PRACTICE OF STATISTICS F/AP EXAM
PRACTICE OF STATISTICS F/AP EXAM
6th Edition
ISBN: 9781319113339
Author: Starnes
Publisher: MAC HIGHER
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Chapter 11.2, Problem 53E

a)

To determine

To state the null and alternative hypotheses.

a)

Expert Solution
Check Mark

Answer to Problem 53E

  H0 : There is no difference in the improvement rate of the two treatment groups.

  Ha : There is difference in the improvement rate of the two treatment groups.

Explanation of Solution

Given:

Claim: We can test the hypothesis of “no difference” in the effectiveness of the treatment in two-ways.

PRACTICE OF STATISTICS F/AP EXAM, Chapter 11.2, Problem 53E , additional homework tip  1

Calculation:

The null and alternative hypotheses:

  H0 : There is no difference in the improvement rate of the two treatment groups.

  Ha : There isdifference in the improvement rate of the two treatment groups.

b)

To determine

To state the conclusion.

b)

Expert Solution
Check Mark

Answer to Problem 53E

There is no convincing evidence that there is a difference in the improvement rate of the two treatment groups.

Explanation of Solution

Given:

PRACTICE OF STATISTICS F/AP EXAM, Chapter 11.2, Problem 53E , additional homework tip  2

Calculation:

The p-value = 0.570. Therefore, we can say, there is 57% chance of obtaining the sample results or more extreme when there is no difference in the improvement rate of the two treatment groups.

Decision: P-value > 0.05, fail to reject H0.

Conclusion: There is no convincing evidence that there is a difference in the improvement rate of the two treatment groups.

c)

To determine

To compare chi-square test and z-test.

c)

Expert Solution
Check Mark

Answer to Problem 53E

The p-value of chi-square test and z-test is same so, conclusion based on two tests are same.

Explanation of Solution

Given:

PRACTICE OF STATISTICS F/AP EXAM, Chapter 11.2, Problem 53E , additional homework tip  3

Calculation:

The p-value = 0.570. The p-value of chi-square test and z-test is same. Therefore, conclusion based on two tests are same.

Decision: P-value > 0.05, fail to reject H0.

Also, chi-square test statistic = 0.322 and z-test statistic = -0.57

We know the property that square of z becomes the chi-square.

So, z2=(-0.57)2 = 0.322=χ2

This property is satisfied.

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PRACTICE OF STATISTICS F/AP EXAM

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