The article “Reaction Modeling and Optimization Using Neural Networks and Genetic Algorithms: Case Study Involving TS-1-Catalyzed Hydroxylation of Benzene” (S. Nandi, P. Mukherjee, et al., Industrial and Engineering Chemistry Research , 2002:2159–2169) presents benzene conversions (in mole percent) for 24 different benzenehydroxylation reactions. The results are 52.3 41.1 28.8 67.8 78.6 72.3 9.1 19.0 30.3 41.0 63.0 80.8 26.8 37.3 38.1 33.6 14.3 30.1 33.4 36.2 34.6 40.0 81.2 59.4. a. Can you conclude that the mean conversion is less than 45? Compute the appropriate test statistic and find the P -value. b. Can you conclude that the mean conversion is greater than 30? Compute the appropriate test statistic and find the P -value. c. Can you conclude that the mean conversion differs from 55? Compute the appropriate test statistic and find the P -value.

Question

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Answer 1

Textbook Question

Chapter 6.9, Problem 3E

The article “Reaction Modeling and Optimization Using Neural Networks and Genetic Algorithms: Case Study Involving TS-1-Catalyzed Hydroxylation of Benzene” (S. Nandi, P. Mukherjee, et al., Industrial and Engineering Chemistry Research, 2002:2159–2169) presents benzene conversions (in mole percent) for 24 different benzenehydroxylation reactions. The results are

52.3 41.1 28.8 67.8 78.6 72.3 9.1 19.0

30.3 41.0 63.0 80.8 26.8 37.3 38.1 33.6

14.3 30.1 33.4 36.2 34.6 40.0 81.2 59.4.

a. Can you conclude that the mean conversion is less than 45? Compute the appropriate test statistic and find the P-value.
b. Can you conclude that the mean conversion is greater than 30? Compute the appropriate test statistic and find the P-value.
c. Can you conclude that the mean conversion differs from 55? Compute the appropriate test statistic and find the P-value.

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

a.

Expert Solution

To determine

Check whether there is evidence to conclude the mean conversion is less than 45.

Answer to Problem 3E

There is no evidence to conclude that the mean conversion is less than 45.

Explanation of Solution

Given info:

The data represents the benzene conversions (in mole percent) for 24 different benzene hydroxylation reactions.

Calculation:

State the test hypotheses.

Null hypothesis:

H0:μ≥45

Alternative hypothesis:

H1:μ<45

Here, the sample size is large. That is, n = 24.

The formula for z-score is,

z=S+−n(n+1)4n(n+1)(2n+1)24

The positive and negative ranks are calculated as follows:

x	x−45	Signed ranks
52.3	7.3	5
41.1	–3.9	–1
28.8	–16.2	–14
67.8	22.8	17
78.6	33.6	21
72.3	27.3	19
9.1	–35.9	–23
19	–26	–18
30.3	–14.7	–12
41	–4	–2
63	18	15
80.8	35.8	22
26.8	–18.2	–16
37.3	–7.7	–6
38.1	–6.9	–4
33.6	–11.4	–9
14.3	–30.7	–20
30.1	–14.9	–13
33.4	–11.6	–10
36.2	–8.8	–7
34.6	–10.4	–8
40	–5	–3
81.2	36.2	24
59.4	14.4	11

From the table, the sum of positive ranks is,

S+=5+17+21+19+15+22+24+11=134

The test statistic is calculated as follows:

z=134−24(24+1)424(24+1)(2(24)+1)24=134−1501,225=−1635=−0.46

Thus, the test statistic is –0.46.

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 ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Left Tail for the region of the curve to shade.
Enter the data value as –0.46.
Click OK.

Output using the MINITAB software is given below:

Statistics for Engineers and Scientists, Chapter 6.9, Problem 3E , additional homework tip 1

From the MINITAB output, the P-value is 0.3228.

Decision rule:

If P-value≤α, then reject the null hypothesis (H0).

If P-value>α, then fail to reject the null hypothesis (H0).

Conclusion:

Here, the P-value is greater than the level of significance, 0.05.

Therefore, the null hypothesis is not rejected.

Hence, there is no evidence to conclude that the mean conversion is less than 45.

b.

Expert Solution

To determine

Check whether there is evidence to conclude that the mean conversion is greater than 30.

Answer to Problem 3E

There is evidence to conclude that the mean conversion is greater than 30.

Explanation of Solution

Calculation:

State the test hypotheses.

Null hypothesis:

H0:μ≤30

Alternative hypothesis:

H1:μ>30

The positive and negative ranks are calculated as follows:

x	x−45	Signed ranks
52.3	22.3	17
41.1	11.1	14
28.8	–1.2	–3
67.8	37.8	20
78.6	48.6	22
72.3	42.3	21
9.1	–20.9	–16
19	–11	–12.5
30.3	0.3	2
41	11	12.5
63	33	19
80.8	50.8	23
26.8	–3.2	–4
37.3	7.3	9
38.1	8.1	10
33.6	3.6	6
14.3	–15.7	–15
30.1	0.1	1
33.4	3.4	5
36.2	6.2	8
34.6	4.6	7
40	10	11
81.2	51.2	24
59.4	29.4	18

From the table, the sum of positive ranks is,

S+=17+14+20+22+21+2+12.5+19+23+9+10+6+1+5+8+7+11+24+18=249.5

The test statistic is calculated as follows:

z=249.5−24(24+1)424(24+1)(2(24)+1)24=249.5−1501,225=99.535=2.84

Thus, the test statistic is 2.84.

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 ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Right Tail for the region of the curve to shade.
Enter the data value as 2.84.
Click OK.

Output using the MINITAB software is given below:

Statistics for Engineers and Scientists, Chapter 6.9, Problem 3E , additional homework tip 2

From the MINITAB output, the P-value is 0.0023.

Conclusion:

Here, the P-value is less than the level of significance, 0.05.

Therefore, the null hypothesis is rejected.

Hence, there is evidence to conclude that the mean conversion is greater than 30.

c.

Expert Solution

To determine

Check whether there is evidence to conclude that the mean conversion differs from 55.

Answer to Problem 3E

There is evidence to conclude that the mean conversion differs from 55.

Explanation of Solution

Calculation:

State the test hypotheses.

Null hypothesis:

H0:μ=55

Alternative hypothesis:

H1:μ≠55

The positive and negative ranks are calculated as follows:

x	x−45	Signed ranks
52.3	–2.7	–1
41.1	–13.9	–5
28.8	–26.2	–19.5
67.8	12.8	4
78.6	23.6	15
72.3	17.3	9
9.1	–45.9	–24
19	–36	–22
30.3	–24.7	–16
41	–14	–6
63	8	3
80.8	25.8	18
26.8	–28.2	–21
37.3	–17.7	–10
38.1	–16.9	–8
33.6	–21.4	–13
14.3	–40.7	–23
30.1	–24.9	–17
33.4	–21.6	–14
36.2	–18.8	–11
34.6	–20.4	–12
40	–15	–7
81.2	26.2	19.5
59.4	4.4	2

From the table, the sum of positive ranks is,

S+=4+15+9+3+18+19.5+2=70.5

The test statistic is calculated as follows:

z=70.5−24(24+1)424(24+1)(2(24)+1)24=70.5−1501,225=−79.535=−2.27

Thus, the test statistic is –2.27.

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 ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Both Tail for the region of the curve to shade.
Enter the data value as –2.27.
Click OK.

Output using the MINITAB software is given below:

Statistics for Engineers and Scientists, Chapter 6.9, Problem 3E , additional homework tip 3

From the MINITAB output, the P-value for two tailed test is, 0.0116+0.0116=0.0232.

Conclusion:

Here, the P-value is less than the level of significance, 0.05.

Therefore, the null hypothesis is rejected.

Hence, there is evidence to conclude that the mean conversion differs from 55.

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Answer 2

Textbook Question

Chapter 6.9, Problem 3E

The article “Reaction Modeling and Optimization Using Neural Networks and Genetic Algorithms: Case Study Involving TS-1-Catalyzed Hydroxylation of Benzene” (S. Nandi, P. Mukherjee, et al., Industrial and Engineering Chemistry Research, 2002:2159–2169) presents benzene conversions (in mole percent) for 24 different benzenehydroxylation reactions. The results are

52.3 41.1 28.8 67.8 78.6 72.3 9.1 19.0

30.3 41.0 63.0 80.8 26.8 37.3 38.1 33.6

14.3 30.1 33.4 36.2 34.6 40.0 81.2 59.4.

a. Can you conclude that the mean conversion is less than 45? Compute the appropriate test statistic and find the P-value.
b. Can you conclude that the mean conversion is greater than 30? Compute the appropriate test statistic and find the P-value.
c. Can you conclude that the mean conversion differs from 55? Compute the appropriate test statistic and find the P-value.

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

a.

Expert Solution

To determine

Check whether there is evidence to conclude the mean conversion is less than 45.

Answer to Problem 3E

There is no evidence to conclude that the mean conversion is less than 45.

Explanation of Solution

Given info:

The data represents the benzene conversions (in mole percent) for 24 different benzene hydroxylation reactions.

Calculation:

State the test hypotheses.

Null hypothesis:

H0:μ≥45

Alternative hypothesis:

H1:μ<45

Here, the sample size is large. That is, n = 24.

The formula for z-score is,

z=S+−n(n+1)4n(n+1)(2n+1)24

The positive and negative ranks are calculated as follows:

x	x−45	Signed ranks
52.3	7.3	5
41.1	–3.9	–1
28.8	–16.2	–14
67.8	22.8	17
78.6	33.6	21
72.3	27.3	19
9.1	–35.9	–23
19	–26	–18
30.3	–14.7	–12
41	–4	–2
63	18	15
80.8	35.8	22
26.8	–18.2	–16
37.3	–7.7	–6
38.1	–6.9	–4
33.6	–11.4	–9
14.3	–30.7	–20
30.1	–14.9	–13
33.4	–11.6	–10
36.2	–8.8	–7
34.6	–10.4	–8
40	–5	–3
81.2	36.2	24
59.4	14.4	11

From the table, the sum of positive ranks is,

S+=5+17+21+19+15+22+24+11=134

The test statistic is calculated as follows:

z=134−24(24+1)424(24+1)(2(24)+1)24=134−1501,225=−1635=−0.46

Thus, the test statistic is –0.46.

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 ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Left Tail for the region of the curve to shade.
Enter the data value as –0.46.
Click OK.

Output using the MINITAB software is given below:

Statistics for Engineers and Scientists, Chapter 6.9, Problem 3E , additional homework tip 1

From the MINITAB output, the P-value is 0.3228.

Decision rule:

If P-value≤α, then reject the null hypothesis (H0).

If P-value>α, then fail to reject the null hypothesis (H0).

Conclusion:

Here, the P-value is greater than the level of significance, 0.05.

Therefore, the null hypothesis is not rejected.

Hence, there is no evidence to conclude that the mean conversion is less than 45.

b.

Expert Solution

To determine

Check whether there is evidence to conclude that the mean conversion is greater than 30.

Answer to Problem 3E

There is evidence to conclude that the mean conversion is greater than 30.

Explanation of Solution

Calculation:

State the test hypotheses.

Null hypothesis:

H0:μ≤30

Alternative hypothesis:

H1:μ>30

The positive and negative ranks are calculated as follows:

x	x−45	Signed ranks
52.3	22.3	17
41.1	11.1	14
28.8	–1.2	–3
67.8	37.8	20
78.6	48.6	22
72.3	42.3	21
9.1	–20.9	–16
19	–11	–12.5
30.3	0.3	2
41	11	12.5
63	33	19
80.8	50.8	23
26.8	–3.2	–4
37.3	7.3	9
38.1	8.1	10
33.6	3.6	6
14.3	–15.7	–15
30.1	0.1	1
33.4	3.4	5
36.2	6.2	8
34.6	4.6	7
40	10	11
81.2	51.2	24
59.4	29.4	18

From the table, the sum of positive ranks is,

S+=17+14+20+22+21+2+12.5+19+23+9+10+6+1+5+8+7+11+24+18=249.5

The test statistic is calculated as follows:

z=249.5−24(24+1)424(24+1)(2(24)+1)24=249.5−1501,225=99.535=2.84

Thus, the test statistic is 2.84.

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 ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Right Tail for the region of the curve to shade.
Enter the data value as 2.84.
Click OK.

Output using the MINITAB software is given below:

Statistics for Engineers and Scientists, Chapter 6.9, Problem 3E , additional homework tip 2

From the MINITAB output, the P-value is 0.0023.

Conclusion:

Here, the P-value is less than the level of significance, 0.05.

Therefore, the null hypothesis is rejected.

Hence, there is evidence to conclude that the mean conversion is greater than 30.

c.

Expert Solution

To determine

Check whether there is evidence to conclude that the mean conversion differs from 55.

Answer to Problem 3E

There is evidence to conclude that the mean conversion differs from 55.

Explanation of Solution

Calculation:

State the test hypotheses.

Null hypothesis:

H0:μ=55

Alternative hypothesis:

H1:μ≠55

The positive and negative ranks are calculated as follows:

x	x−45	Signed ranks
52.3	–2.7	–1
41.1	–13.9	–5
28.8	–26.2	–19.5
67.8	12.8	4
78.6	23.6	15
72.3	17.3	9
9.1	–45.9	–24
19	–36	–22
30.3	–24.7	–16
41	–14	–6
63	8	3
80.8	25.8	18
26.8	–28.2	–21
37.3	–17.7	–10
38.1	–16.9	–8
33.6	–21.4	–13
14.3	–40.7	–23
30.1	–24.9	–17
33.4	–21.6	–14
36.2	–18.8	–11
34.6	–20.4	–12
40	–15	–7
81.2	26.2	19.5
59.4	4.4	2

From the table, the sum of positive ranks is,

S+=4+15+9+3+18+19.5+2=70.5

The test statistic is calculated as follows:

z=70.5−24(24+1)424(24+1)(2(24)+1)24=70.5−1501,225=−79.535=−2.27

Thus, the test statistic is –2.27.

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 ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Both Tail for the region of the curve to shade.
Enter the data value as –2.27.
Click OK.

Output using the MINITAB software is given below:

Statistics for Engineers and Scientists, Chapter 6.9, Problem 3E , additional homework tip 3

From the MINITAB output, the P-value for two tailed test is, 0.0116+0.0116=0.0232.

Conclusion:

Here, the P-value is less than the level of significance, 0.05.

Therefore, the null hypothesis is rejected.

Hence, there is evidence to conclude that the mean conversion differs from 55.

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Answer 3

Textbook Question

Chapter 6.9, Problem 3E

The article “Reaction Modeling and Optimization Using Neural Networks and Genetic Algorithms: Case Study Involving TS-1-Catalyzed Hydroxylation of Benzene” (S. Nandi, P. Mukherjee, et al., Industrial and Engineering Chemistry Research, 2002:2159–2169) presents benzene conversions (in mole percent) for 24 different benzenehydroxylation reactions. The results are

52.3 41.1 28.8 67.8 78.6 72.3 9.1 19.0

30.3 41.0 63.0 80.8 26.8 37.3 38.1 33.6

14.3 30.1 33.4 36.2 34.6 40.0 81.2 59.4.

a. Can you conclude that the mean conversion is less than 45? Compute the appropriate test statistic and find the P-value.
b. Can you conclude that the mean conversion is greater than 30? Compute the appropriate test statistic and find the P-value.
c. Can you conclude that the mean conversion differs from 55? Compute the appropriate test statistic and find the P-value.

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

a.

Expert Solution

To determine

Check whether there is evidence to conclude the mean conversion is less than 45.

Answer to Problem 3E

There is no evidence to conclude that the mean conversion is less than 45.

Explanation of Solution

Given info:

The data represents the benzene conversions (in mole percent) for 24 different benzene hydroxylation reactions.

Calculation:

State the test hypotheses.

Null hypothesis:

H0:μ≥45

Alternative hypothesis:

H1:μ<45

Here, the sample size is large. That is, n = 24.

The formula for z-score is,

z=S+−n(n+1)4n(n+1)(2n+1)24

The positive and negative ranks are calculated as follows:

x	x−45	Signed ranks
52.3	7.3	5
41.1	–3.9	–1
28.8	–16.2	–14
67.8	22.8	17
78.6	33.6	21
72.3	27.3	19
9.1	–35.9	–23
19	–26	–18
30.3	–14.7	–12
41	–4	–2
63	18	15
80.8	35.8	22
26.8	–18.2	–16
37.3	–7.7	–6
38.1	–6.9	–4
33.6	–11.4	–9
14.3	–30.7	–20
30.1	–14.9	–13
33.4	–11.6	–10
36.2	–8.8	–7
34.6	–10.4	–8
40	–5	–3
81.2	36.2	24
59.4	14.4	11

From the table, the sum of positive ranks is,

S+=5+17+21+19+15+22+24+11=134

The test statistic is calculated as follows:

z=134−24(24+1)424(24+1)(2(24)+1)24=134−1501,225=−1635=−0.46

Thus, the test statistic is –0.46.

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 ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Left Tail for the region of the curve to shade.
Enter the data value as –0.46.
Click OK.

Output using the MINITAB software is given below:

Statistics for Engineers and Scientists, Chapter 6.9, Problem 3E , additional homework tip 1

From the MINITAB output, the P-value is 0.3228.

Decision rule:

If P-value≤α, then reject the null hypothesis (H0).

If P-value>α, then fail to reject the null hypothesis (H0).

Conclusion:

Here, the P-value is greater than the level of significance, 0.05.

Therefore, the null hypothesis is not rejected.

Hence, there is no evidence to conclude that the mean conversion is less than 45.

b.

Expert Solution

To determine

Check whether there is evidence to conclude that the mean conversion is greater than 30.

Answer to Problem 3E

There is evidence to conclude that the mean conversion is greater than 30.

Explanation of Solution

Calculation:

State the test hypotheses.

Null hypothesis:

H0:μ≤30

Alternative hypothesis:

H1:μ>30

The positive and negative ranks are calculated as follows:

x	x−45	Signed ranks
52.3	22.3	17
41.1	11.1	14
28.8	–1.2	–3
67.8	37.8	20
78.6	48.6	22
72.3	42.3	21
9.1	–20.9	–16
19	–11	–12.5
30.3	0.3	2
41	11	12.5
63	33	19
80.8	50.8	23
26.8	–3.2	–4
37.3	7.3	9
38.1	8.1	10
33.6	3.6	6
14.3	–15.7	–15
30.1	0.1	1
33.4	3.4	5
36.2	6.2	8
34.6	4.6	7
40	10	11
81.2	51.2	24
59.4	29.4	18

From the table, the sum of positive ranks is,

S+=17+14+20+22+21+2+12.5+19+23+9+10+6+1+5+8+7+11+24+18=249.5

The test statistic is calculated as follows:

z=249.5−24(24+1)424(24+1)(2(24)+1)24=249.5−1501,225=99.535=2.84

Thus, the test statistic is 2.84.

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 ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Right Tail for the region of the curve to shade.
Enter the data value as 2.84.
Click OK.

Output using the MINITAB software is given below:

Statistics for Engineers and Scientists, Chapter 6.9, Problem 3E , additional homework tip 2

From the MINITAB output, the P-value is 0.0023.

Conclusion:

Here, the P-value is less than the level of significance, 0.05.

Therefore, the null hypothesis is rejected.

Hence, there is evidence to conclude that the mean conversion is greater than 30.

c.

Expert Solution

To determine

Check whether there is evidence to conclude that the mean conversion differs from 55.

Answer to Problem 3E

There is evidence to conclude that the mean conversion differs from 55.

Explanation of Solution

Calculation:

State the test hypotheses.

Null hypothesis:

H0:μ=55

Alternative hypothesis:

H1:μ≠55

The positive and negative ranks are calculated as follows:

x	x−45	Signed ranks
52.3	–2.7	–1
41.1	–13.9	–5
28.8	–26.2	–19.5
67.8	12.8	4
78.6	23.6	15
72.3	17.3	9
9.1	–45.9	–24
19	–36	–22
30.3	–24.7	–16
41	–14	–6
63	8	3
80.8	25.8	18
26.8	–28.2	–21
37.3	–17.7	–10
38.1	–16.9	–8
33.6	–21.4	–13
14.3	–40.7	–23
30.1	–24.9	–17
33.4	–21.6	–14
36.2	–18.8	–11
34.6	–20.4	–12
40	–15	–7
81.2	26.2	19.5
59.4	4.4	2

From the table, the sum of positive ranks is,

S+=4+15+9+3+18+19.5+2=70.5

The test statistic is calculated as follows:

z=70.5−24(24+1)424(24+1)(2(24)+1)24=70.5−1501,225=−79.535=−2.27

Thus, the test statistic is –2.27.

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 ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Both Tail for the region of the curve to shade.
Enter the data value as –2.27.
Click OK.

Output using the MINITAB software is given below:

Statistics for Engineers and Scientists, Chapter 6.9, Problem 3E , additional homework tip 3

From the MINITAB output, the P-value for two tailed test is, 0.0116+0.0116=0.0232.

Conclusion:

Here, the P-value is less than the level of significance, 0.05.

Therefore, the null hypothesis is rejected.

Hence, there is evidence to conclude that the mean conversion differs from 55.

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Answer 4

Textbook Question

Chapter 6.9, Problem 3E

The article “Reaction Modeling and Optimization Using Neural Networks and Genetic Algorithms: Case Study Involving TS-1-Catalyzed Hydroxylation of Benzene” (S. Nandi, P. Mukherjee, et al., Industrial and Engineering Chemistry Research, 2002:2159–2169) presents benzene conversions (in mole percent) for 24 different benzenehydroxylation reactions. The results are

52.3 41.1 28.8 67.8 78.6 72.3 9.1 19.0

30.3 41.0 63.0 80.8 26.8 37.3 38.1 33.6

14.3 30.1 33.4 36.2 34.6 40.0 81.2 59.4.

a. Can you conclude that the mean conversion is less than 45? Compute the appropriate test statistic and find the P-value.
b. Can you conclude that the mean conversion is greater than 30? Compute the appropriate test statistic and find the P-value.
c. Can you conclude that the mean conversion differs from 55? Compute the appropriate test statistic and find the P-value.

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

a.

Expert Solution

To determine

Check whether there is evidence to conclude the mean conversion is less than 45.

Answer to Problem 3E

There is no evidence to conclude that the mean conversion is less than 45.

Explanation of Solution

Given info:

The data represents the benzene conversions (in mole percent) for 24 different benzene hydroxylation reactions.

Calculation:

State the test hypotheses.

Null hypothesis:

H0:μ≥45

Alternative hypothesis:

H1:μ<45

Here, the sample size is large. That is, n = 24.

The formula for z-score is,

z=S+−n(n+1)4n(n+1)(2n+1)24

The positive and negative ranks are calculated as follows:

x	x−45	Signed ranks
52.3	7.3	5
41.1	–3.9	–1
28.8	–16.2	–14
67.8	22.8	17
78.6	33.6	21
72.3	27.3	19
9.1	–35.9	–23
19	–26	–18
30.3	–14.7	–12
41	–4	–2
63	18	15
80.8	35.8	22
26.8	–18.2	–16
37.3	–7.7	–6
38.1	–6.9	–4
33.6	–11.4	–9
14.3	–30.7	–20
30.1	–14.9	–13
33.4	–11.6	–10
36.2	–8.8	–7
34.6	–10.4	–8
40	–5	–3
81.2	36.2	24
59.4	14.4	11

From the table, the sum of positive ranks is,

S+=5+17+21+19+15+22+24+11=134

The test statistic is calculated as follows:

z=134−24(24+1)424(24+1)(2(24)+1)24=134−1501,225=−1635=−0.46

Thus, the test statistic is –0.46.

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 ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Left Tail for the region of the curve to shade.
Enter the data value as –0.46.
Click OK.

Output using the MINITAB software is given below:

Statistics for Engineers and Scientists, Chapter 6.9, Problem 3E , additional homework tip 1

From the MINITAB output, the P-value is 0.3228.

Decision rule:

If P-value≤α, then reject the null hypothesis (H0).

If P-value>α, then fail to reject the null hypothesis (H0).

Conclusion:

Here, the P-value is greater than the level of significance, 0.05.

Therefore, the null hypothesis is not rejected.

Hence, there is no evidence to conclude that the mean conversion is less than 45.

b.

Expert Solution

To determine

Check whether there is evidence to conclude that the mean conversion is greater than 30.

Answer to Problem 3E

There is evidence to conclude that the mean conversion is greater than 30.

Explanation of Solution

Calculation:

State the test hypotheses.

Null hypothesis:

H0:μ≤30

Alternative hypothesis:

H1:μ>30

The positive and negative ranks are calculated as follows:

x	x−45	Signed ranks
52.3	22.3	17
41.1	11.1	14
28.8	–1.2	–3
67.8	37.8	20
78.6	48.6	22
72.3	42.3	21
9.1	–20.9	–16
19	–11	–12.5
30.3	0.3	2
41	11	12.5
63	33	19
80.8	50.8	23
26.8	–3.2	–4
37.3	7.3	9
38.1	8.1	10
33.6	3.6	6
14.3	–15.7	–15
30.1	0.1	1
33.4	3.4	5
36.2	6.2	8
34.6	4.6	7
40	10	11
81.2	51.2	24
59.4	29.4	18

From the table, the sum of positive ranks is,

S+=17+14+20+22+21+2+12.5+19+23+9+10+6+1+5+8+7+11+24+18=249.5

The test statistic is calculated as follows:

z=249.5−24(24+1)424(24+1)(2(24)+1)24=249.5−1501,225=99.535=2.84

Thus, the test statistic is 2.84.

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 ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Right Tail for the region of the curve to shade.
Enter the data value as 2.84.
Click OK.

Output using the MINITAB software is given below:

Statistics for Engineers and Scientists, Chapter 6.9, Problem 3E , additional homework tip 2

From the MINITAB output, the P-value is 0.0023.

Conclusion:

Here, the P-value is less than the level of significance, 0.05.

Therefore, the null hypothesis is rejected.

Hence, there is evidence to conclude that the mean conversion is greater than 30.

c.

Expert Solution

To determine

Check whether there is evidence to conclude that the mean conversion differs from 55.

Answer to Problem 3E

There is evidence to conclude that the mean conversion differs from 55.

Explanation of Solution

Calculation:

State the test hypotheses.

Null hypothesis:

H0:μ=55

Alternative hypothesis:

H1:μ≠55

The positive and negative ranks are calculated as follows:

x	x−45	Signed ranks
52.3	–2.7	–1
41.1	–13.9	–5
28.8	–26.2	–19.5
67.8	12.8	4
78.6	23.6	15
72.3	17.3	9
9.1	–45.9	–24
19	–36	–22
30.3	–24.7	–16
41	–14	–6
63	8	3
80.8	25.8	18
26.8	–28.2	–21
37.3	–17.7	–10
38.1	–16.9	–8
33.6	–21.4	–13
14.3	–40.7	–23
30.1	–24.9	–17
33.4	–21.6	–14
36.2	–18.8	–11
34.6	–20.4	–12
40	–15	–7
81.2	26.2	19.5
59.4	4.4	2

From the table, the sum of positive ranks is,

S+=4+15+9+3+18+19.5+2=70.5

The test statistic is calculated as follows:

z=70.5−24(24+1)424(24+1)(2(24)+1)24=70.5−1501,225=−79.535=−2.27

Thus, the test statistic is –2.27.

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 ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Both Tail for the region of the curve to shade.
Enter the data value as –2.27.
Click OK.

Output using the MINITAB software is given below:

Statistics for Engineers and Scientists, Chapter 6.9, Problem 3E , additional homework tip 3

From the MINITAB output, the P-value for two tailed test is, 0.0116+0.0116=0.0232.

Conclusion:

Here, the P-value is less than the level of significance, 0.05.

Therefore, the null hypothesis is rejected.

Hence, there is evidence to conclude that the mean conversion differs from 55.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

a.

Expert Solution

To determine

Check whether there is evidence to conclude the mean conversion is less than 45.

Answer to Problem 3E

There is no evidence to conclude that the mean conversion is less than 45.

Explanation of Solution

Given info:

The data represents the benzene conversions (in mole percent) for 24 different benzene hydroxylation reactions.

Calculation:

State the test hypotheses.

Null hypothesis:

H0:μ≥45

Alternative hypothesis:

H1:μ<45

Here, the sample size is large. That is, n = 24.

The formula for z-score is,

z=S+−n(n+1)4n(n+1)(2n+1)24

The positive and negative ranks are calculated as follows:

x	x−45	Signed ranks
52.3	7.3	5
41.1	–3.9	–1
28.8	–16.2	–14
67.8	22.8	17
78.6	33.6	21
72.3	27.3	19
9.1	–35.9	–23
19	–26	–18
30.3	–14.7	–12
41	–4	–2
63	18	15
80.8	35.8	22
26.8	–18.2	–16
37.3	–7.7	–6
38.1	–6.9	–4
33.6	–11.4	–9
14.3	–30.7	–20
30.1	–14.9	–13
33.4	–11.6	–10
36.2	–8.8	–7
34.6	–10.4	–8
40	–5	–3
81.2	36.2	24
59.4	14.4	11

From the table, the sum of positive ranks is,

S+=5+17+21+19+15+22+24+11=134

The test statistic is calculated as follows:

z=134−24(24+1)424(24+1)(2(24)+1)24=134−1501,225=−1635=−0.46

Thus, the test statistic is –0.46.

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 ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Left Tail for the region of the curve to shade.
Enter the data value as –0.46.
Click OK.

Output using the MINITAB software is given below:

Statistics for Engineers and Scientists, Chapter 6.9, Problem 3E , additional homework tip 1

From the MINITAB output, the P-value is 0.3228.

Decision rule:

If P-value≤α, then reject the null hypothesis (H0).

If P-value>α, then fail to reject the null hypothesis (H0).

Conclusion:

Here, the P-value is greater than the level of significance, 0.05.

Therefore, the null hypothesis is not rejected.

Hence, there is no evidence to conclude that the mean conversion is less than 45.

b.

Expert Solution

To determine

Check whether there is evidence to conclude that the mean conversion is greater than 30.

Answer to Problem 3E

There is evidence to conclude that the mean conversion is greater than 30.

Explanation of Solution

Calculation:

State the test hypotheses.

Null hypothesis:

H0:μ≤30

Alternative hypothesis:

H1:μ>30

The positive and negative ranks are calculated as follows:

x	x−45	Signed ranks
52.3	22.3	17
41.1	11.1	14
28.8	–1.2	–3
67.8	37.8	20
78.6	48.6	22
72.3	42.3	21
9.1	–20.9	–16
19	–11	–12.5
30.3	0.3	2
41	11	12.5
63	33	19
80.8	50.8	23
26.8	–3.2	–4
37.3	7.3	9
38.1	8.1	10
33.6	3.6	6
14.3	–15.7	–15
30.1	0.1	1
33.4	3.4	5
36.2	6.2	8
34.6	4.6	7
40	10	11
81.2	51.2	24
59.4	29.4	18

From the table, the sum of positive ranks is,

S+=17+14+20+22+21+2+12.5+19+23+9+10+6+1+5+8+7+11+24+18=249.5

The test statistic is calculated as follows:

z=249.5−24(24+1)424(24+1)(2(24)+1)24=249.5−1501,225=99.535=2.84

Thus, the test statistic is 2.84.

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 ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Right Tail for the region of the curve to shade.
Enter the data value as 2.84.
Click OK.

Output using the MINITAB software is given below:

Statistics for Engineers and Scientists, Chapter 6.9, Problem 3E , additional homework tip 2

From the MINITAB output, the P-value is 0.0023.

Conclusion:

Here, the P-value is less than the level of significance, 0.05.

Therefore, the null hypothesis is rejected.

Hence, there is evidence to conclude that the mean conversion is greater than 30.

c.

Expert Solution

To determine

Check whether there is evidence to conclude that the mean conversion differs from 55.

Answer to Problem 3E

There is evidence to conclude that the mean conversion differs from 55.

Explanation of Solution

Calculation:

State the test hypotheses.

Null hypothesis:

H0:μ=55

Alternative hypothesis:

H1:μ≠55

The positive and negative ranks are calculated as follows:

x	x−45	Signed ranks
52.3	–2.7	–1
41.1	–13.9	–5
28.8	–26.2	–19.5
67.8	12.8	4
78.6	23.6	15
72.3	17.3	9
9.1	–45.9	–24
19	–36	–22
30.3	–24.7	–16
41	–14	–6
63	8	3
80.8	25.8	18
26.8	–28.2	–21
37.3	–17.7	–10
38.1	–16.9	–8
33.6	–21.4	–13
14.3	–40.7	–23
30.1	–24.9	–17
33.4	–21.6	–14
36.2	–18.8	–11
34.6	–20.4	–12
40	–15	–7
81.2	26.2	19.5
59.4	4.4	2

From the table, the sum of positive ranks is,

S+=4+15+9+3+18+19.5+2=70.5

The test statistic is calculated as follows:

z=70.5−24(24+1)424(24+1)(2(24)+1)24=70.5−1501,225=−79.535=−2.27

Thus, the test statistic is –2.27.

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 ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Both Tail for the region of the curve to shade.
Enter the data value as –2.27.
Click OK.

Output using the MINITAB software is given below:

Statistics for Engineers and Scientists, Chapter 6.9, Problem 3E , additional homework tip 3

From the MINITAB output, the P-value for two tailed test is, 0.0116+0.0116=0.0232.

Conclusion:

Here, the P-value is less than the level of significance, 0.05.

Therefore, the null hypothesis is rejected.

Hence, there is evidence to conclude that the mean conversion differs from 55.

Answer 8

a.

Expert Solution

To determine

Check whether there is evidence to conclude the mean conversion is less than 45.

Answer to Problem 3E

There is no evidence to conclude that the mean conversion is less than 45.

Explanation of Solution

Given info:

The data represents the benzene conversions (in mole percent) for 24 different benzene hydroxylation reactions.

Calculation:

State the test hypotheses.

Null hypothesis:

H0:μ≥45

Alternative hypothesis:

H1:μ<45

Here, the sample size is large. That is, n = 24.

The formula for z-score is,

z=S+−n(n+1)4n(n+1)(2n+1)24

The positive and negative ranks are calculated as follows:

x	x−45	Signed ranks
52.3	7.3	5
41.1	–3.9	–1
28.8	–16.2	–14
67.8	22.8	17
78.6	33.6	21
72.3	27.3	19
9.1	–35.9	–23
19	–26	–18
30.3	–14.7	–12
41	–4	–2
63	18	15
80.8	35.8	22
26.8	–18.2	–16
37.3	–7.7	–6
38.1	–6.9	–4
33.6	–11.4	–9
14.3	–30.7	–20
30.1	–14.9	–13
33.4	–11.6	–10
36.2	–8.8	–7
34.6	–10.4	–8
40	–5	–3
81.2	36.2	24
59.4	14.4	11

From the table, the sum of positive ranks is,

S+=5+17+21+19+15+22+24+11=134

The test statistic is calculated as follows:

z=134−24(24+1)424(24+1)(2(24)+1)24=134−1501,225=−1635=−0.46

Thus, the test statistic is –0.46.

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 ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Left Tail for the region of the curve to shade.
Enter the data value as –0.46.
Click OK.

Output using the MINITAB software is given below:

Statistics for Engineers and Scientists, Chapter 6.9, Problem 3E , additional homework tip 1

From the MINITAB output, the P-value is 0.3228.

Decision rule:

If P-value≤α, then reject the null hypothesis (H0).

If P-value>α, then fail to reject the null hypothesis (H0).

Conclusion:

Here, the P-value is greater than the level of significance, 0.05.

Therefore, the null hypothesis is not rejected.

Hence, there is no evidence to conclude that the mean conversion is less than 45.

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

Check whether there is evidence to conclude the mean conversion is less than 45.

Answer 13

Answer to Problem 3E

There is no evidence to conclude that the mean conversion is less than 45.

Answer 14

Explanation of Solution

Given info:

The data represents the benzene conversions (in mole percent) for 24 different benzene hydroxylation reactions.

Calculation:

State the test hypotheses.

Null hypothesis:

H0:μ≥45

Alternative hypothesis:

H1:μ<45

Here, the sample size is large. That is, n = 24.

The formula for z-score is,

z=S+−n(n+1)4n(n+1)(2n+1)24

The positive and negative ranks are calculated as follows:

x	x−45	Signed ranks
52.3	7.3	5
41.1	–3.9	–1
28.8	–16.2	–14
67.8	22.8	17
78.6	33.6	21
72.3	27.3	19
9.1	–35.9	–23
19	–26	–18
30.3	–14.7	–12
41	–4	–2
63	18	15
80.8	35.8	22
26.8	–18.2	–16
37.3	–7.7	–6
38.1	–6.9	–4
33.6	–11.4	–9
14.3	–30.7	–20
30.1	–14.9	–13
33.4	–11.6	–10
36.2	–8.8	–7
34.6	–10.4	–8
40	–5	–3
81.2	36.2	24
59.4	14.4	11

From the table, the sum of positive ranks is,

S+=5+17+21+19+15+22+24+11=134

The test statistic is calculated as follows:

z=134−24(24+1)424(24+1)(2(24)+1)24=134−1501,225=−1635=−0.46

Thus, the test statistic is –0.46.

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 ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Left Tail for the region of the curve to shade.
Enter the data value as –0.46.
Click OK.

Output using the MINITAB software is given below:

Statistics for Engineers and Scientists, Chapter 6.9, Problem 3E , additional homework tip 1

From the MINITAB output, the P-value is 0.3228.

Decision rule:

If P-value≤α, then reject the null hypothesis (H0).

If P-value>α, then fail to reject the null hypothesis (H0).

Conclusion:

Here, the P-value is greater than the level of significance, 0.05.

Therefore, the null hypothesis is not rejected.

Hence, there is no evidence to conclude that the mean conversion is less than 45.

Answer 15

b.

Expert Solution

To determine

Check whether there is evidence to conclude that the mean conversion is greater than 30.

Answer to Problem 3E

There is evidence to conclude that the mean conversion is greater than 30.

Explanation of Solution

Calculation:

State the test hypotheses.

Null hypothesis:

H0:μ≤30

Alternative hypothesis:

H1:μ>30

The positive and negative ranks are calculated as follows:

x	x−45	Signed ranks
52.3	22.3	17
41.1	11.1	14
28.8	–1.2	–3
67.8	37.8	20
78.6	48.6	22
72.3	42.3	21
9.1	–20.9	–16
19	–11	–12.5
30.3	0.3	2
41	11	12.5
63	33	19
80.8	50.8	23
26.8	–3.2	–4
37.3	7.3	9
38.1	8.1	10
33.6	3.6	6
14.3	–15.7	–15
30.1	0.1	1
33.4	3.4	5
36.2	6.2	8
34.6	4.6	7
40	10	11
81.2	51.2	24
59.4	29.4	18

From the table, the sum of positive ranks is,

S+=17+14+20+22+21+2+12.5+19+23+9+10+6+1+5+8+7+11+24+18=249.5

The test statistic is calculated as follows:

z=249.5−24(24+1)424(24+1)(2(24)+1)24=249.5−1501,225=99.535=2.84

Thus, the test statistic is 2.84.

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 ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Right Tail for the region of the curve to shade.
Enter the data value as 2.84.
Click OK.

Output using the MINITAB software is given below:

Statistics for Engineers and Scientists, Chapter 6.9, Problem 3E , additional homework tip 2

From the MINITAB output, the P-value is 0.0023.

Conclusion:

Here, the P-value is less than the level of significance, 0.05.

Therefore, the null hypothesis is rejected.

Hence, there is evidence to conclude that the mean conversion is greater than 30.

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

Check whether there is evidence to conclude that the mean conversion is greater than 30.

Answer 20

Answer to Problem 3E

There is evidence to conclude that the mean conversion is greater than 30.

Answer 21

Explanation of Solution

Calculation:

State the test hypotheses.

Null hypothesis:

H0:μ≤30

Alternative hypothesis:

H1:μ>30

The positive and negative ranks are calculated as follows:

x	x−45	Signed ranks
52.3	22.3	17
41.1	11.1	14
28.8	–1.2	–3
67.8	37.8	20
78.6	48.6	22
72.3	42.3	21
9.1	–20.9	–16
19	–11	–12.5
30.3	0.3	2
41	11	12.5
63	33	19
80.8	50.8	23
26.8	–3.2	–4
37.3	7.3	9
38.1	8.1	10
33.6	3.6	6
14.3	–15.7	–15
30.1	0.1	1
33.4	3.4	5
36.2	6.2	8
34.6	4.6	7
40	10	11
81.2	51.2	24
59.4	29.4	18

From the table, the sum of positive ranks is,

S+=17+14+20+22+21+2+12.5+19+23+9+10+6+1+5+8+7+11+24+18=249.5

The test statistic is calculated as follows:

z=249.5−24(24+1)424(24+1)(2(24)+1)24=249.5−1501,225=99.535=2.84

Thus, the test statistic is 2.84.

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 ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Right Tail for the region of the curve to shade.
Enter the data value as 2.84.
Click OK.

Output using the MINITAB software is given below:

Statistics for Engineers and Scientists, Chapter 6.9, Problem 3E , additional homework tip 2

From the MINITAB output, the P-value is 0.0023.

Conclusion:

Here, the P-value is less than the level of significance, 0.05.

Therefore, the null hypothesis is rejected.

Hence, there is evidence to conclude that the mean conversion is greater than 30.

Answer 22

c.

Expert Solution

To determine

Check whether there is evidence to conclude that the mean conversion differs from 55.

Answer to Problem 3E

There is evidence to conclude that the mean conversion differs from 55.

Explanation of Solution

Calculation:

State the test hypotheses.

Null hypothesis:

H0:μ=55

Alternative hypothesis:

H1:μ≠55

The positive and negative ranks are calculated as follows:

x	x−45	Signed ranks
52.3	–2.7	–1
41.1	–13.9	–5
28.8	–26.2	–19.5
67.8	12.8	4
78.6	23.6	15
72.3	17.3	9
9.1	–45.9	–24
19	–36	–22
30.3	–24.7	–16
41	–14	–6
63	8	3
80.8	25.8	18
26.8	–28.2	–21
37.3	–17.7	–10
38.1	–16.9	–8
33.6	–21.4	–13
14.3	–40.7	–23
30.1	–24.9	–17
33.4	–21.6	–14
36.2	–18.8	–11
34.6	–20.4	–12
40	–15	–7
81.2	26.2	19.5
59.4	4.4	2

From the table, the sum of positive ranks is,

S+=4+15+9+3+18+19.5+2=70.5

The test statistic is calculated as follows:

z=70.5−24(24+1)424(24+1)(2(24)+1)24=70.5−1501,225=−79.535=−2.27

Thus, the test statistic is –2.27.

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 ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Both Tail for the region of the curve to shade.
Enter the data value as –2.27.
Click OK.

Output using the MINITAB software is given below:

Statistics for Engineers and Scientists, Chapter 6.9, Problem 3E , additional homework tip 3

From the MINITAB output, the P-value for two tailed test is, 0.0116+0.0116=0.0232.

Conclusion:

Here, the P-value is less than the level of significance, 0.05.

Therefore, the null hypothesis is rejected.

Hence, there is evidence to conclude that the mean conversion differs from 55.

Answer 23

c.

Expert Solution

Answer 24

c.

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

Check whether there is evidence to conclude that the mean conversion differs from 55.

Answer 27

Answer to Problem 3E

There is evidence to conclude that the mean conversion differs from 55.

Answer 28

Explanation of Solution

Calculation:

State the test hypotheses.

Null hypothesis:

H0:μ=55

Alternative hypothesis:

H1:μ≠55

The positive and negative ranks are calculated as follows:

x	x−45	Signed ranks
52.3	–2.7	–1
41.1	–13.9	–5
28.8	–26.2	–19.5
67.8	12.8	4
78.6	23.6	15
72.3	17.3	9
9.1	–45.9	–24
19	–36	–22
30.3	–24.7	–16
41	–14	–6
63	8	3
80.8	25.8	18
26.8	–28.2	–21
37.3	–17.7	–10
38.1	–16.9	–8
33.6	–21.4	–13
14.3	–40.7	–23
30.1	–24.9	–17
33.4	–21.6	–14
36.2	–18.8	–11
34.6	–20.4	–12
40	–15	–7
81.2	26.2	19.5
59.4	4.4	2

From the table, the sum of positive ranks is,

S+=4+15+9+3+18+19.5+2=70.5

The test statistic is calculated as follows:

z=70.5−24(24+1)424(24+1)(2(24)+1)24=70.5−1501,225=−79.535=−2.27

Thus, the test statistic is –2.27.

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 ‘Normal’ distribution.
Click the Shaded Area tab.
Choose X Value and Both Tail for the region of the curve to shade.
Enter the data value as –2.27.
Click OK.