Introduction to the Practice of Statistics
Introduction to the Practice of Statistics
9th Edition
ISBN: 9781319013387
Author: David S. Moore, George P. McCabe, Bruce A. Craig
Publisher: W. H. Freeman
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Chapter 7, Problem 138E

(a)

Section1:

To determine

To test: The null hypothesis that there is no significant difference in mean between Athletes and Sedentary students for body fat.

(a)

Section1:

Expert Solution
Check Mark

Answer to Problem 138E

Solution: The null hypothesis gets rejected. So, there is a significant difference in mean between Athletes and Sedentary students for body fat.

Explanation of Solution

Calculation: The null and alternative hypotheses are:

H0:μ1=μ2Ha:μ1μ2

The t-statistic is used to test the significance of the difference in mean. The pooled standard deviation is used for two standard deviation as σmax<2×σmin, which is obtained as follow:

pooled standard deviation=(n11)×s12+(n21)×s22n1+n22=79×5.542+79×8.05280+802=6.90992

Degree of freedom for the two samples can be calculated by following formula:

D.F = n1+n22D.F = 80+802D.F = 158

Now the t-value can be obtained as follow:

t=(x¯1x¯2)sp1n1+1n2=(25.6132.51)6.90992180+180=6.91.092554=6.3155

The P-value can be calculated by using the function =T.DIST.RT(x, deg_freedom, tails) in Excel. The screenshot is given below:

Introduction to the Practice of Statistics, Chapter 7, Problem 138E , additional homework tip  1

Conclusion: The P-value is less than 0.05. Hence the null hypothesis gets rejected, which states that there is a significant difference in mean between Athletes and Sedentary students for body fat.

Section 2:

To determine

To test: The null hypothesis that there is no difference in mean between Athletes and Sedentary students for body mass index.

Section 2:

Expert Solution
Check Mark

Answer to Problem 138E

Solution: The null hypothesis gets rejected. So, there is a significant difference in mean between Athletes and Sedentary students for body mass index.

Explanation of Solution

Calculation: The null and alternative hypotheses are:

H0:μ1=μ2Ha:μ1μ2

The t-statistic is used to test the significance of the difference in mean. The pooled standard deviation is used for two standard deviation as σmax<2×σmin, which is obtained as follow:

pooled standard deviation=(n11)×s12+(n21)×s22n1+n22=79×2.462+79×2.73280+802=2.59851

Degree of freedom for the two samples can be calculated by following formula:

D.F = n1+n22D.F = 80+802D.F = 158

The t-value is obtained as follow:

t=(x¯1x¯2)sp1n1+1n2=(21.626.41)2.59851180+180=11.7071

The P-value can be calculated by using the function =T.DIST.RT(x, deg_freedom, tails) in Excel. The screenshot is given below:

Introduction to the Practice of Statistics, Chapter 7, Problem 138E , additional homework tip  2

Conclusion: The P-value is less than 0.05. Hence the null hypothesis gets rejected, which states that there is a significant difference in mean between Athletes and Sedentary students for body mass index.

Section 3:

To determine

To test: The null hypothesis that there is no significant difference in mean between Athletes and Sedentary students for calcium deficit.

Section 3:

Expert Solution
Check Mark

Answer to Problem 138E

Solution: The null hypothesis gets rejected. So, there is a difference in mean between Athletes and Sedentary students for calcium deficit.

Explanation of Solution

Calculation: The null and alternative hypotheses are:

H0:μ1=μ2Ha:μ1μ2

The t-statistic is used to test the significance of the difference in mean. The pooled standard deviation is used for two standard deviation as σmax<2×σmin, which is obtained as follow:

pooled standard deviation=(n11)×s12+(n21)×s22n1+n22=79×516.632+79×372.77280+802=450.480

Degree of freedom for the two samples can be calculated by following formula:

D.F = n1+n22D.F = 80+802D.F = 158

The t-value is obtained as follow:

t=(x¯1x¯2)sp1n1+1n2=(297.13580.54)450.480180+180=3.979

The P-value can be calculated by using the function =T.DIST.RT(x, deg_freedom, tails) in Excel. The screenshot is given below:

Introduction to the Practice of Statistics, Chapter 7, Problem 138E , additional homework tip  3

Conclusion: The P-value is less than 0.05. Hence the null hypothesis gets rejected, which states that there is a significant difference in mean between Athletes and Sedentary students for calcium deficit.

Section 4:

To determine

To test: The null hypothesis that there is no significant difference in mean between Athletes and Sedentary students for glasses of milk/day.

Section 4:

Expert Solution
Check Mark

Answer to Problem 138E

Solution: The null hypothesis does not get rejected. So, there is no difference in mean between Athletes and Sedentary students for glasses of milk/day.

Explanation of Solution

Calculation: The null and alternative hypotheses are:

H0:μ1=μ2Ha:μ1μ2

The t-statistic is used to test the significance of the difference in mean. The pooled standard deviation is used for two standard deviation as σmax<2×σmin, which is obtained as follow:

pooled standard deviation=(n11)×s12+(n21)×s22n1+n22=79×1.462+79×1.24280+802=1.35447

Degree of freedom for the two samples can be calculated by following formula:

D.F = n1+n22D.F = 80+802D.F = 158

The t-value is obtained as follow:

t=(x¯1x¯2)sp1n1+1n2=(2.211.82)1.35447180+180=1.821

The P-value can be calculated by using the function =T.DIST.RT(x, deg_freedom, tails) in Excel. The screenshot is given below:

Introduction to the Practice of Statistics, Chapter 7, Problem 138E , additional homework tip  4

Conclusion: The P-value is greater than 0.05. So, the null hypothesis does not get rejected, which states that there is no significant difference in mean between Athletes and Sedentary students for glasses of milk/day.

(b)

To determine

The summary of the hypothesis test results.

(b)

Expert Solution
Check Mark

Answer to Problem 138E

Solution: The null hypotheses get rejected for all the characteristic except the Glasses of milk/day.

Explanation of Solution

With reference to part (a), the P-values are:

P-valueBody fat=0<0.05P-valueCalcium deficit=0<0.05P-valueBody mass index=0<0.05P-valueGlasses of milk/day=0.0705>0.05

The P-value are less than 0.05 for all the characteristic except for Glasses of milk/day. The null hypotheses are rejected for all except for Glasses of milk/day.

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Chapter 7 Solutions

Introduction to the Practice of Statistics

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