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

Question

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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.

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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.

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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.

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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.

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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

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