The difference in mean and standard error for body weight.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 7, Problem 131E

(a)

Section 1:

To determine

To find: The difference in mean and standard error for body weight.

(a)

Section 1:

Expert Solution

Answer to Problem 131E

Solution: The difference in mean for body weight is −0.7_ and standard error is 2.29_.

Explanation of Solution

Calculation: The difference of mean can be calculated as follows:

x¯1−x¯2=0.4−1.1=−0.7

The standard error (SE) of difference of mean can be calculated as follows:

SE=sn=8.614=2.298447≈2.29

Section 2:

To determine

To find: The difference in mean and standard error for calorie intake.

Section 2:

Expert Solution

Answer to Problem 131E

Solution: The difference in mean for body weight is 14_ and standard error is 56.125_.

Explanation of Solution

Calculation:

The difference of mean can be calculated as follows:

x¯1−x¯2=2589−2575=14

The standard error (SE) of difference of mean can be calculated as follows:

SE=sn=21014=56.1248≈56.125

(b)

Section 1:

To determine

To test: The significant differences in body weight.

(b)

Section 1:

Expert Solution

Answer to Problem 131E

Solution: There is no significant difference in body weight.

The t-statistic for the test hypothesis is obtained as t-value =−0.305.

The P-statistic for the test hypothesis is obtained as P-value =0.76418.

Explanation of Solution

Calculation: The hypothesis for study is defined as

H0: The difference of body weight is = 0.

Ha: The difference of body weight is ≠0.

To test the hypothesis that there is no significant difference in body weight, t-statistic is used to determine the significance of the difference. The t-value is obtained as follows:

t=(x¯−μ)sn=(−0.7−0)8.614=−0.72.29845=−0.30455

≈0.305

Now, the P-value can be obtained by using the standard normal table for t=−0.305 and the P-value =0.76418 for two-tailed t-statistic.

Conclusion: The P-value for t-test is greater than 0.05. So, the null hypothesis cannot be rejected significantly, which states that there is no significant difference in body weight.

Section 2:

To determine

To test: The significant differences in calorie intake.

Section 2:

Expert Solution

Answer to Problem 131E

Solution: There is no significant difference in calorie intake.

The t-statistic for the test hypothesis is obtained as t-value =0.249.

The P-statistic for the test hypothesis is obtained as P-value =0.80258.

Explanation of Solution

Calculation: The hypothesis for study is defined as

H0: The difference of calorie intake is =0.

Ha: The difference of calorie intake is ≠0.

To test the hypothesis, t-statistic is used to determine the significance of the difference. The t- value is obtained as follow:

t=(x¯−μ)sn=(14−0)21014=1456.12486=0.24944

≈0.249

Now, the P-value can be obtained by using the standard normal table for t=0.24944. The P-value is 0.80258 for two-tailed t-statistic.

Conclusion: The P-value for t-test is greater than 0.05. So, the null hypothesis cannot be rejected significantly, which states that there is no significant difference in calorie intake.

(c)

Section 1:

To determine

To find: The 95% confidence interval for the difference of body weight.

(c)

Section 1:

Expert Solution

Answer to Problem 131E

Solution: The required 95% confidence interval is confidence interval=(−5.66,4.26)_.

Explanation of Solution

Calculation: The confidence interval for the difference of body weight can be obtained by first calculating the margin of error. The margin of error is obtained as follow:

Margin of error=t*×sn=2.16×8.614=2.16×2.298=4.965

Now, the confidence interval can be obtained as follows:

Confidence interval=x¯±Margin of error= −0.7±4.96=(−5.66,4.26)

Interpretation: As the hypothesized mean of 0 lies inside the 95% confidence interval, the null hypothesis cannot be rejected, which states that difference of body weight is within the 95% confidence interval significantly.

Section 2:

To determine

To find: The 95% confidence interval for the difference of calorie intake.

Section 2:

Expert Solution

Answer to Problem 131E

Solution: The required 95% confidence interval is (−107.23,135.23)_.

Explanation of Solution

Calculation: The confidence interval for the difference of calorie intake can be obtained by first calculating the margin of error. The margin of error is obtained as follow:

Margin of error=t*×sn=2.16×21014=2.16×56.125=121.23

Now, the confidence interval can be obtained as follows:

Confidence interval=x¯±Margin of error=14±121.23(−107.23,135.23)

Interpretation: As the hypothesized mean of 0 lies inside the 95% confidence interval, the null hypothesis cannot be rejected, which states that difference of calorie intake is within the 95% confidence interval significantly.

(d)

To determine

The summary of the results.

(d)

Expert Solution

Answer to Problem 131E

Solution: The result cannot be generalized and the outcome of study will also have affected due to the violation of instruction by three subjects as the sample size is too small.

Explanation of Solution

Here, the study is biased for a specific city, the model cannot be generalized for a whole population, and there are three violations out of fourteen subjects. The effect could be a serious concern in the study due to small sample size.

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!

Students have asked these similar questions

The daily sales (in hundreds of dollars) for a store in one month (30 days) are: 15, 22, 18, 25, 19 12, 17, 24, 20, 23 30, 28, 26, 31, 35 21, 19, 27, 18, 20 16, 15, 32, 30, 17 24, 29, 22, 33, 25 a. Construct a grouped frequency distribution with class intervals of width 5 starting from 12. b. Draw a histogram and state whether the data is symmetric, skewed left, or skewed right. Instruction: 1. Please answer the question given above for your tutorial participation mark. 2. Please upload your hand-written answers (pdf format).

Don’t solve i will dislike ?

What is the difference between population and sample in statistics?

Answer 2

Question

Chapter 7, Problem 131E

(a)

Section 1:

To determine

To find: The difference in mean and standard error for body weight.

(a)

Section 1:

Expert Solution

Answer to Problem 131E

Solution: The difference in mean for body weight is −0.7_ and standard error is 2.29_.

Explanation of Solution

Calculation: The difference of mean can be calculated as follows:

x¯1−x¯2=0.4−1.1=−0.7

The standard error (SE) of difference of mean can be calculated as follows:

SE=sn=8.614=2.298447≈2.29

Section 2:

To determine

To find: The difference in mean and standard error for calorie intake.

Section 2:

Expert Solution

Answer to Problem 131E

Solution: The difference in mean for body weight is 14_ and standard error is 56.125_.

Explanation of Solution

Calculation:

The difference of mean can be calculated as follows:

x¯1−x¯2=2589−2575=14

The standard error (SE) of difference of mean can be calculated as follows:

SE=sn=21014=56.1248≈56.125

(b)

Section 1:

To determine

To test: The significant differences in body weight.

(b)

Section 1:

Expert Solution

Answer to Problem 131E

Solution: There is no significant difference in body weight.

The t-statistic for the test hypothesis is obtained as t-value =−0.305.

The P-statistic for the test hypothesis is obtained as P-value =0.76418.

Explanation of Solution

Calculation: The hypothesis for study is defined as

H0: The difference of body weight is = 0.

Ha: The difference of body weight is ≠0.

To test the hypothesis that there is no significant difference in body weight, t-statistic is used to determine the significance of the difference. The t-value is obtained as follows:

t=(x¯−μ)sn=(−0.7−0)8.614=−0.72.29845=−0.30455

≈0.305

Now, the P-value can be obtained by using the standard normal table for t=−0.305 and the P-value =0.76418 for two-tailed t-statistic.

Conclusion: The P-value for t-test is greater than 0.05. So, the null hypothesis cannot be rejected significantly, which states that there is no significant difference in body weight.

Section 2:

To determine

To test: The significant differences in calorie intake.

Section 2:

Expert Solution

Answer to Problem 131E

Solution: There is no significant difference in calorie intake.

The t-statistic for the test hypothesis is obtained as t-value =0.249.

The P-statistic for the test hypothesis is obtained as P-value =0.80258.

Explanation of Solution

Calculation: The hypothesis for study is defined as

H0: The difference of calorie intake is =0.

Ha: The difference of calorie intake is ≠0.

To test the hypothesis, t-statistic is used to determine the significance of the difference. The t- value is obtained as follow:

t=(x¯−μ)sn=(14−0)21014=1456.12486=0.24944

≈0.249

Now, the P-value can be obtained by using the standard normal table for t=0.24944. The P-value is 0.80258 for two-tailed t-statistic.

Conclusion: The P-value for t-test is greater than 0.05. So, the null hypothesis cannot be rejected significantly, which states that there is no significant difference in calorie intake.

(c)

Section 1:

To determine

To find: The 95% confidence interval for the difference of body weight.

(c)

Section 1:

Expert Solution

Answer to Problem 131E

Solution: The required 95% confidence interval is confidence interval=(−5.66,4.26)_.

Explanation of Solution

Calculation: The confidence interval for the difference of body weight can be obtained by first calculating the margin of error. The margin of error is obtained as follow:

Margin of error=t*×sn=2.16×8.614=2.16×2.298=4.965

Now, the confidence interval can be obtained as follows:

Confidence interval=x¯±Margin of error= −0.7±4.96=(−5.66,4.26)

Interpretation: As the hypothesized mean of 0 lies inside the 95% confidence interval, the null hypothesis cannot be rejected, which states that difference of body weight is within the 95% confidence interval significantly.

Section 2:

To determine

To find: The 95% confidence interval for the difference of calorie intake.

Section 2:

Expert Solution

Answer to Problem 131E

Solution: The required 95% confidence interval is (−107.23,135.23)_.

Explanation of Solution

Calculation: The confidence interval for the difference of calorie intake can be obtained by first calculating the margin of error. The margin of error is obtained as follow:

Margin of error=t*×sn=2.16×21014=2.16×56.125=121.23

Now, the confidence interval can be obtained as follows:

Confidence interval=x¯±Margin of error=14±121.23(−107.23,135.23)

Interpretation: As the hypothesized mean of 0 lies inside the 95% confidence interval, the null hypothesis cannot be rejected, which states that difference of calorie intake is within the 95% confidence interval significantly.

(d)

To determine

The summary of the results.

(d)

Expert Solution

Answer to Problem 131E

Solution: The result cannot be generalized and the outcome of study will also have affected due to the violation of instruction by three subjects as the sample size is too small.

Explanation of Solution

Here, the study is biased for a specific city, the model cannot be generalized for a whole population, and there are three violations out of fourteen subjects. The effect could be a serious concern in the study due to small sample size.

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 3

Question

Chapter 7, Problem 131E

(a)

Section 1:

To determine

To find: The difference in mean and standard error for body weight.

(a)

Section 1:

Expert Solution

Answer to Problem 131E

Solution: The difference in mean for body weight is −0.7_ and standard error is 2.29_.

Explanation of Solution

Calculation: The difference of mean can be calculated as follows:

x¯1−x¯2=0.4−1.1=−0.7

The standard error (SE) of difference of mean can be calculated as follows:

SE=sn=8.614=2.298447≈2.29

Section 2:

To determine

To find: The difference in mean and standard error for calorie intake.

Section 2:

Expert Solution

Answer to Problem 131E

Solution: The difference in mean for body weight is 14_ and standard error is 56.125_.

Explanation of Solution

Calculation:

The difference of mean can be calculated as follows:

x¯1−x¯2=2589−2575=14

The standard error (SE) of difference of mean can be calculated as follows:

SE=sn=21014=56.1248≈56.125

(b)

Section 1:

To determine

To test: The significant differences in body weight.

(b)

Section 1:

Expert Solution

Answer to Problem 131E

Solution: There is no significant difference in body weight.

The t-statistic for the test hypothesis is obtained as t-value =−0.305.

The P-statistic for the test hypothesis is obtained as P-value =0.76418.

Explanation of Solution

Calculation: The hypothesis for study is defined as

H0: The difference of body weight is = 0.

Ha: The difference of body weight is ≠0.

To test the hypothesis that there is no significant difference in body weight, t-statistic is used to determine the significance of the difference. The t-value is obtained as follows:

t=(x¯−μ)sn=(−0.7−0)8.614=−0.72.29845=−0.30455

≈0.305

Now, the P-value can be obtained by using the standard normal table for t=−0.305 and the P-value =0.76418 for two-tailed t-statistic.

Conclusion: The P-value for t-test is greater than 0.05. So, the null hypothesis cannot be rejected significantly, which states that there is no significant difference in body weight.

Section 2:

To determine

To test: The significant differences in calorie intake.

Section 2:

Expert Solution

Answer to Problem 131E

Solution: There is no significant difference in calorie intake.

The t-statistic for the test hypothesis is obtained as t-value =0.249.

The P-statistic for the test hypothesis is obtained as P-value =0.80258.

Explanation of Solution

Calculation: The hypothesis for study is defined as

H0: The difference of calorie intake is =0.

Ha: The difference of calorie intake is ≠0.

To test the hypothesis, t-statistic is used to determine the significance of the difference. The t- value is obtained as follow:

t=(x¯−μ)sn=(14−0)21014=1456.12486=0.24944

≈0.249

Now, the P-value can be obtained by using the standard normal table for t=0.24944. The P-value is 0.80258 for two-tailed t-statistic.

Conclusion: The P-value for t-test is greater than 0.05. So, the null hypothesis cannot be rejected significantly, which states that there is no significant difference in calorie intake.

(c)

Section 1:

To determine

To find: The 95% confidence interval for the difference of body weight.

(c)

Section 1:

Expert Solution

Answer to Problem 131E

Solution: The required 95% confidence interval is confidence interval=(−5.66,4.26)_.

Explanation of Solution

Calculation: The confidence interval for the difference of body weight can be obtained by first calculating the margin of error. The margin of error is obtained as follow:

Margin of error=t*×sn=2.16×8.614=2.16×2.298=4.965

Now, the confidence interval can be obtained as follows:

Confidence interval=x¯±Margin of error= −0.7±4.96=(−5.66,4.26)

Interpretation: As the hypothesized mean of 0 lies inside the 95% confidence interval, the null hypothesis cannot be rejected, which states that difference of body weight is within the 95% confidence interval significantly.

Section 2:

To determine

To find: The 95% confidence interval for the difference of calorie intake.

Section 2:

Expert Solution

Answer to Problem 131E

Solution: The required 95% confidence interval is (−107.23,135.23)_.

Explanation of Solution

Calculation: The confidence interval for the difference of calorie intake can be obtained by first calculating the margin of error. The margin of error is obtained as follow:

Margin of error=t*×sn=2.16×21014=2.16×56.125=121.23

Now, the confidence interval can be obtained as follows:

Confidence interval=x¯±Margin of error=14±121.23(−107.23,135.23)

Interpretation: As the hypothesized mean of 0 lies inside the 95% confidence interval, the null hypothesis cannot be rejected, which states that difference of calorie intake is within the 95% confidence interval significantly.

(d)

To determine

The summary of the results.

(d)

Expert Solution

Answer to Problem 131E

Solution: The result cannot be generalized and the outcome of study will also have affected due to the violation of instruction by three subjects as the sample size is too small.

Explanation of Solution

Here, the study is biased for a specific city, the model cannot be generalized for a whole population, and there are three violations out of fourteen subjects. The effect could be a serious concern in the study due to small sample size.

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

Chapter 7, Problem 131E

(a)

Section 1:

To determine

To find: The difference in mean and standard error for body weight.

(a)

Section 1:

Expert Solution

Answer to Problem 131E

Solution: The difference in mean for body weight is −0.7_ and standard error is 2.29_.

Explanation of Solution

Calculation: The difference of mean can be calculated as follows:

x¯1−x¯2=0.4−1.1=−0.7

The standard error (SE) of difference of mean can be calculated as follows:

SE=sn=8.614=2.298447≈2.29

Section 2:

To determine

To find: The difference in mean and standard error for calorie intake.

Section 2:

Expert Solution

Answer to Problem 131E

Solution: The difference in mean for body weight is 14_ and standard error is 56.125_.

Explanation of Solution

Calculation:

The difference of mean can be calculated as follows:

x¯1−x¯2=2589−2575=14

The standard error (SE) of difference of mean can be calculated as follows:

SE=sn=21014=56.1248≈56.125

(b)

Section 1:

To determine

To test: The significant differences in body weight.

(b)

Section 1:

Expert Solution

Answer to Problem 131E

Solution: There is no significant difference in body weight.

The t-statistic for the test hypothesis is obtained as t-value =−0.305.

The P-statistic for the test hypothesis is obtained as P-value =0.76418.

Explanation of Solution

Calculation: The hypothesis for study is defined as

H0: The difference of body weight is = 0.

Ha: The difference of body weight is ≠0.

To test the hypothesis that there is no significant difference in body weight, t-statistic is used to determine the significance of the difference. The t-value is obtained as follows:

t=(x¯−μ)sn=(−0.7−0)8.614=−0.72.29845=−0.30455

≈0.305

Now, the P-value can be obtained by using the standard normal table for t=−0.305 and the P-value =0.76418 for two-tailed t-statistic.

Conclusion: The P-value for t-test is greater than 0.05. So, the null hypothesis cannot be rejected significantly, which states that there is no significant difference in body weight.

Section 2:

To determine

To test: The significant differences in calorie intake.

Section 2:

Expert Solution

Answer to Problem 131E

Solution: There is no significant difference in calorie intake.

The t-statistic for the test hypothesis is obtained as t-value =0.249.

The P-statistic for the test hypothesis is obtained as P-value =0.80258.

Explanation of Solution

Calculation: The hypothesis for study is defined as

H0: The difference of calorie intake is =0.

Ha: The difference of calorie intake is ≠0.

To test the hypothesis, t-statistic is used to determine the significance of the difference. The t- value is obtained as follow:

t=(x¯−μ)sn=(14−0)21014=1456.12486=0.24944

≈0.249

Now, the P-value can be obtained by using the standard normal table for t=0.24944. The P-value is 0.80258 for two-tailed t-statistic.

Conclusion: The P-value for t-test is greater than 0.05. So, the null hypothesis cannot be rejected significantly, which states that there is no significant difference in calorie intake.

(c)

Section 1:

To determine

To find: The 95% confidence interval for the difference of body weight.

(c)

Section 1:

Expert Solution

Answer to Problem 131E

Solution: The required 95% confidence interval is confidence interval=(−5.66,4.26)_.

Explanation of Solution

Calculation: The confidence interval for the difference of body weight can be obtained by first calculating the margin of error. The margin of error is obtained as follow:

Margin of error=t*×sn=2.16×8.614=2.16×2.298=4.965

Now, the confidence interval can be obtained as follows:

Confidence interval=x¯±Margin of error= −0.7±4.96=(−5.66,4.26)

Interpretation: As the hypothesized mean of 0 lies inside the 95% confidence interval, the null hypothesis cannot be rejected, which states that difference of body weight is within the 95% confidence interval significantly.

Section 2:

To determine

To find: The 95% confidence interval for the difference of calorie intake.

Section 2:

Expert Solution

Answer to Problem 131E

Solution: The required 95% confidence interval is (−107.23,135.23)_.

Explanation of Solution

Calculation: The confidence interval for the difference of calorie intake can be obtained by first calculating the margin of error. The margin of error is obtained as follow:

Margin of error=t*×sn=2.16×21014=2.16×56.125=121.23

Now, the confidence interval can be obtained as follows:

Confidence interval=x¯±Margin of error=14±121.23(−107.23,135.23)

Interpretation: As the hypothesized mean of 0 lies inside the 95% confidence interval, the null hypothesis cannot be rejected, which states that difference of calorie intake is within the 95% confidence interval significantly.

(d)

To determine

The summary of the results.

(d)

Expert Solution

Answer to Problem 131E

Solution: The result cannot be generalized and the outcome of study will also have affected due to the violation of instruction by three subjects as the sample size is too small.

Explanation of Solution

Here, the study is biased for a specific city, the model cannot be generalized for a whole population, and there are three violations out of fourteen subjects. The effect could be a serious concern in the study due to small sample size.

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 5

Chapter 7, Problem 131E

(a)

Section 1:

To determine

To find: The difference in mean and standard error for body weight.

(a)

Section 1:

Expert Solution

Answer to Problem 131E

Solution: The difference in mean for body weight is −0.7_ and standard error is 2.29_.

Explanation of Solution

Calculation: The difference of mean can be calculated as follows:

x¯1−x¯2=0.4−1.1=−0.7

The standard error (SE) of difference of mean can be calculated as follows:

SE=sn=8.614=2.298447≈2.29

Section 2:

To determine

To find: The difference in mean and standard error for calorie intake.

Section 2:

Expert Solution

Answer to Problem 131E

Solution: The difference in mean for body weight is 14_ and standard error is 56.125_.

Explanation of Solution

Calculation:

The difference of mean can be calculated as follows:

x¯1−x¯2=2589−2575=14

The standard error (SE) of difference of mean can be calculated as follows:

SE=sn=21014=56.1248≈56.125

(b)

Section 1:

To determine

To test: The significant differences in body weight.

(b)

Section 1:

Expert Solution

Answer to Problem 131E

Solution: There is no significant difference in body weight.

The t-statistic for the test hypothesis is obtained as t-value =−0.305.

The P-statistic for the test hypothesis is obtained as P-value =0.76418.

Explanation of Solution

Calculation: The hypothesis for study is defined as

H0: The difference of body weight is = 0.

Ha: The difference of body weight is ≠0.

To test the hypothesis that there is no significant difference in body weight, t-statistic is used to determine the significance of the difference. The t-value is obtained as follows:

t=(x¯−μ)sn=(−0.7−0)8.614=−0.72.29845=−0.30455

≈0.305

Now, the P-value can be obtained by using the standard normal table for t=−0.305 and the P-value =0.76418 for two-tailed t-statistic.

Conclusion: The P-value for t-test is greater than 0.05. So, the null hypothesis cannot be rejected significantly, which states that there is no significant difference in body weight.

Section 2:

To determine

To test: The significant differences in calorie intake.

Section 2:

Expert Solution

Answer to Problem 131E

Solution: There is no significant difference in calorie intake.

The t-statistic for the test hypothesis is obtained as t-value =0.249.

The P-statistic for the test hypothesis is obtained as P-value =0.80258.

Explanation of Solution

Calculation: The hypothesis for study is defined as

H0: The difference of calorie intake is =0.

Ha: The difference of calorie intake is ≠0.

To test the hypothesis, t-statistic is used to determine the significance of the difference. The t- value is obtained as follow:

t=(x¯−μ)sn=(14−0)21014=1456.12486=0.24944

≈0.249

Now, the P-value can be obtained by using the standard normal table for t=0.24944. The P-value is 0.80258 for two-tailed t-statistic.

Conclusion: The P-value for t-test is greater than 0.05. So, the null hypothesis cannot be rejected significantly, which states that there is no significant difference in calorie intake.

(c)

Section 1:

To determine

To find: The 95% confidence interval for the difference of body weight.

(c)

Section 1:

Expert Solution

Answer to Problem 131E

Solution: The required 95% confidence interval is confidence interval=(−5.66,4.26)_.

Explanation of Solution

Calculation: The confidence interval for the difference of body weight can be obtained by first calculating the margin of error. The margin of error is obtained as follow:

Margin of error=t*×sn=2.16×8.614=2.16×2.298=4.965

Now, the confidence interval can be obtained as follows:

Confidence interval=x¯±Margin of error= −0.7±4.96=(−5.66,4.26)

Interpretation: As the hypothesized mean of 0 lies inside the 95% confidence interval, the null hypothesis cannot be rejected, which states that difference of body weight is within the 95% confidence interval significantly.

Section 2:

To determine

To find: The 95% confidence interval for the difference of calorie intake.

Section 2:

Expert Solution

Answer to Problem 131E

Solution: The required 95% confidence interval is (−107.23,135.23)_.

Explanation of Solution

Calculation: The confidence interval for the difference of calorie intake can be obtained by first calculating the margin of error. The margin of error is obtained as follow:

Margin of error=t*×sn=2.16×21014=2.16×56.125=121.23

Now, the confidence interval can be obtained as follows:

Confidence interval=x¯±Margin of error=14±121.23(−107.23,135.23)

Interpretation: As the hypothesized mean of 0 lies inside the 95% confidence interval, the null hypothesis cannot be rejected, which states that difference of calorie intake is within the 95% confidence interval significantly.

(d)

To determine

The summary of the results.

(d)

Expert Solution

Answer to Problem 131E

Solution: The result cannot be generalized and the outcome of study will also have affected due to the violation of instruction by three subjects as the sample size is too small.

Explanation of Solution

Here, the study is biased for a specific city, the model cannot be generalized for a whole population, and there are three violations out of fourteen subjects. The effect could be a serious concern in the study due to small sample size.

Answer 6

Chapter 7, Problem 131E

(a)

Section 1:

To determine

To find: The difference in mean and standard error for body weight.

(a)

Section 1:

Expert Solution

Answer to Problem 131E

Solution: The difference in mean for body weight is −0.7_ and standard error is 2.29_.

Explanation of Solution

Calculation: The difference of mean can be calculated as follows:

x¯1−x¯2=0.4−1.1=−0.7

The standard error (SE) of difference of mean can be calculated as follows:

SE=sn=8.614=2.298447≈2.29

Answer 7

Chapter 7, Problem 131E

(a)

Section 1:

To determine

To find: The difference in mean and standard error for body weight.

Answer 8

(a)

Section 1:

Expert Solution

Answer to Problem 131E

Solution: The difference in mean for body weight is −0.7_ and standard error is 2.29_.

Answer 9

(a)

Section 1:

Expert Solution

Answer 10

(a)

Section 1:

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Calculation: The difference of mean can be calculated as follows:

x¯1−x¯2=0.4−1.1=−0.7

The standard error (SE) of difference of mean can be calculated as follows:

SE=sn=8.614=2.298447≈2.29

Answer 13

Section 2:

To determine

To find: The difference in mean and standard error for calorie intake.

Section 2:

Expert Solution

Answer to Problem 131E

Solution: The difference in mean for body weight is 14_ and standard error is 56.125_.

Explanation of Solution

Calculation:

The difference of mean can be calculated as follows:

x¯1−x¯2=2589−2575=14

The standard error (SE) of difference of mean can be calculated as follows:

SE=sn=21014=56.1248≈56.125

Answer 14

Section 2:

To determine

To find: The difference in mean and standard error for calorie intake.

Answer 15

Section 2:

Expert Solution

Answer to Problem 131E

Solution: The difference in mean for body weight is 14_ and standard error is 56.125_.

Answer 16

Section 2:

Expert Solution

Answer 17

Section 2:

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Calculation:

The difference of mean can be calculated as follows:

x¯1−x¯2=2589−2575=14

The standard error (SE) of difference of mean can be calculated as follows:

SE=sn=21014=56.1248≈56.125

Answer 20

(b)

Section 1:

To determine

To test: The significant differences in body weight.

(b)

Section 1:

Expert Solution

Answer to Problem 131E

Solution: There is no significant difference in body weight.

The t-statistic for the test hypothesis is obtained as t-value =−0.305.

The P-statistic for the test hypothesis is obtained as P-value =0.76418.

Explanation of Solution

Calculation: The hypothesis for study is defined as

H0: The difference of body weight is = 0.

Ha: The difference of body weight is ≠0.

To test the hypothesis that there is no significant difference in body weight, t-statistic is used to determine the significance of the difference. The t-value is obtained as follows:

t=(x¯−μ)sn=(−0.7−0)8.614=−0.72.29845=−0.30455

≈0.305

Now, the P-value can be obtained by using the standard normal table for t=−0.305 and the P-value =0.76418 for two-tailed t-statistic.

Conclusion: The P-value for t-test is greater than 0.05. So, the null hypothesis cannot be rejected significantly, which states that there is no significant difference in body weight.

Answer 21

(b)

Section 1:

To determine

To test: The significant differences in body weight.

Answer 22

(b)

Section 1:

Expert Solution

Answer to Problem 131E

Solution: There is no significant difference in body weight.

The t-statistic for the test hypothesis is obtained as t-value =−0.305.

The P-statistic for the test hypothesis is obtained as P-value =0.76418.

Answer 23

(b)

Section 1:

Expert Solution

Answer 24

(b)

Section 1:

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Calculation: The hypothesis for study is defined as

H0: The difference of body weight is = 0.

Ha: The difference of body weight is ≠0.

To test the hypothesis that there is no significant difference in body weight, t-statistic is used to determine the significance of the difference. The t-value is obtained as follows:

t=(x¯−μ)sn=(−0.7−0)8.614=−0.72.29845=−0.30455

≈0.305

Now, the P-value can be obtained by using the standard normal table for t=−0.305 and the P-value =0.76418 for two-tailed t-statistic.

Conclusion: The P-value for t-test is greater than 0.05. So, the null hypothesis cannot be rejected significantly, which states that there is no significant difference in body weight.

Answer 27

Section 2:

To determine

To test: The significant differences in calorie intake.

Section 2:

Expert Solution

Answer to Problem 131E

Solution: There is no significant difference in calorie intake.

The t-statistic for the test hypothesis is obtained as t-value =0.249.

The P-statistic for the test hypothesis is obtained as P-value =0.80258.

Explanation of Solution

Calculation: The hypothesis for study is defined as

H0: The difference of calorie intake is =0.

Ha: The difference of calorie intake is ≠0.

To test the hypothesis, t-statistic is used to determine the significance of the difference. The t- value is obtained as follow:

t=(x¯−μ)sn=(14−0)21014=1456.12486=0.24944

≈0.249

Now, the P-value can be obtained by using the standard normal table for t=0.24944. The P-value is 0.80258 for two-tailed t-statistic.

Conclusion: The P-value for t-test is greater than 0.05. So, the null hypothesis cannot be rejected significantly, which states that there is no significant difference in calorie intake.

Answer 28

Section 2:

To determine

To test: The significant differences in calorie intake.

Answer 29

Section 2:

Expert Solution

Answer to Problem 131E

Solution: There is no significant difference in calorie intake.

The t-statistic for the test hypothesis is obtained as t-value =0.249.

The P-statistic for the test hypothesis is obtained as P-value =0.80258.

Answer 30

Section 2:

Expert Solution

Answer 31

Section 2:

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Calculation: The hypothesis for study is defined as

H0: The difference of calorie intake is =0.

Ha: The difference of calorie intake is ≠0.

To test the hypothesis, t-statistic is used to determine the significance of the difference. The t- value is obtained as follow:

t=(x¯−μ)sn=(14−0)21014=1456.12486=0.24944

≈0.249

Now, the P-value can be obtained by using the standard normal table for t=0.24944. The P-value is 0.80258 for two-tailed t-statistic.

Conclusion: The P-value for t-test is greater than 0.05. So, the null hypothesis cannot be rejected significantly, which states that there is no significant difference in calorie intake.

Answer 34

(c)

Section 1:

To determine

To find: The 95% confidence interval for the difference of body weight.

(c)

Section 1:

Expert Solution

Answer to Problem 131E

Solution: The required 95% confidence interval is confidence interval=(−5.66,4.26)_.

Explanation of Solution

Calculation: The confidence interval for the difference of body weight can be obtained by first calculating the margin of error. The margin of error is obtained as follow:

Margin of error=t*×sn=2.16×8.614=2.16×2.298=4.965

Now, the confidence interval can be obtained as follows:

Confidence interval=x¯±Margin of error= −0.7±4.96=(−5.66,4.26)

Interpretation: As the hypothesized mean of 0 lies inside the 95% confidence interval, the null hypothesis cannot be rejected, which states that difference of body weight is within the 95% confidence interval significantly.

Answer 35

(c)

Section 1:

To determine

To find: The 95% confidence interval for the difference of body weight.

Answer 36

(c)

Section 1:

Expert Solution

Answer to Problem 131E

Solution: The required 95% confidence interval is confidence interval=(−5.66,4.26)_.

Answer 37

(c)

Section 1:

Expert Solution

Answer 38

(c)

Section 1:

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

Calculation: The confidence interval for the difference of body weight can be obtained by first calculating the margin of error. The margin of error is obtained as follow:

Margin of error=t*×sn=2.16×8.614=2.16×2.298=4.965

Now, the confidence interval can be obtained as follows:

Confidence interval=x¯±Margin of error= −0.7±4.96=(−5.66,4.26)

Interpretation: As the hypothesized mean of 0 lies inside the 95% confidence interval, the null hypothesis cannot be rejected, which states that difference of body weight is within the 95% confidence interval significantly.

Answer 41

Section 2:

To determine

To find: The 95% confidence interval for the difference of calorie intake.

Section 2:

Expert Solution

Answer to Problem 131E

Solution: The required 95% confidence interval is (−107.23,135.23)_.

Explanation of Solution

Calculation: The confidence interval for the difference of calorie intake can be obtained by first calculating the margin of error. The margin of error is obtained as follow:

Margin of error=t*×sn=2.16×21014=2.16×56.125=121.23

Now, the confidence interval can be obtained as follows:

Confidence interval=x¯±Margin of error=14±121.23(−107.23,135.23)

Interpretation: As the hypothesized mean of 0 lies inside the 95% confidence interval, the null hypothesis cannot be rejected, which states that difference of calorie intake is within the 95% confidence interval significantly.

Answer 42

Section 2:

To determine

To find: The 95% confidence interval for the difference of calorie intake.

Answer 43

Section 2:

Expert Solution

Answer to Problem 131E

Solution: The required 95% confidence interval is (−107.23,135.23)_.

Answer 44

Section 2:

Expert Solution

Answer 45

Section 2:

Expert Solution

Answer 46

Expert Solution

Answer 47

Explanation of Solution

Calculation: The confidence interval for the difference of calorie intake can be obtained by first calculating the margin of error. The margin of error is obtained as follow:

Margin of error=t*×sn=2.16×21014=2.16×56.125=121.23

Now, the confidence interval can be obtained as follows:

Confidence interval=x¯±Margin of error=14±121.23(−107.23,135.23)

Interpretation: As the hypothesized mean of 0 lies inside the 95% confidence interval, the null hypothesis cannot be rejected, which states that difference of calorie intake is within the 95% confidence interval significantly.

Answer 48

(d)

To determine

The summary of the results.

(d)

Expert Solution

Answer to Problem 131E

Solution: The result cannot be generalized and the outcome of study will also have affected due to the violation of instruction by three subjects as the sample size is too small.

Explanation of Solution

Here, the study is biased for a specific city, the model cannot be generalized for a whole population, and there are three violations out of fourteen subjects. The effect could be a serious concern in the study due to small sample size.

Answer 49

(d)

To determine

The summary of the results.

Answer 50

(d)

Expert Solution

Answer to Problem 131E

Solution: The result cannot be generalized and the outcome of study will also have affected due to the violation of instruction by three subjects as the sample size is too small.

Answer 51

(d)

Expert Solution

Answer 52

(d)

Expert Solution

Answer 53

Expert Solution

Answer 54

Explanation of Solution

Here, the study is biased for a specific city, the model cannot be generalized for a whole population, and there are three violations out of fourteen subjects. The effect could be a serious concern in the study due to small sample size.