To test: The comparison of means of the two samples using significance test of pooled methods.

Question

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Answer 1

Question

Chapter 7.2, Problem 92E

To determine

To test: The comparison of means of the two samples using significance test of pooled methods.

Expert Solution

Answer to Problem 92E

Solution: The data strongly suggests that there is no significant difference between the means of early eaters and late eaters in terms of consumption of fat.

Explanation of Solution

Given: The data on total fats for early eaters and late eaters are provided. The summary of statistics fats consumption by the two groups are provided as:

x¯1=23.1x¯2=21.4s1=12.5s2=8.2

n1=202 n2=200

Explanation:

Calculation: The significance test is to compare the means of two groups in terms of consumption of fats. Hence, the null hypothesis is that the consumption of fats in the two groups is the same as against the alternative that the consumption of fats in the two groups is not same.

Therefore, the hypotheses are formulated as:

H0:μEarly=μLateHa:μEarly≠μLate

In the above hypothesis, μEarly represents the average fat consumption of early eaters and μLate represents the average fat consumption of late eaters.

The two-sample t- test statistic for pooled methods is defined as:

t=(x¯1−x¯2)−(μ1−μ2)sp1n1+1n2

where,

x¯1=Sample mean of the fat consumption by early eatersx¯2=Sample mean of the fat consumption by late eaterss1=Sample standard deviation of fat consumption by early eaterss2=Sample standard deviation of fat consumption by late eaters

n1=Sample size of fat consumption by early eatersn2=Sample size of fat consumption by late eatersμ1−μ2=Difference of true population meanssp=The pooled sample standard deviation

First, determine the pooled sample standard deviation. The formula for pooled sample standard deviation is defined as:

sp=(n1−1)s12+(n2−1)s22n1+n2−2

Substitute the provided values and determine the pooled sample variance:

sp=[(202−1)×(12.52)]+[(200−1)×(8.22)]202+200−2=44787.01400=10.58

Now, determine the t- statistic. The difference of means is assumed as 0 in the null hypothesis. Substitute the provided values in the above-defined formula to compute the two sample t-statistic. So,

t=(x¯1−x¯2)−(μ1−μ2)sp1n1+1n2=(23.1−21.4)−0(10.58)×1202+1200=1.71.06=1.60

The p-value for the provided two-sided test is calculated as P(T≥1.60). For the pooled two-sample t-statistic, the degrees of freedom are n1+n2−2. So, the degrees of freedom are calculated as:

Degrees of freedom=n1+n2−2 =202+200−2=400

So, the degrees of freedom are 400.

The Excel function to determine the p- value from t-test statistic is displayed in the screenshot below:

Introduction to the Practice of Statistics: w/CrunchIt/EESEE Access Card, Chapter 7.2, Problem 92E

Conclusion: Therefore, the p-value is obtained as 0.1104, which is more than 0.05. So, do not reject the null hypothesis and hence it is concluded that the data strongly suggests that there is no significant difference between the means of the two groups in terms of fats consumption.

To determine

To find: A 95% confidence interval for the difference of means between early eaters and late eaters in terms of consumption of fats using pooled methods.

Expert Solution

Answer to Problem 92E

Solution: A 95% confidence interval is (−0.3747,3.7747)_.

Explanation of Solution

Calculation: The confidence interval is calculated as:

Confidence interval=(x¯1−x¯2)±t*(sp1n1+1n2)

where,

x¯1=Sample mean of the fat consumption by early eatersx¯2=Sample mean of the fat consumption by late eaterss1=Sample standard deviation of fat consumption by early eaterss2=Sample standard deviation of fat consumption by late eaters

n1=Sample size of fat consumption by early eatersn2=Sample size of fat consumption by late eatersμ1−μ2=Difference of true population meanssp=The pooled sample standard deviation

The pooled sample variance is obtained as 10.58 in the previous part. The critical value of t for 95% confidence level and 400 degrees of freedom is 1.9659. Substitute the provided values in the above-defined formula to determine the 95% confidence interval for the difference between the mean consumption of fats of two groups. So,

(x¯1−x¯2)±t*(sp1n1+1n2)=(23.1−21.4)±[(1.9659)×(10.58)×1202+1200]=(1.7±2.0747)=(1.7−2.0747,1.7+2.0747)=(−0.3747,3.7747)

Interpretation: Therefore, the 95% confidence interval for the difference between the means is obtained as (−0.3747,3.7747).

To determine

To explain: The results obtained in the provided problem with those obtained in Exercise 7.71.

Expert Solution

Answer to Problem 92E

Solution: The results are almost similar to those obtained in Exercise 7.71.

Explanation of Solution

The results of Exercise 7.71 are obtained as:

t=1.62p−value=0.1081.

The results obtained in this provided problem are:

t=1.62p−value=0.1104

Therefore, the results are almost similar.

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Answer 2

Question

Chapter 7.2, Problem 92E

To determine

To test: The comparison of means of the two samples using significance test of pooled methods.

Expert Solution

Answer to Problem 92E

Solution: The data strongly suggests that there is no significant difference between the means of early eaters and late eaters in terms of consumption of fat.

Explanation of Solution

Given: The data on total fats for early eaters and late eaters are provided. The summary of statistics fats consumption by the two groups are provided as:

x¯1=23.1x¯2=21.4s1=12.5s2=8.2

n1=202 n2=200

Explanation:

Calculation: The significance test is to compare the means of two groups in terms of consumption of fats. Hence, the null hypothesis is that the consumption of fats in the two groups is the same as against the alternative that the consumption of fats in the two groups is not same.

Therefore, the hypotheses are formulated as:

H0:μEarly=μLateHa:μEarly≠μLate

In the above hypothesis, μEarly represents the average fat consumption of early eaters and μLate represents the average fat consumption of late eaters.

The two-sample t- test statistic for pooled methods is defined as:

t=(x¯1−x¯2)−(μ1−μ2)sp1n1+1n2

where,

x¯1=Sample mean of the fat consumption by early eatersx¯2=Sample mean of the fat consumption by late eaterss1=Sample standard deviation of fat consumption by early eaterss2=Sample standard deviation of fat consumption by late eaters

n1=Sample size of fat consumption by early eatersn2=Sample size of fat consumption by late eatersμ1−μ2=Difference of true population meanssp=The pooled sample standard deviation

First, determine the pooled sample standard deviation. The formula for pooled sample standard deviation is defined as:

sp=(n1−1)s12+(n2−1)s22n1+n2−2

Substitute the provided values and determine the pooled sample variance:

sp=[(202−1)×(12.52)]+[(200−1)×(8.22)]202+200−2=44787.01400=10.58

Now, determine the t- statistic. The difference of means is assumed as 0 in the null hypothesis. Substitute the provided values in the above-defined formula to compute the two sample t-statistic. So,

t=(x¯1−x¯2)−(μ1−μ2)sp1n1+1n2=(23.1−21.4)−0(10.58)×1202+1200=1.71.06=1.60

The p-value for the provided two-sided test is calculated as P(T≥1.60). For the pooled two-sample t-statistic, the degrees of freedom are n1+n2−2. So, the degrees of freedom are calculated as:

Degrees of freedom=n1+n2−2 =202+200−2=400

So, the degrees of freedom are 400.

The Excel function to determine the p- value from t-test statistic is displayed in the screenshot below:

Introduction to the Practice of Statistics: w/CrunchIt/EESEE Access Card, Chapter 7.2, Problem 92E

Conclusion: Therefore, the p-value is obtained as 0.1104, which is more than 0.05. So, do not reject the null hypothesis and hence it is concluded that the data strongly suggests that there is no significant difference between the means of the two groups in terms of fats consumption.

To determine

To find: A 95% confidence interval for the difference of means between early eaters and late eaters in terms of consumption of fats using pooled methods.

Expert Solution

Answer to Problem 92E

Solution: A 95% confidence interval is (−0.3747,3.7747)_.

Explanation of Solution

Calculation: The confidence interval is calculated as:

Confidence interval=(x¯1−x¯2)±t*(sp1n1+1n2)

where,

x¯1=Sample mean of the fat consumption by early eatersx¯2=Sample mean of the fat consumption by late eaterss1=Sample standard deviation of fat consumption by early eaterss2=Sample standard deviation of fat consumption by late eaters

n1=Sample size of fat consumption by early eatersn2=Sample size of fat consumption by late eatersμ1−μ2=Difference of true population meanssp=The pooled sample standard deviation

The pooled sample variance is obtained as 10.58 in the previous part. The critical value of t for 95% confidence level and 400 degrees of freedom is 1.9659. Substitute the provided values in the above-defined formula to determine the 95% confidence interval for the difference between the mean consumption of fats of two groups. So,

(x¯1−x¯2)±t*(sp1n1+1n2)=(23.1−21.4)±[(1.9659)×(10.58)×1202+1200]=(1.7±2.0747)=(1.7−2.0747,1.7+2.0747)=(−0.3747,3.7747)

Interpretation: Therefore, the 95% confidence interval for the difference between the means is obtained as (−0.3747,3.7747).

To determine

To explain: The results obtained in the provided problem with those obtained in Exercise 7.71.

Expert Solution

Answer to Problem 92E

Solution: The results are almost similar to those obtained in Exercise 7.71.

Explanation of Solution

The results of Exercise 7.71 are obtained as:

t=1.62p−value=0.1081.

The results obtained in this provided problem are:

t=1.62p−value=0.1104

Therefore, the results are almost similar.

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Answer 3

Question

Chapter 7.2, Problem 92E

To determine

To test: The comparison of means of the two samples using significance test of pooled methods.

Expert Solution

Answer to Problem 92E

Solution: The data strongly suggests that there is no significant difference between the means of early eaters and late eaters in terms of consumption of fat.

Explanation of Solution

Given: The data on total fats for early eaters and late eaters are provided. The summary of statistics fats consumption by the two groups are provided as:

x¯1=23.1x¯2=21.4s1=12.5s2=8.2

n1=202 n2=200

Explanation:

Calculation: The significance test is to compare the means of two groups in terms of consumption of fats. Hence, the null hypothesis is that the consumption of fats in the two groups is the same as against the alternative that the consumption of fats in the two groups is not same.

Therefore, the hypotheses are formulated as:

H0:μEarly=μLateHa:μEarly≠μLate

In the above hypothesis, μEarly represents the average fat consumption of early eaters and μLate represents the average fat consumption of late eaters.

The two-sample t- test statistic for pooled methods is defined as:

t=(x¯1−x¯2)−(μ1−μ2)sp1n1+1n2

where,

x¯1=Sample mean of the fat consumption by early eatersx¯2=Sample mean of the fat consumption by late eaterss1=Sample standard deviation of fat consumption by early eaterss2=Sample standard deviation of fat consumption by late eaters

n1=Sample size of fat consumption by early eatersn2=Sample size of fat consumption by late eatersμ1−μ2=Difference of true population meanssp=The pooled sample standard deviation

First, determine the pooled sample standard deviation. The formula for pooled sample standard deviation is defined as:

sp=(n1−1)s12+(n2−1)s22n1+n2−2

Substitute the provided values and determine the pooled sample variance:

sp=[(202−1)×(12.52)]+[(200−1)×(8.22)]202+200−2=44787.01400=10.58

Now, determine the t- statistic. The difference of means is assumed as 0 in the null hypothesis. Substitute the provided values in the above-defined formula to compute the two sample t-statistic. So,

t=(x¯1−x¯2)−(μ1−μ2)sp1n1+1n2=(23.1−21.4)−0(10.58)×1202+1200=1.71.06=1.60

The p-value for the provided two-sided test is calculated as P(T≥1.60). For the pooled two-sample t-statistic, the degrees of freedom are n1+n2−2. So, the degrees of freedom are calculated as:

Degrees of freedom=n1+n2−2 =202+200−2=400

So, the degrees of freedom are 400.

The Excel function to determine the p- value from t-test statistic is displayed in the screenshot below:

Introduction to the Practice of Statistics: w/CrunchIt/EESEE Access Card, Chapter 7.2, Problem 92E

Conclusion: Therefore, the p-value is obtained as 0.1104, which is more than 0.05. So, do not reject the null hypothesis and hence it is concluded that the data strongly suggests that there is no significant difference between the means of the two groups in terms of fats consumption.

To determine

To find: A 95% confidence interval for the difference of means between early eaters and late eaters in terms of consumption of fats using pooled methods.

Expert Solution

Answer to Problem 92E

Solution: A 95% confidence interval is (−0.3747,3.7747)_.

Explanation of Solution

Calculation: The confidence interval is calculated as:

Confidence interval=(x¯1−x¯2)±t*(sp1n1+1n2)

where,

x¯1=Sample mean of the fat consumption by early eatersx¯2=Sample mean of the fat consumption by late eaterss1=Sample standard deviation of fat consumption by early eaterss2=Sample standard deviation of fat consumption by late eaters

n1=Sample size of fat consumption by early eatersn2=Sample size of fat consumption by late eatersμ1−μ2=Difference of true population meanssp=The pooled sample standard deviation

The pooled sample variance is obtained as 10.58 in the previous part. The critical value of t for 95% confidence level and 400 degrees of freedom is 1.9659. Substitute the provided values in the above-defined formula to determine the 95% confidence interval for the difference between the mean consumption of fats of two groups. So,

(x¯1−x¯2)±t*(sp1n1+1n2)=(23.1−21.4)±[(1.9659)×(10.58)×1202+1200]=(1.7±2.0747)=(1.7−2.0747,1.7+2.0747)=(−0.3747,3.7747)

Interpretation: Therefore, the 95% confidence interval for the difference between the means is obtained as (−0.3747,3.7747).

To determine

To explain: The results obtained in the provided problem with those obtained in Exercise 7.71.

Expert Solution

Answer to Problem 92E

Solution: The results are almost similar to those obtained in Exercise 7.71.

Explanation of Solution

The results of Exercise 7.71 are obtained as:

t=1.62p−value=0.1081.

The results obtained in this provided problem are:

t=1.62p−value=0.1104

Therefore, the results are almost similar.

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

Answer 4

Question

Chapter 7.2, Problem 92E

To determine

To test: The comparison of means of the two samples using significance test of pooled methods.

Expert Solution

Answer to Problem 92E

Solution: The data strongly suggests that there is no significant difference between the means of early eaters and late eaters in terms of consumption of fat.

Explanation of Solution

Given: The data on total fats for early eaters and late eaters are provided. The summary of statistics fats consumption by the two groups are provided as:

x¯1=23.1x¯2=21.4s1=12.5s2=8.2

n1=202 n2=200

Explanation:

Calculation: The significance test is to compare the means of two groups in terms of consumption of fats. Hence, the null hypothesis is that the consumption of fats in the two groups is the same as against the alternative that the consumption of fats in the two groups is not same.

Therefore, the hypotheses are formulated as:

H0:μEarly=μLateHa:μEarly≠μLate

In the above hypothesis, μEarly represents the average fat consumption of early eaters and μLate represents the average fat consumption of late eaters.

The two-sample t- test statistic for pooled methods is defined as:

t=(x¯1−x¯2)−(μ1−μ2)sp1n1+1n2

where,

x¯1=Sample mean of the fat consumption by early eatersx¯2=Sample mean of the fat consumption by late eaterss1=Sample standard deviation of fat consumption by early eaterss2=Sample standard deviation of fat consumption by late eaters

n1=Sample size of fat consumption by early eatersn2=Sample size of fat consumption by late eatersμ1−μ2=Difference of true population meanssp=The pooled sample standard deviation

First, determine the pooled sample standard deviation. The formula for pooled sample standard deviation is defined as:

sp=(n1−1)s12+(n2−1)s22n1+n2−2

Substitute the provided values and determine the pooled sample variance:

sp=[(202−1)×(12.52)]+[(200−1)×(8.22)]202+200−2=44787.01400=10.58

Now, determine the t- statistic. The difference of means is assumed as 0 in the null hypothesis. Substitute the provided values in the above-defined formula to compute the two sample t-statistic. So,

t=(x¯1−x¯2)−(μ1−μ2)sp1n1+1n2=(23.1−21.4)−0(10.58)×1202+1200=1.71.06=1.60

The p-value for the provided two-sided test is calculated as P(T≥1.60). For the pooled two-sample t-statistic, the degrees of freedom are n1+n2−2. So, the degrees of freedom are calculated as:

Degrees of freedom=n1+n2−2 =202+200−2=400

So, the degrees of freedom are 400.

The Excel function to determine the p- value from t-test statistic is displayed in the screenshot below:

Introduction to the Practice of Statistics: w/CrunchIt/EESEE Access Card, Chapter 7.2, Problem 92E

Conclusion: Therefore, the p-value is obtained as 0.1104, which is more than 0.05. So, do not reject the null hypothesis and hence it is concluded that the data strongly suggests that there is no significant difference between the means of the two groups in terms of fats consumption.

To determine

To find: A 95% confidence interval for the difference of means between early eaters and late eaters in terms of consumption of fats using pooled methods.

Expert Solution

Answer to Problem 92E

Solution: A 95% confidence interval is (−0.3747,3.7747)_.

Explanation of Solution

Calculation: The confidence interval is calculated as:

Confidence interval=(x¯1−x¯2)±t*(sp1n1+1n2)

where,

x¯1=Sample mean of the fat consumption by early eatersx¯2=Sample mean of the fat consumption by late eaterss1=Sample standard deviation of fat consumption by early eaterss2=Sample standard deviation of fat consumption by late eaters

n1=Sample size of fat consumption by early eatersn2=Sample size of fat consumption by late eatersμ1−μ2=Difference of true population meanssp=The pooled sample standard deviation

The pooled sample variance is obtained as 10.58 in the previous part. The critical value of t for 95% confidence level and 400 degrees of freedom is 1.9659. Substitute the provided values in the above-defined formula to determine the 95% confidence interval for the difference between the mean consumption of fats of two groups. So,

(x¯1−x¯2)±t*(sp1n1+1n2)=(23.1−21.4)±[(1.9659)×(10.58)×1202+1200]=(1.7±2.0747)=(1.7−2.0747,1.7+2.0747)=(−0.3747,3.7747)

Interpretation: Therefore, the 95% confidence interval for the difference between the means is obtained as (−0.3747,3.7747).

To determine

To explain: The results obtained in the provided problem with those obtained in Exercise 7.71.

Expert Solution

Answer to Problem 92E

Solution: The results are almost similar to those obtained in Exercise 7.71.

Explanation of Solution

The results of Exercise 7.71 are obtained as:

t=1.62p−value=0.1081.

The results obtained in this provided problem are:

t=1.62p−value=0.1104

Therefore, the results are almost similar.

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

Answer 5

Chapter 7.2, Problem 92E

To determine

To test: The comparison of means of the two samples using significance test of pooled methods.

Expert Solution

Answer to Problem 92E

Solution: The data strongly suggests that there is no significant difference between the means of early eaters and late eaters in terms of consumption of fat.

Explanation of Solution

Given: The data on total fats for early eaters and late eaters are provided. The summary of statistics fats consumption by the two groups are provided as:

x¯1=23.1x¯2=21.4s1=12.5s2=8.2

n1=202 n2=200

Explanation:

Calculation: The significance test is to compare the means of two groups in terms of consumption of fats. Hence, the null hypothesis is that the consumption of fats in the two groups is the same as against the alternative that the consumption of fats in the two groups is not same.

Therefore, the hypotheses are formulated as:

H0:μEarly=μLateHa:μEarly≠μLate

In the above hypothesis, μEarly represents the average fat consumption of early eaters and μLate represents the average fat consumption of late eaters.

The two-sample t- test statistic for pooled methods is defined as:

t=(x¯1−x¯2)−(μ1−μ2)sp1n1+1n2

where,

x¯1=Sample mean of the fat consumption by early eatersx¯2=Sample mean of the fat consumption by late eaterss1=Sample standard deviation of fat consumption by early eaterss2=Sample standard deviation of fat consumption by late eaters

n1=Sample size of fat consumption by early eatersn2=Sample size of fat consumption by late eatersμ1−μ2=Difference of true population meanssp=The pooled sample standard deviation

First, determine the pooled sample standard deviation. The formula for pooled sample standard deviation is defined as:

sp=(n1−1)s12+(n2−1)s22n1+n2−2

Substitute the provided values and determine the pooled sample variance:

sp=[(202−1)×(12.52)]+[(200−1)×(8.22)]202+200−2=44787.01400=10.58

Now, determine the t- statistic. The difference of means is assumed as 0 in the null hypothesis. Substitute the provided values in the above-defined formula to compute the two sample t-statistic. So,

t=(x¯1−x¯2)−(μ1−μ2)sp1n1+1n2=(23.1−21.4)−0(10.58)×1202+1200=1.71.06=1.60

The p-value for the provided two-sided test is calculated as P(T≥1.60). For the pooled two-sample t-statistic, the degrees of freedom are n1+n2−2. So, the degrees of freedom are calculated as:

Degrees of freedom=n1+n2−2 =202+200−2=400

So, the degrees of freedom are 400.

The Excel function to determine the p- value from t-test statistic is displayed in the screenshot below:

Introduction to the Practice of Statistics: w/CrunchIt/EESEE Access Card, Chapter 7.2, Problem 92E

Conclusion: Therefore, the p-value is obtained as 0.1104, which is more than 0.05. So, do not reject the null hypothesis and hence it is concluded that the data strongly suggests that there is no significant difference between the means of the two groups in terms of fats consumption.

To determine

To find: A 95% confidence interval for the difference of means between early eaters and late eaters in terms of consumption of fats using pooled methods.

Expert Solution

Answer to Problem 92E

Solution: A 95% confidence interval is (−0.3747,3.7747)_.

Explanation of Solution

Calculation: The confidence interval is calculated as:

Confidence interval=(x¯1−x¯2)±t*(sp1n1+1n2)

where,

x¯1=Sample mean of the fat consumption by early eatersx¯2=Sample mean of the fat consumption by late eaterss1=Sample standard deviation of fat consumption by early eaterss2=Sample standard deviation of fat consumption by late eaters

n1=Sample size of fat consumption by early eatersn2=Sample size of fat consumption by late eatersμ1−μ2=Difference of true population meanssp=The pooled sample standard deviation

The pooled sample variance is obtained as 10.58 in the previous part. The critical value of t for 95% confidence level and 400 degrees of freedom is 1.9659. Substitute the provided values in the above-defined formula to determine the 95% confidence interval for the difference between the mean consumption of fats of two groups. So,

(x¯1−x¯2)±t*(sp1n1+1n2)=(23.1−21.4)±[(1.9659)×(10.58)×1202+1200]=(1.7±2.0747)=(1.7−2.0747,1.7+2.0747)=(−0.3747,3.7747)

Interpretation: Therefore, the 95% confidence interval for the difference between the means is obtained as (−0.3747,3.7747).

To determine

To explain: The results obtained in the provided problem with those obtained in Exercise 7.71.

Expert Solution

Answer to Problem 92E

Solution: The results are almost similar to those obtained in Exercise 7.71.

Explanation of Solution

The results of Exercise 7.71 are obtained as:

t=1.62p−value=0.1081.

The results obtained in this provided problem are:

t=1.62p−value=0.1104

Therefore, the results are almost similar.

Answer 6

Chapter 7.2, Problem 92E

To determine

To test: The comparison of means of the two samples using significance test of pooled methods.

Expert Solution

Answer to Problem 92E

Solution: The data strongly suggests that there is no significant difference between the means of early eaters and late eaters in terms of consumption of fat.

Explanation of Solution

Given: The data on total fats for early eaters and late eaters are provided. The summary of statistics fats consumption by the two groups are provided as:

x¯1=23.1x¯2=21.4s1=12.5s2=8.2

n1=202 n2=200

Explanation:

Calculation: The significance test is to compare the means of two groups in terms of consumption of fats. Hence, the null hypothesis is that the consumption of fats in the two groups is the same as against the alternative that the consumption of fats in the two groups is not same.

Therefore, the hypotheses are formulated as:

H0:μEarly=μLateHa:μEarly≠μLate

In the above hypothesis, μEarly represents the average fat consumption of early eaters and μLate represents the average fat consumption of late eaters.

The two-sample t- test statistic for pooled methods is defined as:

t=(x¯1−x¯2)−(μ1−μ2)sp1n1+1n2

where,

x¯1=Sample mean of the fat consumption by early eatersx¯2=Sample mean of the fat consumption by late eaterss1=Sample standard deviation of fat consumption by early eaterss2=Sample standard deviation of fat consumption by late eaters

n1=Sample size of fat consumption by early eatersn2=Sample size of fat consumption by late eatersμ1−μ2=Difference of true population meanssp=The pooled sample standard deviation

First, determine the pooled sample standard deviation. The formula for pooled sample standard deviation is defined as:

sp=(n1−1)s12+(n2−1)s22n1+n2−2

Substitute the provided values and determine the pooled sample variance:

sp=[(202−1)×(12.52)]+[(200−1)×(8.22)]202+200−2=44787.01400=10.58

Now, determine the t- statistic. The difference of means is assumed as 0 in the null hypothesis. Substitute the provided values in the above-defined formula to compute the two sample t-statistic. So,

t=(x¯1−x¯2)−(μ1−μ2)sp1n1+1n2=(23.1−21.4)−0(10.58)×1202+1200=1.71.06=1.60

The p-value for the provided two-sided test is calculated as P(T≥1.60). For the pooled two-sample t-statistic, the degrees of freedom are n1+n2−2. So, the degrees of freedom are calculated as:

Degrees of freedom=n1+n2−2 =202+200−2=400

So, the degrees of freedom are 400.

The Excel function to determine the p- value from t-test statistic is displayed in the screenshot below:

Introduction to the Practice of Statistics: w/CrunchIt/EESEE Access Card, Chapter 7.2, Problem 92E

Conclusion: Therefore, the p-value is obtained as 0.1104, which is more than 0.05. So, do not reject the null hypothesis and hence it is concluded that the data strongly suggests that there is no significant difference between the means of the two groups in terms of fats consumption.

Answer 7

Chapter 7.2, Problem 92E

To determine

To test: The comparison of means of the two samples using significance test of pooled methods.

Answer 8

Expert Solution

Answer to Problem 92E

Solution: The data strongly suggests that there is no significant difference between the means of early eaters and late eaters in terms of consumption of fat.

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given: The data on total fats for early eaters and late eaters are provided. The summary of statistics fats consumption by the two groups are provided as:

x¯1=23.1x¯2=21.4s1=12.5s2=8.2

n1=202 n2=200

Explanation:

Calculation: The significance test is to compare the means of two groups in terms of consumption of fats. Hence, the null hypothesis is that the consumption of fats in the two groups is the same as against the alternative that the consumption of fats in the two groups is not same.

Therefore, the hypotheses are formulated as:

H0:μEarly=μLateHa:μEarly≠μLate

In the above hypothesis, μEarly represents the average fat consumption of early eaters and μLate represents the average fat consumption of late eaters.

The two-sample t- test statistic for pooled methods is defined as:

t=(x¯1−x¯2)−(μ1−μ2)sp1n1+1n2

where,

x¯1=Sample mean of the fat consumption by early eatersx¯2=Sample mean of the fat consumption by late eaterss1=Sample standard deviation of fat consumption by early eaterss2=Sample standard deviation of fat consumption by late eaters

n1=Sample size of fat consumption by early eatersn2=Sample size of fat consumption by late eatersμ1−μ2=Difference of true population meanssp=The pooled sample standard deviation

First, determine the pooled sample standard deviation. The formula for pooled sample standard deviation is defined as:

sp=(n1−1)s12+(n2−1)s22n1+n2−2

Substitute the provided values and determine the pooled sample variance:

sp=[(202−1)×(12.52)]+[(200−1)×(8.22)]202+200−2=44787.01400=10.58

Now, determine the t- statistic. The difference of means is assumed as 0 in the null hypothesis. Substitute the provided values in the above-defined formula to compute the two sample t-statistic. So,

t=(x¯1−x¯2)−(μ1−μ2)sp1n1+1n2=(23.1−21.4)−0(10.58)×1202+1200=1.71.06=1.60

The p-value for the provided two-sided test is calculated as P(T≥1.60). For the pooled two-sample t-statistic, the degrees of freedom are n1+n2−2. So, the degrees of freedom are calculated as:

Degrees of freedom=n1+n2−2 =202+200−2=400

So, the degrees of freedom are 400.

The Excel function to determine the p- value from t-test statistic is displayed in the screenshot below:

Introduction to the Practice of Statistics: w/CrunchIt/EESEE Access Card, Chapter 7.2, Problem 92E

Conclusion: Therefore, the p-value is obtained as 0.1104, which is more than 0.05. So, do not reject the null hypothesis and hence it is concluded that the data strongly suggests that there is no significant difference between the means of the two groups in terms of fats consumption.

Answer 13

To determine

To find: A 95% confidence interval for the difference of means between early eaters and late eaters in terms of consumption of fats using pooled methods.

Expert Solution

Answer to Problem 92E

Solution: A 95% confidence interval is (−0.3747,3.7747)_.

Explanation of Solution

Calculation: The confidence interval is calculated as:

Confidence interval=(x¯1−x¯2)±t*(sp1n1+1n2)

where,

x¯1=Sample mean of the fat consumption by early eatersx¯2=Sample mean of the fat consumption by late eaterss1=Sample standard deviation of fat consumption by early eaterss2=Sample standard deviation of fat consumption by late eaters

n1=Sample size of fat consumption by early eatersn2=Sample size of fat consumption by late eatersμ1−μ2=Difference of true population meanssp=The pooled sample standard deviation

The pooled sample variance is obtained as 10.58 in the previous part. The critical value of t for 95% confidence level and 400 degrees of freedom is 1.9659. Substitute the provided values in the above-defined formula to determine the 95% confidence interval for the difference between the mean consumption of fats of two groups. So,

(x¯1−x¯2)±t*(sp1n1+1n2)=(23.1−21.4)±[(1.9659)×(10.58)×1202+1200]=(1.7±2.0747)=(1.7−2.0747,1.7+2.0747)=(−0.3747,3.7747)

Interpretation: Therefore, the 95% confidence interval for the difference between the means is obtained as (−0.3747,3.7747).

Answer 14

To determine

To find: A 95% confidence interval for the difference of means between early eaters and late eaters in terms of consumption of fats using pooled methods.

Answer 15

Expert Solution

Answer to Problem 92E

Solution: A 95% confidence interval is (−0.3747,3.7747)_.

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Calculation: The confidence interval is calculated as:

Confidence interval=(x¯1−x¯2)±t*(sp1n1+1n2)

where,

x¯1=Sample mean of the fat consumption by early eatersx¯2=Sample mean of the fat consumption by late eaterss1=Sample standard deviation of fat consumption by early eaterss2=Sample standard deviation of fat consumption by late eaters

n1=Sample size of fat consumption by early eatersn2=Sample size of fat consumption by late eatersμ1−μ2=Difference of true population meanssp=The pooled sample standard deviation

The pooled sample variance is obtained as 10.58 in the previous part. The critical value of t for 95% confidence level and 400 degrees of freedom is 1.9659. Substitute the provided values in the above-defined formula to determine the 95% confidence interval for the difference between the mean consumption of fats of two groups. So,

(x¯1−x¯2)±t*(sp1n1+1n2)=(23.1−21.4)±[(1.9659)×(10.58)×1202+1200]=(1.7±2.0747)=(1.7−2.0747,1.7+2.0747)=(−0.3747,3.7747)

Interpretation: Therefore, the 95% confidence interval for the difference between the means is obtained as (−0.3747,3.7747).

Answer 20

To determine

To explain: The results obtained in the provided problem with those obtained in Exercise 7.71.

Expert Solution

Answer to Problem 92E

Solution: The results are almost similar to those obtained in Exercise 7.71.

Explanation of Solution

The results of Exercise 7.71 are obtained as:

t=1.62p−value=0.1081.

The results obtained in this provided problem are:

t=1.62p−value=0.1104

Therefore, the results are almost similar.

Answer 21

To determine

To explain: The results obtained in the provided problem with those obtained in Exercise 7.71.

Answer 22

Expert Solution

Answer to Problem 92E

Solution: The results are almost similar to those obtained in Exercise 7.71.

Answer 23

Expert Solution

Answer 24

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

The results of Exercise 7.71 are obtained as:

t=1.62p−value=0.1081.

The results obtained in this provided problem are:

t=1.62p−value=0.1104

Therefore, the results are almost similar.

Answer 27

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 comparison of means of the two samples using significance test of pooled methods.

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To test: The comparison of means of the two samples using significance test of pooled methods.

Answer to Problem 92E

Explanation of Solution

To find: A 95% confidence interval for the difference of means between early eaters and late eaters in terms of consumption of fats using pooled methods.

Answer to Problem 92E

Explanation of Solution

To explain: The results obtained in the provided problem with those obtained in Exercise 7.71.

Answer to Problem 92E

Explanation of Solution

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