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

Question

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Answer 1

Question

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

Chapter 7, Problem 134E

(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

Answer to Problem 134E

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:μ1≠μ2

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=(n1−1)×s12+(n2−1)×s22n1+n2−2=79×5.542+79×8.05280+80−2=6.90992

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

D.F = n1+n2−2D.F = 80+80−2D.F = 158

Now the t-value can be obtained as follow:

t=(x¯1−x¯2)sp1n1+1n2=(25.61−32.51)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: w/CrunchIt/EESEE Access Card, Chapter 7, Problem 134E , 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

Answer to Problem 134E

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:μ1≠μ2

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=(n1−1)×s12+(n2−1)×s22n1+n2−2=79×2.462+79×2.73280+80−2=2.59851

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

D.F = n1+n2−2D.F = 80+80−2D.F = 158

The t-value is obtained as follow:

t=(x¯1−x¯2)sp1n1+1n2=(21.6−26.41)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: w/CrunchIt/EESEE Access Card, Chapter 7, Problem 134E , 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

Answer to Problem 134E

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:μ1≠μ2

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=(n1−1)×s12+(n2−1)×s22n1+n2−2=79×516.632+79×372.77280+80−2=450.480

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

D.F = n1+n2−2D.F = 80+80−2D.F = 158

The t-value is obtained as follow:

t=(x¯1−x¯2)sp1n1+1n2=(297.13−580.54)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: w/CrunchIt/EESEE Access Card, Chapter 7, Problem 134E , 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

Answer to Problem 134E

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:μ1≠μ2

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=(n1−1)×s12+(n2−1)×s22n1+n2−2=79×1.462+79×1.24280+80−2=1.35447

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

D.F = n1+n2−2D.F = 80+80−2D.F = 158

The t-value is obtained as follow:

t=(x¯1−x¯2)sp1n1+1n2=(2.21−1.82)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: w/CrunchIt/EESEE Access Card, Chapter 7, Problem 134E , 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

Answer to Problem 134E

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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Answer 2

Question

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

Chapter 7, Problem 134E

(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

Answer to Problem 134E

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:μ1≠μ2

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=(n1−1)×s12+(n2−1)×s22n1+n2−2=79×5.542+79×8.05280+80−2=6.90992

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

D.F = n1+n2−2D.F = 80+80−2D.F = 158

Now the t-value can be obtained as follow:

t=(x¯1−x¯2)sp1n1+1n2=(25.61−32.51)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: w/CrunchIt/EESEE Access Card, Chapter 7, Problem 134E , 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

Answer to Problem 134E

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:μ1≠μ2

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=(n1−1)×s12+(n2−1)×s22n1+n2−2=79×2.462+79×2.73280+80−2=2.59851

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

D.F = n1+n2−2D.F = 80+80−2D.F = 158

The t-value is obtained as follow:

t=(x¯1−x¯2)sp1n1+1n2=(21.6−26.41)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: w/CrunchIt/EESEE Access Card, Chapter 7, Problem 134E , 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

Answer to Problem 134E

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:μ1≠μ2

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=(n1−1)×s12+(n2−1)×s22n1+n2−2=79×516.632+79×372.77280+80−2=450.480

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

D.F = n1+n2−2D.F = 80+80−2D.F = 158

The t-value is obtained as follow:

t=(x¯1−x¯2)sp1n1+1n2=(297.13−580.54)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: w/CrunchIt/EESEE Access Card, Chapter 7, Problem 134E , 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

Answer to Problem 134E

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:μ1≠μ2

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=(n1−1)×s12+(n2−1)×s22n1+n2−2=79×1.462+79×1.24280+80−2=1.35447

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

D.F = n1+n2−2D.F = 80+80−2D.F = 158

The t-value is obtained as follow:

t=(x¯1−x¯2)sp1n1+1n2=(2.21−1.82)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: w/CrunchIt/EESEE Access Card, Chapter 7, Problem 134E , 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

Answer to Problem 134E

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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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

Chapter 7, Problem 134E

(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

Answer to Problem 134E

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:μ1≠μ2

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=(n1−1)×s12+(n2−1)×s22n1+n2−2=79×5.542+79×8.05280+80−2=6.90992

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

D.F = n1+n2−2D.F = 80+80−2D.F = 158

Now the t-value can be obtained as follow:

t=(x¯1−x¯2)sp1n1+1n2=(25.61−32.51)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: w/CrunchIt/EESEE Access Card, Chapter 7, Problem 134E , 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

Answer to Problem 134E

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:μ1≠μ2

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=(n1−1)×s12+(n2−1)×s22n1+n2−2=79×2.462+79×2.73280+80−2=2.59851

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

D.F = n1+n2−2D.F = 80+80−2D.F = 158

The t-value is obtained as follow:

t=(x¯1−x¯2)sp1n1+1n2=(21.6−26.41)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: w/CrunchIt/EESEE Access Card, Chapter 7, Problem 134E , 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

Answer to Problem 134E

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:μ1≠μ2

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=(n1−1)×s12+(n2−1)×s22n1+n2−2=79×516.632+79×372.77280+80−2=450.480

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

D.F = n1+n2−2D.F = 80+80−2D.F = 158

The t-value is obtained as follow:

t=(x¯1−x¯2)sp1n1+1n2=(297.13−580.54)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: w/CrunchIt/EESEE Access Card, Chapter 7, Problem 134E , 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

Answer to Problem 134E

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:μ1≠μ2

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=(n1−1)×s12+(n2−1)×s22n1+n2−2=79×1.462+79×1.24280+80−2=1.35447

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

D.F = n1+n2−2D.F = 80+80−2D.F = 158

The t-value is obtained as follow:

t=(x¯1−x¯2)sp1n1+1n2=(2.21−1.82)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: w/CrunchIt/EESEE Access Card, Chapter 7, Problem 134E , 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

Answer to Problem 134E

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

Chapter 7, Problem 134E

(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

Answer to Problem 134E

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:μ1≠μ2

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=(n1−1)×s12+(n2−1)×s22n1+n2−2=79×5.542+79×8.05280+80−2=6.90992

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

D.F = n1+n2−2D.F = 80+80−2D.F = 158

Now the t-value can be obtained as follow:

t=(x¯1−x¯2)sp1n1+1n2=(25.61−32.51)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: w/CrunchIt/EESEE Access Card, Chapter 7, Problem 134E , 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

Answer to Problem 134E

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:μ1≠μ2

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=(n1−1)×s12+(n2−1)×s22n1+n2−2=79×2.462+79×2.73280+80−2=2.59851

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

D.F = n1+n2−2D.F = 80+80−2D.F = 158

The t-value is obtained as follow:

t=(x¯1−x¯2)sp1n1+1n2=(21.6−26.41)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: w/CrunchIt/EESEE Access Card, Chapter 7, Problem 134E , 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

Answer to Problem 134E

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:μ1≠μ2

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=(n1−1)×s12+(n2−1)×s22n1+n2−2=79×516.632+79×372.77280+80−2=450.480

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

D.F = n1+n2−2D.F = 80+80−2D.F = 158

The t-value is obtained as follow:

t=(x¯1−x¯2)sp1n1+1n2=(297.13−580.54)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: w/CrunchIt/EESEE Access Card, Chapter 7, Problem 134E , 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

Answer to Problem 134E

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:μ1≠μ2

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=(n1−1)×s12+(n2−1)×s22n1+n2−2=79×1.462+79×1.24280+80−2=1.35447

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

D.F = n1+n2−2D.F = 80+80−2D.F = 158

The t-value is obtained as follow:

t=(x¯1−x¯2)sp1n1+1n2=(2.21−1.82)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: w/CrunchIt/EESEE Access Card, Chapter 7, Problem 134E , 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

Answer to Problem 134E

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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Answer 5

Chapter 7, Problem 134E

(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

Answer to Problem 134E

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:μ1≠μ2

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=(n1−1)×s12+(n2−1)×s22n1+n2−2=79×5.542+79×8.05280+80−2=6.90992

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

D.F = n1+n2−2D.F = 80+80−2D.F = 158

Now the t-value can be obtained as follow:

t=(x¯1−x¯2)sp1n1+1n2=(25.61−32.51)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: w/CrunchIt/EESEE Access Card, Chapter 7, Problem 134E , 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

Answer to Problem 134E

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:μ1≠μ2

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=(n1−1)×s12+(n2−1)×s22n1+n2−2=79×2.462+79×2.73280+80−2=2.59851

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

D.F = n1+n2−2D.F = 80+80−2D.F = 158

The t-value is obtained as follow:

t=(x¯1−x¯2)sp1n1+1n2=(21.6−26.41)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: w/CrunchIt/EESEE Access Card, Chapter 7, Problem 134E , 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

Answer to Problem 134E

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:μ1≠μ2

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=(n1−1)×s12+(n2−1)×s22n1+n2−2=79×516.632+79×372.77280+80−2=450.480

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

D.F = n1+n2−2D.F = 80+80−2D.F = 158

The t-value is obtained as follow:

t=(x¯1−x¯2)sp1n1+1n2=(297.13−580.54)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: w/CrunchIt/EESEE Access Card, Chapter 7, Problem 134E , 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

Answer to Problem 134E

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:μ1≠μ2

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=(n1−1)×s12+(n2−1)×s22n1+n2−2=79×1.462+79×1.24280+80−2=1.35447

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

D.F = n1+n2−2D.F = 80+80−2D.F = 158

The t-value is obtained as follow:

t=(x¯1−x¯2)sp1n1+1n2=(2.21−1.82)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: w/CrunchIt/EESEE Access Card, Chapter 7, Problem 134E , 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

Answer to Problem 134E

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.

Answer 6

Chapter 7, Problem 134E

(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

Answer to Problem 134E

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:μ1≠μ2

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=(n1−1)×s12+(n2−1)×s22n1+n2−2=79×5.542+79×8.05280+80−2=6.90992

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

D.F = n1+n2−2D.F = 80+80−2D.F = 158

Now the t-value can be obtained as follow:

t=(x¯1−x¯2)sp1n1+1n2=(25.61−32.51)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: w/CrunchIt/EESEE Access Card, Chapter 7, Problem 134E , 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.

Answer 7

Chapter 7, Problem 134E

(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.

Answer 8

(a)

Section1:

Expert Solution

Answer to Problem 134E

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

Answer 9

(a)

Section1:

Expert Solution

Answer 10

(a)

Section1:

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Calculation: The null and alternative hypotheses are:

H0:μ1=μ2Ha:μ1≠μ2

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=(n1−1)×s12+(n2−1)×s22n1+n2−2=79×5.542+79×8.05280+80−2=6.90992

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

D.F = n1+n2−2D.F = 80+80−2D.F = 158

Now the t-value can be obtained as follow:

t=(x¯1−x¯2)sp1n1+1n2=(25.61−32.51)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: w/CrunchIt/EESEE Access Card, Chapter 7, Problem 134E , 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.

Answer 13

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

Answer to Problem 134E

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:μ1≠μ2

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=(n1−1)×s12+(n2−1)×s22n1+n2−2=79×2.462+79×2.73280+80−2=2.59851

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

D.F = n1+n2−2D.F = 80+80−2D.F = 158

The t-value is obtained as follow:

t=(x¯1−x¯2)sp1n1+1n2=(21.6−26.41)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: w/CrunchIt/EESEE Access Card, Chapter 7, Problem 134E , 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.

Answer 14

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.

Answer 15

Section 2:

Expert Solution

Answer to Problem 134E

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

Answer 16

Section 2:

Expert Solution

Answer 17

Section 2:

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Calculation: The null and alternative hypotheses are:

H0:μ1=μ2Ha:μ1≠μ2

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=(n1−1)×s12+(n2−1)×s22n1+n2−2=79×2.462+79×2.73280+80−2=2.59851

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

D.F = n1+n2−2D.F = 80+80−2D.F = 158

The t-value is obtained as follow:

t=(x¯1−x¯2)sp1n1+1n2=(21.6−26.41)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: w/CrunchIt/EESEE Access Card, Chapter 7, Problem 134E , 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.

Answer 20

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

Answer to Problem 134E

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:μ1≠μ2

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=(n1−1)×s12+(n2−1)×s22n1+n2−2=79×516.632+79×372.77280+80−2=450.480

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

D.F = n1+n2−2D.F = 80+80−2D.F = 158

The t-value is obtained as follow:

t=(x¯1−x¯2)sp1n1+1n2=(297.13−580.54)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: w/CrunchIt/EESEE Access Card, Chapter 7, Problem 134E , 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.

Answer 21

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.

Answer 22

Section 3:

Expert Solution

Answer to Problem 134E

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

Answer 23

Section 3:

Expert Solution

Answer 24

Section 3:

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Calculation: The null and alternative hypotheses are:

H0:μ1=μ2Ha:μ1≠μ2

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=(n1−1)×s12+(n2−1)×s22n1+n2−2=79×516.632+79×372.77280+80−2=450.480

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

D.F = n1+n2−2D.F = 80+80−2D.F = 158

The t-value is obtained as follow:

t=(x¯1−x¯2)sp1n1+1n2=(297.13−580.54)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: w/CrunchIt/EESEE Access Card, Chapter 7, Problem 134E , 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.

Answer 27

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

Answer to Problem 134E

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:μ1≠μ2

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=(n1−1)×s12+(n2−1)×s22n1+n2−2=79×1.462+79×1.24280+80−2=1.35447

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

D.F = n1+n2−2D.F = 80+80−2D.F = 158

The t-value is obtained as follow:

t=(x¯1−x¯2)sp1n1+1n2=(2.21−1.82)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: w/CrunchIt/EESEE Access Card, Chapter 7, Problem 134E , 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.

Answer 28

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.

Answer 29

Section 4:

Expert Solution

Answer to Problem 134E

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.

Answer 30

Section 4:

Expert Solution

Answer 31

Section 4:

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Calculation: The null and alternative hypotheses are:

H0:μ1=μ2Ha:μ1≠μ2

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=(n1−1)×s12+(n2−1)×s22n1+n2−2=79×1.462+79×1.24280+80−2=1.35447

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

D.F = n1+n2−2D.F = 80+80−2D.F = 158

The t-value is obtained as follow:

t=(x¯1−x¯2)sp1n1+1n2=(2.21−1.82)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: w/CrunchIt/EESEE Access Card, Chapter 7, Problem 134E , 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.

Answer 34

(b)

To determine

The summary of the hypothesis test results.

(b)

Expert Solution

Answer to Problem 134E

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.

Answer 35

(b)

To determine

The summary of the hypothesis test results.

Answer 36

(b)

Expert Solution

Answer to Problem 134E

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

Answer 37

(b)

Expert Solution

Answer 38

(b)

Expert Solution

Answer 39

Expert Solution

Answer 40

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.

Answer 41

Ch. 7.1 - Prob. 1UYK Ch. 7.1 - Prob. 2UYK Ch. 7.1 - Prob. 3UYK Ch. 7.1 - Prob. 4UYK Ch. 7.1 - Prob. 5UYK Ch. 7.1 - Prob. 6UYK Ch. 7.1 - Prob. 7UYK Ch. 7.1 - Prob. 8UYK Ch. 7.1 - Prob. 9UYK Ch. 7.1 - Prob. 10UYK

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

Concept explainers

Videos

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

Answer to Problem 134E

Explanation of Solution

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

Answer to Problem 134E

Explanation of Solution

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

Answer to Problem 134E

Explanation of Solution

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

Answer to Problem 134E

Explanation of Solution

The summary of the hypothesis test results.

Answer to Problem 134E

Explanation of Solution

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Chapter 7 Solutions