prob. 4CRCh. 11.CR - Prob. 5CRCh. 11.CR - Prob. 6CRCh. 11.CR - Prob. 7CRCh. 11.CR - Prob. 8CRCh. 11.CR - Prob. 9CRCh. 11.CR - Prob. 10CR
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