Compute the differences in samples of matched pairs Sample-1– Sample-2.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 9.3, Problem 10E

a.

To determine

Compute the differences in samples of matched pairs Sample-1– Sample-2.

a.

Expert Solution

Answer to Problem 10E

The differences in samples of matched pairs Sample-1– Sample-2 is,

S.no	Difference
1	–6
2	–1
3	–9
4	–1
5	–5
6	–1
7	–4
8	–8
9	–2
10	–9

Explanation of Solution

Calculation:

The data represents the sample of a five matched pairs.

Let μd=μ1−μ2 where μ1 be the population mean-1 is and μ2 be the population mean-2.

The differences in samples of matched pairs Sample-1– Sample-2 is,

S.no	Sample-1	Sample-2	Difference = Sample1 − Sample2
1	28	34	28−34=−6
2	29	30	29−30=−1
3	22	31	22−31=−9
4	25	26	25−26=−1
5	26	31	26−31=−5
6	29	30	29−30=−1
7	27	31	27−31=−4
8	24	32	24−32=−8
9	27	29	27−29=−2
10	28	37	28−37=−9

b.

To determine

Find the value of t-test statistic.

b.

Expert Solution

Answer to Problem 10E

The value of test statistic is –4.3948.

Explanation of Solution

Calculation:

The test hypotheses are given below:

Null hypothesis:

H0:μd=0

That is, there is no significant difference between the population mean1 and population mean2.

Alternate hypothesis:

H1:μd≠0

That is, there is a significant difference between the population mean1 and population mean2.

Mean and standard deviation of differences:

Software procedure:

Step-by-step procedure to obtain the mean and standard deviation using the MINITAB software:

Choose Stat > Basic statistic > Display descriptive statistics.
In Variables, enter the column of Differences.
In Statistics, select mean, standard deviation and N total.
Click OK.

Output using the MINITAB software is given below:

Essential Statistics, Chapter 9.3, Problem 10E , additional homework tip 1

From the MINITAB output, the mean and standard deviation are –4.6 and 3.31.

The mean and standard deviation of the differences is –4.6 and 3.31.

The test statistic is obtained as follows,

t=d¯−0(sdn)=−4.6−0(3.3110)=−4.61.0467=−4.3948

Thus, the test statistic is –4.3948.

c.

To determine

Check whether the null hypothesis is rejected at α=0.05 level of significance.

c.

Expert Solution

Answer to Problem 10E

There is enough evidence to reject the null hypothesis H0 at α=0.05 level of significance.

Explanation of Solution

Degrees of freedom:

The degrees of freedom for the test statistic is,

d.f=n−1=10−1=9

Thus, the degree of freedom is 9.

P-value:

Software procedure:

Step-by-step procedure to obtain the P-value using the MINITAB software:

Choose Graph > Probability Distribution Plot.
Choose View Probability > OK.
From Distribution, choose ‘t’ distribution.
In Degrees of freedom, enter 9.
Click the Shaded Area tab.
Choose X value and Both Tail for the region of the curve to shade.
In X-value enter –4.3948.
Click OK.

Output using the MINITAB software is given below:

Essential Statistics, Chapter 9.3, Problem 10E , additional homework tip 2

From the MINITAB output, the P-value is 2×0.0008667=0.0017 .

Thus, the P-value is 0.0017.

Decision rule based on P-value:

If P−value≤α, then reject the null hypothesis H0.

If P−value>α, then fail to reject the null hypothesis H0.

Here, the level of significance is α=0.05.

Conclusion based on P-value approach:

The P-value is 0.0017 and α value is 0.05.

Here, P-value is less than the α value.

That is, 0.0017(=P−value)<0.05(=α).

By the rejection rule, reject the null hypothesis.

Thus, there is enough evidence to reject the null hypothesis H0.

d.

To determine

Check whether the null hypothesis is rejected at α=0.01 level of significance.

d.

Expert Solution

Answer to Problem 10E

There is enough evidence to reject the null hypothesis H0 at α=0.01 level of significance.

Explanation of Solution

From part (c), the P-value is 0.0017.

Decision rule based on P-value:

If P−value≤α, then reject the null hypothesis H0.

If P−value>α, then fail to reject the null hypothesis H0.

Here, the level of significance is α=0.01.

Conclusion based on P-value approach:

The P-value is 0.0017 and α value is 0.01.

Here, P-value is less than the α value.

That is, 0.0017(=P−value)<0.01(=α).

By the rejection rule, reject the null hypothesis.

Thus, there is enough evidence to reject the null hypothesis H0.

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

In order to find probability, you can use this formula in Microsoft Excel: The best way to understand and solve these problems is by first drawing a bell curve and marking key points such as x, the mean, and the areas of interest. Once marked on the bell curve, figure out what calculations are needed to find the area of interest. =NORM.DIST(x, Mean, Standard Dev., TRUE). When the question mentions “greater than” you may have to subtract your answer from 1. When the question mentions “between (two values)”, you need to do separate calculation for both values and then subtract their results to get the answer. 1. Compute the probability of a value between 44.0 and 55.0. (The question requires finding probability value between 44 and 55. Solve it in 3 steps. In the first step, use the above formula and x = 44, calculate probability value. In the second step repeat the first step with the only difference that x=55. In the third step, subtract the answer of the first part from the…

If a uniform distribution is defined over the interval from 6 to 10, then answer the followings: What is the mean of this uniform distribution? Show that the probability of any value between 6 and 10 is equal to 1.0 Find the probability of a value more than 7. Find the probability of a value between 7 and 9. The closing price of Schnur Sporting Goods Inc. common stock is uniformly distributed between $20 and $30 per share. What is the probability that the stock price will be: More than $27? Less than or equal to $24? The April rainfall in Flagstaff, Arizona, follows a uniform distribution between 0.5 and 3.00 inches. What is the mean amount of rainfall for the month? What is the probability of less than an inch of rain for the month? What is the probability of exactly 1.00 inch of rain? What is the probability of more than 1.50 inches of rain for the month? The best way to solve this problem is begin by creating a chart. Clearly mark the range, identifying the lower and upper…

Problem 1: The mean hourly pay of an American Airlines flight attendant is normally distributed with a mean of 40 per hour and a standard deviation of 3.00 per hour. What is the probability that the hourly pay of a randomly selected flight attendant is: Between the mean and $45 per hour? More than $45 per hour? Less than $32 per hour? Problem 2: The mean of a normal probability distribution is 400 pounds. The standard deviation is 10 pounds. What is the area between 415 pounds and the mean of 400 pounds? What is the area between the mean and 395 pounds? What is the probability of randomly selecting a value less than 395 pounds? Problem 3: In New York State, the mean salary for high school teachers in 2022 was 81,410 with a standard deviation of 9,500. Only Alaska’s mean salary was higher. Assume New York’s state salaries follow a normal distribution. What percent of New York State high school teachers earn between 70,000 and 75,000? What percent of New York State high school…

Answer 2

Question

Chapter 9.3, Problem 10E

a.

To determine

Compute the differences in samples of matched pairs Sample-1– Sample-2.

a.

Expert Solution

Answer to Problem 10E

The differences in samples of matched pairs Sample-1– Sample-2 is,

S.no	Difference
1	–6
2	–1
3	–9
4	–1
5	–5
6	–1
7	–4
8	–8
9	–2
10	–9

Explanation of Solution

Calculation:

The data represents the sample of a five matched pairs.

Let μd=μ1−μ2 where μ1 be the population mean-1 is and μ2 be the population mean-2.

The differences in samples of matched pairs Sample-1– Sample-2 is,

S.no	Sample-1	Sample-2	Difference = Sample1 − Sample2
1	28	34	28−34=−6
2	29	30	29−30=−1
3	22	31	22−31=−9
4	25	26	25−26=−1
5	26	31	26−31=−5
6	29	30	29−30=−1
7	27	31	27−31=−4
8	24	32	24−32=−8
9	27	29	27−29=−2
10	28	37	28−37=−9

b.

To determine

Find the value of t-test statistic.

b.

Expert Solution

Answer to Problem 10E

The value of test statistic is –4.3948.

Explanation of Solution

Calculation:

The test hypotheses are given below:

Null hypothesis:

H0:μd=0

That is, there is no significant difference between the population mean1 and population mean2.

Alternate hypothesis:

H1:μd≠0

That is, there is a significant difference between the population mean1 and population mean2.

Mean and standard deviation of differences:

Software procedure:

Step-by-step procedure to obtain the mean and standard deviation using the MINITAB software:

Choose Stat > Basic statistic > Display descriptive statistics.
In Variables, enter the column of Differences.
In Statistics, select mean, standard deviation and N total.
Click OK.

Output using the MINITAB software is given below:

Essential Statistics, Chapter 9.3, Problem 10E , additional homework tip 1

From the MINITAB output, the mean and standard deviation are –4.6 and 3.31.

The mean and standard deviation of the differences is –4.6 and 3.31.

The test statistic is obtained as follows,

t=d¯−0(sdn)=−4.6−0(3.3110)=−4.61.0467=−4.3948

Thus, the test statistic is –4.3948.

c.

To determine

Check whether the null hypothesis is rejected at α=0.05 level of significance.

c.

Expert Solution

Answer to Problem 10E

There is enough evidence to reject the null hypothesis H0 at α=0.05 level of significance.

Explanation of Solution

Degrees of freedom:

The degrees of freedom for the test statistic is,

d.f=n−1=10−1=9

Thus, the degree of freedom is 9.

P-value:

Software procedure:

Step-by-step procedure to obtain the P-value using the MINITAB software:

Choose Graph > Probability Distribution Plot.
Choose View Probability > OK.
From Distribution, choose ‘t’ distribution.
In Degrees of freedom, enter 9.
Click the Shaded Area tab.
Choose X value and Both Tail for the region of the curve to shade.
In X-value enter –4.3948.
Click OK.

Output using the MINITAB software is given below:

Essential Statistics, Chapter 9.3, Problem 10E , additional homework tip 2

From the MINITAB output, the P-value is 2×0.0008667=0.0017 .

Thus, the P-value is 0.0017.

Decision rule based on P-value:

If P−value≤α, then reject the null hypothesis H0.

If P−value>α, then fail to reject the null hypothesis H0.

Here, the level of significance is α=0.05.

Conclusion based on P-value approach:

The P-value is 0.0017 and α value is 0.05.

Here, P-value is less than the α value.

That is, 0.0017(=P−value)<0.05(=α).

By the rejection rule, reject the null hypothesis.

Thus, there is enough evidence to reject the null hypothesis H0.

d.

To determine

Check whether the null hypothesis is rejected at α=0.01 level of significance.

d.

Expert Solution

Answer to Problem 10E

There is enough evidence to reject the null hypothesis H0 at α=0.01 level of significance.

Explanation of Solution

From part (c), the P-value is 0.0017.

Decision rule based on P-value:

If P−value≤α, then reject the null hypothesis H0.

If P−value>α, then fail to reject the null hypothesis H0.

Here, the level of significance is α=0.01.

Conclusion based on P-value approach:

The P-value is 0.0017 and α value is 0.01.

Here, P-value is less than the α value.

That is, 0.0017(=P−value)<0.01(=α).

By the rejection rule, reject the null hypothesis.

Thus, there is enough evidence to reject the null hypothesis H0.

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 9.3, Problem 10E

a.

To determine

Compute the differences in samples of matched pairs Sample-1– Sample-2.

a.

Expert Solution

Answer to Problem 10E

The differences in samples of matched pairs Sample-1– Sample-2 is,

S.no	Difference
1	–6
2	–1
3	–9
4	–1
5	–5
6	–1
7	–4
8	–8
9	–2
10	–9

Explanation of Solution

Calculation:

The data represents the sample of a five matched pairs.

Let μd=μ1−μ2 where μ1 be the population mean-1 is and μ2 be the population mean-2.

The differences in samples of matched pairs Sample-1– Sample-2 is,

S.no	Sample-1	Sample-2	Difference = Sample1 − Sample2
1	28	34	28−34=−6
2	29	30	29−30=−1
3	22	31	22−31=−9
4	25	26	25−26=−1
5	26	31	26−31=−5
6	29	30	29−30=−1
7	27	31	27−31=−4
8	24	32	24−32=−8
9	27	29	27−29=−2
10	28	37	28−37=−9

b.

To determine

Find the value of t-test statistic.

b.

Expert Solution

Answer to Problem 10E

The value of test statistic is –4.3948.

Explanation of Solution

Calculation:

The test hypotheses are given below:

Null hypothesis:

H0:μd=0

That is, there is no significant difference between the population mean1 and population mean2.

Alternate hypothesis:

H1:μd≠0

That is, there is a significant difference between the population mean1 and population mean2.

Mean and standard deviation of differences:

Software procedure:

Step-by-step procedure to obtain the mean and standard deviation using the MINITAB software:

Choose Stat > Basic statistic > Display descriptive statistics.
In Variables, enter the column of Differences.
In Statistics, select mean, standard deviation and N total.
Click OK.

Output using the MINITAB software is given below:

Essential Statistics, Chapter 9.3, Problem 10E , additional homework tip 1

From the MINITAB output, the mean and standard deviation are –4.6 and 3.31.

The mean and standard deviation of the differences is –4.6 and 3.31.

The test statistic is obtained as follows,

t=d¯−0(sdn)=−4.6−0(3.3110)=−4.61.0467=−4.3948

Thus, the test statistic is –4.3948.

c.

To determine

Check whether the null hypothesis is rejected at α=0.05 level of significance.

c.

Expert Solution

Answer to Problem 10E

There is enough evidence to reject the null hypothesis H0 at α=0.05 level of significance.

Explanation of Solution

Degrees of freedom:

The degrees of freedom for the test statistic is,

d.f=n−1=10−1=9

Thus, the degree of freedom is 9.

P-value:

Software procedure:

Step-by-step procedure to obtain the P-value using the MINITAB software:

Choose Graph > Probability Distribution Plot.
Choose View Probability > OK.
From Distribution, choose ‘t’ distribution.
In Degrees of freedom, enter 9.
Click the Shaded Area tab.
Choose X value and Both Tail for the region of the curve to shade.
In X-value enter –4.3948.
Click OK.

Output using the MINITAB software is given below:

Essential Statistics, Chapter 9.3, Problem 10E , additional homework tip 2

From the MINITAB output, the P-value is 2×0.0008667=0.0017 .

Thus, the P-value is 0.0017.

Decision rule based on P-value:

If P−value≤α, then reject the null hypothesis H0.

If P−value>α, then fail to reject the null hypothesis H0.

Here, the level of significance is α=0.05.

Conclusion based on P-value approach:

The P-value is 0.0017 and α value is 0.05.

Here, P-value is less than the α value.

That is, 0.0017(=P−value)<0.05(=α).

By the rejection rule, reject the null hypothesis.

Thus, there is enough evidence to reject the null hypothesis H0.

d.

To determine

Check whether the null hypothesis is rejected at α=0.01 level of significance.

d.

Expert Solution

Answer to Problem 10E

There is enough evidence to reject the null hypothesis H0 at α=0.01 level of significance.

Explanation of Solution

From part (c), the P-value is 0.0017.

Decision rule based on P-value:

If P−value≤α, then reject the null hypothesis H0.

If P−value>α, then fail to reject the null hypothesis H0.

Here, the level of significance is α=0.01.

Conclusion based on P-value approach:

The P-value is 0.0017 and α value is 0.01.

Here, P-value is less than the α value.

That is, 0.0017(=P−value)<0.01(=α).

By the rejection rule, reject the null hypothesis.

Thus, there is enough evidence to reject the null hypothesis H0.

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 9.3, Problem 10E

a.

To determine

Compute the differences in samples of matched pairs Sample-1– Sample-2.

a.

Expert Solution

Answer to Problem 10E

The differences in samples of matched pairs Sample-1– Sample-2 is,

S.no	Difference
1	–6
2	–1
3	–9
4	–1
5	–5
6	–1
7	–4
8	–8
9	–2
10	–9

Explanation of Solution

Calculation:

The data represents the sample of a five matched pairs.

Let μd=μ1−μ2 where μ1 be the population mean-1 is and μ2 be the population mean-2.

The differences in samples of matched pairs Sample-1– Sample-2 is,

S.no	Sample-1	Sample-2	Difference = Sample1 − Sample2
1	28	34	28−34=−6
2	29	30	29−30=−1
3	22	31	22−31=−9
4	25	26	25−26=−1
5	26	31	26−31=−5
6	29	30	29−30=−1
7	27	31	27−31=−4
8	24	32	24−32=−8
9	27	29	27−29=−2
10	28	37	28−37=−9

b.

To determine

Find the value of t-test statistic.

b.

Expert Solution

Answer to Problem 10E

The value of test statistic is –4.3948.

Explanation of Solution

Calculation:

The test hypotheses are given below:

Null hypothesis:

H0:μd=0

That is, there is no significant difference between the population mean1 and population mean2.

Alternate hypothesis:

H1:μd≠0

That is, there is a significant difference between the population mean1 and population mean2.

Mean and standard deviation of differences:

Software procedure:

Step-by-step procedure to obtain the mean and standard deviation using the MINITAB software:

Choose Stat > Basic statistic > Display descriptive statistics.
In Variables, enter the column of Differences.
In Statistics, select mean, standard deviation and N total.
Click OK.

Output using the MINITAB software is given below:

Essential Statistics, Chapter 9.3, Problem 10E , additional homework tip 1

From the MINITAB output, the mean and standard deviation are –4.6 and 3.31.

The mean and standard deviation of the differences is –4.6 and 3.31.

The test statistic is obtained as follows,

t=d¯−0(sdn)=−4.6−0(3.3110)=−4.61.0467=−4.3948

Thus, the test statistic is –4.3948.

c.

To determine

Check whether the null hypothesis is rejected at α=0.05 level of significance.

c.

Expert Solution

Answer to Problem 10E

There is enough evidence to reject the null hypothesis H0 at α=0.05 level of significance.

Explanation of Solution

Degrees of freedom:

The degrees of freedom for the test statistic is,

d.f=n−1=10−1=9

Thus, the degree of freedom is 9.

P-value:

Software procedure:

Step-by-step procedure to obtain the P-value using the MINITAB software:

Choose Graph > Probability Distribution Plot.
Choose View Probability > OK.
From Distribution, choose ‘t’ distribution.
In Degrees of freedom, enter 9.
Click the Shaded Area tab.
Choose X value and Both Tail for the region of the curve to shade.
In X-value enter –4.3948.
Click OK.

Output using the MINITAB software is given below:

Essential Statistics, Chapter 9.3, Problem 10E , additional homework tip 2

From the MINITAB output, the P-value is 2×0.0008667=0.0017 .

Thus, the P-value is 0.0017.

Decision rule based on P-value:

If P−value≤α, then reject the null hypothesis H0.

If P−value>α, then fail to reject the null hypothesis H0.

Here, the level of significance is α=0.05.

Conclusion based on P-value approach:

The P-value is 0.0017 and α value is 0.05.

Here, P-value is less than the α value.

That is, 0.0017(=P−value)<0.05(=α).

By the rejection rule, reject the null hypothesis.

Thus, there is enough evidence to reject the null hypothesis H0.

d.

To determine

Check whether the null hypothesis is rejected at α=0.01 level of significance.

d.

Expert Solution

Answer to Problem 10E

There is enough evidence to reject the null hypothesis H0 at α=0.01 level of significance.

Explanation of Solution

From part (c), the P-value is 0.0017.

Decision rule based on P-value:

If P−value≤α, then reject the null hypothesis H0.

If P−value>α, then fail to reject the null hypothesis H0.

Here, the level of significance is α=0.01.

Conclusion based on P-value approach:

The P-value is 0.0017 and α value is 0.01.

Here, P-value is less than the α value.

That is, 0.0017(=P−value)<0.01(=α).

By the rejection rule, reject the null hypothesis.

Thus, there is enough evidence to reject the null hypothesis H0.

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 9.3, Problem 10E

a.

To determine

Compute the differences in samples of matched pairs Sample-1– Sample-2.

a.

Expert Solution

Answer to Problem 10E

The differences in samples of matched pairs Sample-1– Sample-2 is,

S.no	Difference
1	–6
2	–1
3	–9
4	–1
5	–5
6	–1
7	–4
8	–8
9	–2
10	–9

Explanation of Solution

Calculation:

The data represents the sample of a five matched pairs.

Let μd=μ1−μ2 where μ1 be the population mean-1 is and μ2 be the population mean-2.

The differences in samples of matched pairs Sample-1– Sample-2 is,

S.no	Sample-1	Sample-2	Difference = Sample1 − Sample2
1	28	34	28−34=−6
2	29	30	29−30=−1
3	22	31	22−31=−9
4	25	26	25−26=−1
5	26	31	26−31=−5
6	29	30	29−30=−1
7	27	31	27−31=−4
8	24	32	24−32=−8
9	27	29	27−29=−2
10	28	37	28−37=−9

b.

To determine

Find the value of t-test statistic.

b.

Expert Solution

Answer to Problem 10E

The value of test statistic is –4.3948.

Explanation of Solution

Calculation:

The test hypotheses are given below:

Null hypothesis:

H0:μd=0

That is, there is no significant difference between the population mean1 and population mean2.

Alternate hypothesis:

H1:μd≠0

That is, there is a significant difference between the population mean1 and population mean2.

Mean and standard deviation of differences:

Software procedure:

Step-by-step procedure to obtain the mean and standard deviation using the MINITAB software:

Choose Stat > Basic statistic > Display descriptive statistics.
In Variables, enter the column of Differences.
In Statistics, select mean, standard deviation and N total.
Click OK.

Output using the MINITAB software is given below:

Essential Statistics, Chapter 9.3, Problem 10E , additional homework tip 1

From the MINITAB output, the mean and standard deviation are –4.6 and 3.31.

The mean and standard deviation of the differences is –4.6 and 3.31.

The test statistic is obtained as follows,

t=d¯−0(sdn)=−4.6−0(3.3110)=−4.61.0467=−4.3948

Thus, the test statistic is –4.3948.

c.

To determine

Check whether the null hypothesis is rejected at α=0.05 level of significance.

c.

Expert Solution

Answer to Problem 10E

There is enough evidence to reject the null hypothesis H0 at α=0.05 level of significance.

Explanation of Solution

Degrees of freedom:

The degrees of freedom for the test statistic is,

d.f=n−1=10−1=9

Thus, the degree of freedom is 9.

P-value:

Software procedure:

Step-by-step procedure to obtain the P-value using the MINITAB software:

Choose Graph > Probability Distribution Plot.
Choose View Probability > OK.
From Distribution, choose ‘t’ distribution.
In Degrees of freedom, enter 9.
Click the Shaded Area tab.
Choose X value and Both Tail for the region of the curve to shade.
In X-value enter –4.3948.
Click OK.

Output using the MINITAB software is given below:

Essential Statistics, Chapter 9.3, Problem 10E , additional homework tip 2

From the MINITAB output, the P-value is 2×0.0008667=0.0017 .

Thus, the P-value is 0.0017.

Decision rule based on P-value:

If P−value≤α, then reject the null hypothesis H0.

If P−value>α, then fail to reject the null hypothesis H0.

Here, the level of significance is α=0.05.

Conclusion based on P-value approach:

The P-value is 0.0017 and α value is 0.05.

Here, P-value is less than the α value.

That is, 0.0017(=P−value)<0.05(=α).

By the rejection rule, reject the null hypothesis.

Thus, there is enough evidence to reject the null hypothesis H0.

d.

To determine

Check whether the null hypothesis is rejected at α=0.01 level of significance.

d.

Expert Solution

Answer to Problem 10E

There is enough evidence to reject the null hypothesis H0 at α=0.01 level of significance.

Explanation of Solution

From part (c), the P-value is 0.0017.

Decision rule based on P-value:

If P−value≤α, then reject the null hypothesis H0.

If P−value>α, then fail to reject the null hypothesis H0.

Here, the level of significance is α=0.01.

Conclusion based on P-value approach:

The P-value is 0.0017 and α value is 0.01.

Here, P-value is less than the α value.

That is, 0.0017(=P−value)<0.01(=α).

By the rejection rule, reject the null hypothesis.

Thus, there is enough evidence to reject the null hypothesis H0.

Answer 6

Chapter 9.3, Problem 10E

a.

To determine

Compute the differences in samples of matched pairs Sample-1– Sample-2.

a.

Expert Solution

Answer to Problem 10E

The differences in samples of matched pairs Sample-1– Sample-2 is,

S.no	Difference
1	–6
2	–1
3	–9
4	–1
5	–5
6	–1
7	–4
8	–8
9	–2
10	–9

Explanation of Solution

Calculation:

The data represents the sample of a five matched pairs.

Let μd=μ1−μ2 where μ1 be the population mean-1 is and μ2 be the population mean-2.

The differences in samples of matched pairs Sample-1– Sample-2 is,

S.no	Sample-1	Sample-2	Difference = Sample1 − Sample2
1	28	34	28−34=−6
2	29	30	29−30=−1
3	22	31	22−31=−9
4	25	26	25−26=−1
5	26	31	26−31=−5
6	29	30	29−30=−1
7	27	31	27−31=−4
8	24	32	24−32=−8
9	27	29	27−29=−2
10	28	37	28−37=−9

Answer 7

Chapter 9.3, Problem 10E

a.

To determine

Compute the differences in samples of matched pairs Sample-1– Sample-2.

Answer 8

a.

Expert Solution

Answer to Problem 10E

The differences in samples of matched pairs Sample-1– Sample-2 is,

S.no	Difference
1	–6
2	–1
3	–9
4	–1
5	–5
6	–1
7	–4
8	–8
9	–2
10	–9

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Calculation:

The data represents the sample of a five matched pairs.

Let μd=μ1−μ2 where μ1 be the population mean-1 is and μ2 be the population mean-2.

The differences in samples of matched pairs Sample-1– Sample-2 is,

S.no	Sample-1	Sample-2	Difference = Sample1 − Sample2
1	28	34	28−34=−6
2	29	30	29−30=−1
3	22	31	22−31=−9
4	25	26	25−26=−1
5	26	31	26−31=−5
6	29	30	29−30=−1
7	27	31	27−31=−4
8	24	32	24−32=−8
9	27	29	27−29=−2
10	28	37	28−37=−9

Answer 13

b.

To determine

Find the value of t-test statistic.

b.

Expert Solution

Answer to Problem 10E

The value of test statistic is –4.3948.

Explanation of Solution

Calculation:

The test hypotheses are given below:

Null hypothesis:

H0:μd=0

That is, there is no significant difference between the population mean1 and population mean2.

Alternate hypothesis:

H1:μd≠0

That is, there is a significant difference between the population mean1 and population mean2.

Mean and standard deviation of differences:

Software procedure:

Step-by-step procedure to obtain the mean and standard deviation using the MINITAB software:

Choose Stat > Basic statistic > Display descriptive statistics.
In Variables, enter the column of Differences.
In Statistics, select mean, standard deviation and N total.
Click OK.

Output using the MINITAB software is given below:

Essential Statistics, Chapter 9.3, Problem 10E , additional homework tip 1

From the MINITAB output, the mean and standard deviation are –4.6 and 3.31.

The mean and standard deviation of the differences is –4.6 and 3.31.

The test statistic is obtained as follows,

t=d¯−0(sdn)=−4.6−0(3.3110)=−4.61.0467=−4.3948

Thus, the test statistic is –4.3948.

Answer 14

b.

To determine

Find the value of t-test statistic.

Answer 15

b.

Expert Solution

Answer to Problem 10E

The value of test statistic is –4.3948.

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Calculation:

The test hypotheses are given below:

Null hypothesis:

H0:μd=0

That is, there is no significant difference between the population mean1 and population mean2.

Alternate hypothesis:

H1:μd≠0

That is, there is a significant difference between the population mean1 and population mean2.

Mean and standard deviation of differences:

Software procedure:

Step-by-step procedure to obtain the mean and standard deviation using the MINITAB software:

Choose Stat > Basic statistic > Display descriptive statistics.
In Variables, enter the column of Differences.
In Statistics, select mean, standard deviation and N total.
Click OK.

Output using the MINITAB software is given below:

Essential Statistics, Chapter 9.3, Problem 10E , additional homework tip 1

From the MINITAB output, the mean and standard deviation are –4.6 and 3.31.

The mean and standard deviation of the differences is –4.6 and 3.31.

The test statistic is obtained as follows,

t=d¯−0(sdn)=−4.6−0(3.3110)=−4.61.0467=−4.3948

Thus, the test statistic is –4.3948.

Answer 20

c.

To determine

Check whether the null hypothesis is rejected at α=0.05 level of significance.

c.

Expert Solution

Answer to Problem 10E

There is enough evidence to reject the null hypothesis H0 at α=0.05 level of significance.

Explanation of Solution

Degrees of freedom:

The degrees of freedom for the test statistic is,

d.f=n−1=10−1=9

Thus, the degree of freedom is 9.

P-value:

Software procedure:

Step-by-step procedure to obtain the P-value using the MINITAB software:

Choose Graph > Probability Distribution Plot.
Choose View Probability > OK.
From Distribution, choose ‘t’ distribution.
In Degrees of freedom, enter 9.
Click the Shaded Area tab.
Choose X value and Both Tail for the region of the curve to shade.
In X-value enter –4.3948.
Click OK.

Output using the MINITAB software is given below:

Essential Statistics, Chapter 9.3, Problem 10E , additional homework tip 2

From the MINITAB output, the P-value is 2×0.0008667=0.0017 .

Thus, the P-value is 0.0017.

Decision rule based on P-value:

If P−value≤α, then reject the null hypothesis H0.

If P−value>α, then fail to reject the null hypothesis H0.

Here, the level of significance is α=0.05.

Conclusion based on P-value approach:

The P-value is 0.0017 and α value is 0.05.

Here, P-value is less than the α value.

That is, 0.0017(=P−value)<0.05(=α).

By the rejection rule, reject the null hypothesis.

Thus, there is enough evidence to reject the null hypothesis H0.

Answer 21

c.

To determine

Check whether the null hypothesis is rejected at α=0.05 level of significance.

Answer 22

c.

Expert Solution

Answer to Problem 10E

There is enough evidence to reject the null hypothesis H0 at α=0.05 level of significance.

Answer 23

c.

Expert Solution

Answer 24

c.

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Degrees of freedom:

The degrees of freedom for the test statistic is,

d.f=n−1=10−1=9

Thus, the degree of freedom is 9.

P-value:

Software procedure:

Step-by-step procedure to obtain the P-value using the MINITAB software:

Choose Graph > Probability Distribution Plot.
Choose View Probability > OK.
From Distribution, choose ‘t’ distribution.
In Degrees of freedom, enter 9.
Click the Shaded Area tab.
Choose X value and Both Tail for the region of the curve to shade.
In X-value enter –4.3948.
Click OK.

Output using the MINITAB software is given below:

Essential Statistics, Chapter 9.3, Problem 10E , additional homework tip 2

From the MINITAB output, the P-value is 2×0.0008667=0.0017 .

Thus, the P-value is 0.0017.

Decision rule based on P-value:

If P−value≤α, then reject the null hypothesis H0.

If P−value>α, then fail to reject the null hypothesis H0.

Here, the level of significance is α=0.05.

Conclusion based on P-value approach:

The P-value is 0.0017 and α value is 0.05.

Here, P-value is less than the α value.

That is, 0.0017(=P−value)<0.05(=α).

By the rejection rule, reject the null hypothesis.

Thus, there is enough evidence to reject the null hypothesis H0.

Answer 27

d.

To determine

Check whether the null hypothesis is rejected at α=0.01 level of significance.

d.

Expert Solution

Answer to Problem 10E

There is enough evidence to reject the null hypothesis H0 at α=0.01 level of significance.

Explanation of Solution

From part (c), the P-value is 0.0017.

Decision rule based on P-value:

If P−value≤α, then reject the null hypothesis H0.

If P−value>α, then fail to reject the null hypothesis H0.

Here, the level of significance is α=0.01.

Conclusion based on P-value approach:

The P-value is 0.0017 and α value is 0.01.

Here, P-value is less than the α value.

That is, 0.0017(=P−value)<0.01(=α).

By the rejection rule, reject the null hypothesis.

Thus, there is enough evidence to reject the null hypothesis H0.

Answer 28

d.

To determine

Check whether the null hypothesis is rejected at α=0.01 level of significance.

Answer 29

d.

Expert Solution

Answer to Problem 10E

There is enough evidence to reject the null hypothesis H0 at α=0.01 level of significance.

Answer 30

d.

Expert Solution

Answer 31

d.

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

From part (c), the P-value is 0.0017.

Decision rule based on P-value:

If P−value≤α, then reject the null hypothesis H0.

If P−value>α, then fail to reject the null hypothesis H0.

Here, the level of significance is α=0.01.

Conclusion based on P-value approach:

The P-value is 0.0017 and α value is 0.01.

Here, P-value is less than the α value.

That is, 0.0017(=P−value)<0.01(=α).

By the rejection rule, reject the null hypothesis.

Thus, there is enough evidence to reject the null hypothesis H0.

Answer 34

Ch. 9.1 - Prob. 1CYU Ch. 9.1 - Prob. 2CYU Ch. 9.1 - Prob. 3CYU Ch. 9.1 - Prob. 4CYU Ch. 9.1 - Prob. 5CYU Ch. 9.1 - Prob. 6CYU Ch. 9.1 - Prob. 7E Ch. 9.1 - Prob. 8E Ch. 9.1 - Prob. 9E Ch. 9.1 - Prob. 10E

Compute the differences in samples of matched pairs Sample-1– Sample-2.

Compute the differences in samples of matched pairs Sample-1– Sample-2.

Answer to Problem 10E

Explanation of Solution

Find the value of t-test statistic.

Answer to Problem 10E

Explanation of Solution

Check whether the null hypothesis is rejected at α=0.05 level of significance.

Answer to Problem 10E

Explanation of Solution

Check whether the null hypothesis is rejected at α=0.01 level of significance.

Answer to Problem 10E

Explanation of Solution

Want to see more full solutions like this?

Chapter 9 Solutions