The difference between the TBBMC recorded for Operator 1 and the TBBMC for Operator 2.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 7.1, Problem 43E

(a)

Section 1:

To determine

To find: The difference between the TBBMC recorded for Operator 1 and the TBBMC for Operator 2.

(a)

Section 1:

Expert Solution

Answer to Problem 43E

Solution: The table for difference is as follows:

Operator 1	Operator 2	Difference
1.328	1.323	0.005
1.342	1.322	0.02
1.075	1.073	0.002
1.228	1.233	−0.005
0.939	0.934	0.005
1.004	1.019	−0.015
1.178	1.184	−0.006
1.286	1.304	−0.018

Explanation of Solution

Calculation:

To calculate the difference, follow the below mentioned steps in Minitab:

Step 1: Enter the variable name as “Operator 1” in column C1 and enter the data for “Operator 1” and enter the variable name as “Operator 2” in column C2 and enter the data for “Operator 2.”

Step 2: Enter variable name as “Difference” in column C3.

Step 3: Go to “Calc”→“Calculator”.

Step 4: In the dialog box that appears, select “Difference” in “Store result in variable.”

Step 5: Enter “Operator 1”-“Operator 2” in “Expression.” Click on “Ok” on the dialog box.

From Minitab result, the table for difference is as follows:

Operator 1	Operator 2	Difference
1.328	1.323	0.005
1.342	1.322	0.02
1.075	1.073	0.002
1.228	1.233	−0.005
0.939	0.934	0.005
1.004	1.019	−0.015
1.178	1.184	−0.006
1.286	1.304	−0.018

Interpretation: It is clear from the table that the range for difference is −0.018 to 0.02.

Section 2:

To determine

To test: Whether t- methods can be used or not.

Section 2:

Expert Solution

Answer to Problem 43E

Solution: The t- method can be used.

Explanation of Solution

Calculation:

To determine if the t methods can be used or not, a probability plot is drawn in Minitab by following the below mentioned steps:

Step 1: Follow steps 1 and 2 similar to that in Section 1.

Step 2: Go to “Graph” option and click on “Probability Plot.” In the dialog box that appears, select “Single” and click on OK.

Step 3: Then enter the name of the column containing the data of the differences in the “Graph variables” field and click on OK.

The following probability plot is obtained:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 7.1, Problem 43E

Conclusion: The probability plot shows that the data is approximately normal as all the data points fall within the range and, hence, it is appropriate to use t procedures.

(b)

To determine

To test: The significance of the difference between the means of two operators.

(b)

Expert Solution

Answer to Problem 43E

Solution: The test statistic value is −0.35 with degree of freedom 7_ and the p- value 0.739, and the means of the two operators do not differ significantly.

Explanation of Solution

Calculation:

To test the null hypothesis, follow the below mentioned steps in Minitab:

Step 1: Follow the Steps 1 to 5 performed in section 1 of part (a).

Step 2: Go to Stat→Basic statistics→Paired t.

Step 3: In the dialogue box that appears, select samples in columns.

Step 4: Enter “Operator 1” under the field marked as “First sample” and “Operator 2” under the field marked as “Second sample.”

Step 5: In the dialogue box that appears, Click on “Option.” Enter 95.0 in the field of Confidence interval, 0.0 in “Test mean,” and “not equal” in “Alternative.”

From Minitab results, the value of test statistic is −0.35 and the p- value is 0.739.

D.F = n –.

D.F = 8 –.

D.F =.

Conclusion: If the level of significance is 5% then as the p- value is greater than 0.05 the null hypothesis will not be rejected and it is concluded that the two means do not differ significantly at 95% confidence level.

(c)

To determine

To find: The confidence interval for difference.

(c)

Expert Solution

Answer to Problem 43E

Solution: The required confidence interval is (−0.01172, 0.00872)_.

Explanation of Solution

Calculation: To compute the confidence interval, follow the below mentioned steps in Minitab:

Step 1: Follow the steps 1 to 5 performed in section 1 of part (a).

Step 2: Go to Stat→Basic statistics→Paired t.

Step 3: In the dialogue box that appears, select samples in columns.

Step 4: Enter “Operator 1” under the field marked as “First sample” and “Operator 2” under the field marked as “Second sample.”

Step 5: In the dialogue box that appears, Click on “Option.” Enter 95.0 in the field of Confidence interval, 0.0 in “Test mean,” and “not equal” in “Alternative.”

From the Minitab results, the confidence interval is (−0.01172, 0.00872)_.

Interpretation: The confidence interval denotes that the differences recorded will lie in between −0.01172 and 0.00872 with a possible chance of 5% error.

(d)

To determine

The validity of the results if the samples are not random.

(d)

Expert Solution

Answer to Problem 43E

Solution: The testing results and the confidence interval may not be valid if the samples are non-random.

Explanation of Solution

The selection of the sample must be random to construct the confidence interval and to perform testing. The t tests assume the samples to simple random. But, in this case, the sample is not taken at random. So, the results may deviate from the actual ones.

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

Microsoft Excel snapshot for random sampling: Also note the formula used for the last column 02 x✓ fx =INDEX(5852:58551, RANK(C2, $C$2:$C$51)) A B 1 No. States 2 1 ALABAMA Rand No. 0.925957526 3 2 ALASKA 0.372999976 4 3 ARIZONA 0.941323044 5 4 ARKANSAS 0.071266381 Random Sample CALIFORNIA NORTH CAROLINA ARKANSAS WASHINGTON G7 Microsoft Excel snapshot for systematic sampling: xfx INDEX(SD52:50551, F7) A B E F G 1 No. States Rand No. Random Sample population 50 2 1 ALABAMA 0.5296685 NEW HAMPSHIRE sample 10 3 2 ALASKA 0.4493186 OKLAHOMA k 5 4 3 ARIZONA 0.707914 KANSAS 5 4 ARKANSAS 0.4831379 NORTH DAKOTA 6 5 CALIFORNIA 0.7277162 INDIANA Random Sample Sample Name 7 6 COLORADO 0.5865002 MISSISSIPPI 8 7:ONNECTICU 0.7640596 ILLINOIS 9 8 DELAWARE 0.5783029 MISSOURI 525 10 15 INDIANA MARYLAND COLORADO

Suppose the Internal Revenue Service reported that the mean tax refund for the year 2022 was $3401. Assume the standard deviation is $82.5 and that the amounts refunded follow a normal probability distribution. Solve the following three parts? (For the answer to question 14, 15, and 16, start with making a bell curve. Identify on the bell curve where is mean, X, and area(s) to be determined. 1.What percent of the refunds are more than $3,500? 2. What percent of the refunds are more than $3500 but less than $3579? 3. What percent of the refunds are more than $3325 but less than $3579?

A normal distribution has a mean of 50 and a standard deviation of 4. Solve the following three parts? 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 answer of the second part.) 2. Compute the probability of a value greater than 55.0. Use the same formula, x=55 and subtract the answer from 1. 3. Compute the probability of a value between 52.0 and 55.0. (The question requires finding probability value between 52 and 55. Solve it in 3 steps. In the first step, use the above formula and x = 52, 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…

Answer 2

Question

Chapter 7.1, Problem 43E

(a)

Section 1:

To determine

To find: The difference between the TBBMC recorded for Operator 1 and the TBBMC for Operator 2.

(a)

Section 1:

Expert Solution

Answer to Problem 43E

Solution: The table for difference is as follows:

Operator 1	Operator 2	Difference
1.328	1.323	0.005
1.342	1.322	0.02
1.075	1.073	0.002
1.228	1.233	−0.005
0.939	0.934	0.005
1.004	1.019	−0.015
1.178	1.184	−0.006
1.286	1.304	−0.018

Explanation of Solution

Calculation:

To calculate the difference, follow the below mentioned steps in Minitab:

Step 1: Enter the variable name as “Operator 1” in column C1 and enter the data for “Operator 1” and enter the variable name as “Operator 2” in column C2 and enter the data for “Operator 2.”

Step 2: Enter variable name as “Difference” in column C3.

Step 3: Go to “Calc”→“Calculator”.

Step 4: In the dialog box that appears, select “Difference” in “Store result in variable.”

Step 5: Enter “Operator 1”-“Operator 2” in “Expression.” Click on “Ok” on the dialog box.

From Minitab result, the table for difference is as follows:

Operator 1	Operator 2	Difference
1.328	1.323	0.005
1.342	1.322	0.02
1.075	1.073	0.002
1.228	1.233	−0.005
0.939	0.934	0.005
1.004	1.019	−0.015
1.178	1.184	−0.006
1.286	1.304	−0.018

Interpretation: It is clear from the table that the range for difference is −0.018 to 0.02.

Section 2:

To determine

To test: Whether t- methods can be used or not.

Section 2:

Expert Solution

Answer to Problem 43E

Solution: The t- method can be used.

Explanation of Solution

Calculation:

To determine if the t methods can be used or not, a probability plot is drawn in Minitab by following the below mentioned steps:

Step 1: Follow steps 1 and 2 similar to that in Section 1.

Step 2: Go to “Graph” option and click on “Probability Plot.” In the dialog box that appears, select “Single” and click on OK.

Step 3: Then enter the name of the column containing the data of the differences in the “Graph variables” field and click on OK.

The following probability plot is obtained:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 7.1, Problem 43E

Conclusion: The probability plot shows that the data is approximately normal as all the data points fall within the range and, hence, it is appropriate to use t procedures.

(b)

To determine

To test: The significance of the difference between the means of two operators.

(b)

Expert Solution

Answer to Problem 43E

Solution: The test statistic value is −0.35 with degree of freedom 7_ and the p- value 0.739, and the means of the two operators do not differ significantly.

Explanation of Solution

Calculation:

To test the null hypothesis, follow the below mentioned steps in Minitab:

Step 1: Follow the Steps 1 to 5 performed in section 1 of part (a).

Step 2: Go to Stat→Basic statistics→Paired t.

Step 3: In the dialogue box that appears, select samples in columns.

Step 4: Enter “Operator 1” under the field marked as “First sample” and “Operator 2” under the field marked as “Second sample.”

Step 5: In the dialogue box that appears, Click on “Option.” Enter 95.0 in the field of Confidence interval, 0.0 in “Test mean,” and “not equal” in “Alternative.”

From Minitab results, the value of test statistic is −0.35 and the p- value is 0.739.

D.F = n –.

D.F = 8 –.

D.F =.

Conclusion: If the level of significance is 5% then as the p- value is greater than 0.05 the null hypothesis will not be rejected and it is concluded that the two means do not differ significantly at 95% confidence level.

(c)

To determine

To find: The confidence interval for difference.

(c)

Expert Solution

Answer to Problem 43E

Solution: The required confidence interval is (−0.01172, 0.00872)_.

Explanation of Solution

Calculation: To compute the confidence interval, follow the below mentioned steps in Minitab:

Step 1: Follow the steps 1 to 5 performed in section 1 of part (a).

Step 2: Go to Stat→Basic statistics→Paired t.

Step 3: In the dialogue box that appears, select samples in columns.

Step 4: Enter “Operator 1” under the field marked as “First sample” and “Operator 2” under the field marked as “Second sample.”

Step 5: In the dialogue box that appears, Click on “Option.” Enter 95.0 in the field of Confidence interval, 0.0 in “Test mean,” and “not equal” in “Alternative.”

From the Minitab results, the confidence interval is (−0.01172, 0.00872)_.

Interpretation: The confidence interval denotes that the differences recorded will lie in between −0.01172 and 0.00872 with a possible chance of 5% error.

(d)

To determine

The validity of the results if the samples are not random.

(d)

Expert Solution

Answer to Problem 43E

Solution: The testing results and the confidence interval may not be valid if the samples are non-random.

Explanation of Solution

The selection of the sample must be random to construct the confidence interval and to perform testing. The t tests assume the samples to simple random. But, in this case, the sample is not taken at random. So, the results may deviate from the actual ones.

Want to see more full solutions like this?

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

Answer 3

Question

Chapter 7.1, Problem 43E

(a)

Section 1:

To determine

To find: The difference between the TBBMC recorded for Operator 1 and the TBBMC for Operator 2.

(a)

Section 1:

Expert Solution

Answer to Problem 43E

Solution: The table for difference is as follows:

Operator 1	Operator 2	Difference
1.328	1.323	0.005
1.342	1.322	0.02
1.075	1.073	0.002
1.228	1.233	−0.005
0.939	0.934	0.005
1.004	1.019	−0.015
1.178	1.184	−0.006
1.286	1.304	−0.018

Explanation of Solution

Calculation:

To calculate the difference, follow the below mentioned steps in Minitab:

Step 1: Enter the variable name as “Operator 1” in column C1 and enter the data for “Operator 1” and enter the variable name as “Operator 2” in column C2 and enter the data for “Operator 2.”

Step 2: Enter variable name as “Difference” in column C3.

Step 3: Go to “Calc”→“Calculator”.

Step 4: In the dialog box that appears, select “Difference” in “Store result in variable.”

Step 5: Enter “Operator 1”-“Operator 2” in “Expression.” Click on “Ok” on the dialog box.

From Minitab result, the table for difference is as follows:

Operator 1	Operator 2	Difference
1.328	1.323	0.005
1.342	1.322	0.02
1.075	1.073	0.002
1.228	1.233	−0.005
0.939	0.934	0.005
1.004	1.019	−0.015
1.178	1.184	−0.006
1.286	1.304	−0.018

Interpretation: It is clear from the table that the range for difference is −0.018 to 0.02.

Section 2:

To determine

To test: Whether t- methods can be used or not.

Section 2:

Expert Solution

Answer to Problem 43E

Solution: The t- method can be used.

Explanation of Solution

Calculation:

To determine if the t methods can be used or not, a probability plot is drawn in Minitab by following the below mentioned steps:

Step 1: Follow steps 1 and 2 similar to that in Section 1.

Step 2: Go to “Graph” option and click on “Probability Plot.” In the dialog box that appears, select “Single” and click on OK.

Step 3: Then enter the name of the column containing the data of the differences in the “Graph variables” field and click on OK.

The following probability plot is obtained:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 7.1, Problem 43E

Conclusion: The probability plot shows that the data is approximately normal as all the data points fall within the range and, hence, it is appropriate to use t procedures.

(b)

To determine

To test: The significance of the difference between the means of two operators.

(b)

Expert Solution

Answer to Problem 43E

Solution: The test statistic value is −0.35 with degree of freedom 7_ and the p- value 0.739, and the means of the two operators do not differ significantly.

Explanation of Solution

Calculation:

To test the null hypothesis, follow the below mentioned steps in Minitab:

Step 1: Follow the Steps 1 to 5 performed in section 1 of part (a).

Step 2: Go to Stat→Basic statistics→Paired t.

Step 3: In the dialogue box that appears, select samples in columns.

Step 4: Enter “Operator 1” under the field marked as “First sample” and “Operator 2” under the field marked as “Second sample.”

Step 5: In the dialogue box that appears, Click on “Option.” Enter 95.0 in the field of Confidence interval, 0.0 in “Test mean,” and “not equal” in “Alternative.”

From Minitab results, the value of test statistic is −0.35 and the p- value is 0.739.

D.F = n –.

D.F = 8 –.

D.F =.

Conclusion: If the level of significance is 5% then as the p- value is greater than 0.05 the null hypothesis will not be rejected and it is concluded that the two means do not differ significantly at 95% confidence level.

(c)

To determine

To find: The confidence interval for difference.

(c)

Expert Solution

Answer to Problem 43E

Solution: The required confidence interval is (−0.01172, 0.00872)_.

Explanation of Solution

Calculation: To compute the confidence interval, follow the below mentioned steps in Minitab:

Step 1: Follow the steps 1 to 5 performed in section 1 of part (a).

Step 2: Go to Stat→Basic statistics→Paired t.

Step 3: In the dialogue box that appears, select samples in columns.

Step 4: Enter “Operator 1” under the field marked as “First sample” and “Operator 2” under the field marked as “Second sample.”

Step 5: In the dialogue box that appears, Click on “Option.” Enter 95.0 in the field of Confidence interval, 0.0 in “Test mean,” and “not equal” in “Alternative.”

From the Minitab results, the confidence interval is (−0.01172, 0.00872)_.

Interpretation: The confidence interval denotes that the differences recorded will lie in between −0.01172 and 0.00872 with a possible chance of 5% error.

(d)

To determine

The validity of the results if the samples are not random.

(d)

Expert Solution

Answer to Problem 43E

Solution: The testing results and the confidence interval may not be valid if the samples are non-random.

Explanation of Solution

The selection of the sample must be random to construct the confidence interval and to perform testing. The t tests assume the samples to simple random. But, in this case, the sample is not taken at random. So, the results may deviate from the actual ones.

Want to see more full solutions like this?

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

Answer 4

Question

Chapter 7.1, Problem 43E

(a)

Section 1:

To determine

To find: The difference between the TBBMC recorded for Operator 1 and the TBBMC for Operator 2.

(a)

Section 1:

Expert Solution

Answer to Problem 43E

Solution: The table for difference is as follows:

Operator 1	Operator 2	Difference
1.328	1.323	0.005
1.342	1.322	0.02
1.075	1.073	0.002
1.228	1.233	−0.005
0.939	0.934	0.005
1.004	1.019	−0.015
1.178	1.184	−0.006
1.286	1.304	−0.018

Explanation of Solution

Calculation:

To calculate the difference, follow the below mentioned steps in Minitab:

Step 1: Enter the variable name as “Operator 1” in column C1 and enter the data for “Operator 1” and enter the variable name as “Operator 2” in column C2 and enter the data for “Operator 2.”

Step 2: Enter variable name as “Difference” in column C3.

Step 3: Go to “Calc”→“Calculator”.

Step 4: In the dialog box that appears, select “Difference” in “Store result in variable.”

Step 5: Enter “Operator 1”-“Operator 2” in “Expression.” Click on “Ok” on the dialog box.

From Minitab result, the table for difference is as follows:

Operator 1	Operator 2	Difference
1.328	1.323	0.005
1.342	1.322	0.02
1.075	1.073	0.002
1.228	1.233	−0.005
0.939	0.934	0.005
1.004	1.019	−0.015
1.178	1.184	−0.006
1.286	1.304	−0.018

Interpretation: It is clear from the table that the range for difference is −0.018 to 0.02.

Section 2:

To determine

To test: Whether t- methods can be used or not.

Section 2:

Expert Solution

Answer to Problem 43E

Solution: The t- method can be used.

Explanation of Solution

Calculation:

To determine if the t methods can be used or not, a probability plot is drawn in Minitab by following the below mentioned steps:

Step 1: Follow steps 1 and 2 similar to that in Section 1.

Step 2: Go to “Graph” option and click on “Probability Plot.” In the dialog box that appears, select “Single” and click on OK.

Step 3: Then enter the name of the column containing the data of the differences in the “Graph variables” field and click on OK.

The following probability plot is obtained:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 7.1, Problem 43E

Conclusion: The probability plot shows that the data is approximately normal as all the data points fall within the range and, hence, it is appropriate to use t procedures.

(b)

To determine

To test: The significance of the difference between the means of two operators.

(b)

Expert Solution

Answer to Problem 43E

Solution: The test statistic value is −0.35 with degree of freedom 7_ and the p- value 0.739, and the means of the two operators do not differ significantly.

Explanation of Solution

Calculation:

To test the null hypothesis, follow the below mentioned steps in Minitab:

Step 1: Follow the Steps 1 to 5 performed in section 1 of part (a).

Step 2: Go to Stat→Basic statistics→Paired t.

Step 3: In the dialogue box that appears, select samples in columns.

Step 4: Enter “Operator 1” under the field marked as “First sample” and “Operator 2” under the field marked as “Second sample.”

Step 5: In the dialogue box that appears, Click on “Option.” Enter 95.0 in the field of Confidence interval, 0.0 in “Test mean,” and “not equal” in “Alternative.”

From Minitab results, the value of test statistic is −0.35 and the p- value is 0.739.

D.F = n –.

D.F = 8 –.

D.F =.

Conclusion: If the level of significance is 5% then as the p- value is greater than 0.05 the null hypothesis will not be rejected and it is concluded that the two means do not differ significantly at 95% confidence level.

(c)

To determine

To find: The confidence interval for difference.

(c)

Expert Solution

Answer to Problem 43E

Solution: The required confidence interval is (−0.01172, 0.00872)_.

Explanation of Solution

Calculation: To compute the confidence interval, follow the below mentioned steps in Minitab:

Step 1: Follow the steps 1 to 5 performed in section 1 of part (a).

Step 2: Go to Stat→Basic statistics→Paired t.

Step 3: In the dialogue box that appears, select samples in columns.

Step 4: Enter “Operator 1” under the field marked as “First sample” and “Operator 2” under the field marked as “Second sample.”

Step 5: In the dialogue box that appears, Click on “Option.” Enter 95.0 in the field of Confidence interval, 0.0 in “Test mean,” and “not equal” in “Alternative.”

From the Minitab results, the confidence interval is (−0.01172, 0.00872)_.

Interpretation: The confidence interval denotes that the differences recorded will lie in between −0.01172 and 0.00872 with a possible chance of 5% error.

(d)

To determine

The validity of the results if the samples are not random.

(d)

Expert Solution

Answer to Problem 43E

Solution: The testing results and the confidence interval may not be valid if the samples are non-random.

Explanation of Solution

The selection of the sample must be random to construct the confidence interval and to perform testing. The t tests assume the samples to simple random. But, in this case, the sample is not taken at random. So, the results may deviate from the actual ones.

Want to see more full solutions like this?

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

Answer 5

Chapter 7.1, Problem 43E

(a)

Section 1:

To determine

To find: The difference between the TBBMC recorded for Operator 1 and the TBBMC for Operator 2.

(a)

Section 1:

Expert Solution

Answer to Problem 43E

Solution: The table for difference is as follows:

Operator 1	Operator 2	Difference
1.328	1.323	0.005
1.342	1.322	0.02
1.075	1.073	0.002
1.228	1.233	−0.005
0.939	0.934	0.005
1.004	1.019	−0.015
1.178	1.184	−0.006
1.286	1.304	−0.018

Explanation of Solution

Calculation:

To calculate the difference, follow the below mentioned steps in Minitab:

Step 1: Enter the variable name as “Operator 1” in column C1 and enter the data for “Operator 1” and enter the variable name as “Operator 2” in column C2 and enter the data for “Operator 2.”

Step 2: Enter variable name as “Difference” in column C3.

Step 3: Go to “Calc”→“Calculator”.

Step 4: In the dialog box that appears, select “Difference” in “Store result in variable.”

Step 5: Enter “Operator 1”-“Operator 2” in “Expression.” Click on “Ok” on the dialog box.

From Minitab result, the table for difference is as follows:

Operator 1	Operator 2	Difference
1.328	1.323	0.005
1.342	1.322	0.02
1.075	1.073	0.002
1.228	1.233	−0.005
0.939	0.934	0.005
1.004	1.019	−0.015
1.178	1.184	−0.006
1.286	1.304	−0.018

Interpretation: It is clear from the table that the range for difference is −0.018 to 0.02.

Section 2:

To determine

To test: Whether t- methods can be used or not.

Section 2:

Expert Solution

Answer to Problem 43E

Solution: The t- method can be used.

Explanation of Solution

Calculation:

To determine if the t methods can be used or not, a probability plot is drawn in Minitab by following the below mentioned steps:

Step 1: Follow steps 1 and 2 similar to that in Section 1.

Step 2: Go to “Graph” option and click on “Probability Plot.” In the dialog box that appears, select “Single” and click on OK.

Step 3: Then enter the name of the column containing the data of the differences in the “Graph variables” field and click on OK.

The following probability plot is obtained:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 7.1, Problem 43E

Conclusion: The probability plot shows that the data is approximately normal as all the data points fall within the range and, hence, it is appropriate to use t procedures.

(b)

To determine

To test: The significance of the difference between the means of two operators.

(b)

Expert Solution

Answer to Problem 43E

Solution: The test statistic value is −0.35 with degree of freedom 7_ and the p- value 0.739, and the means of the two operators do not differ significantly.

Explanation of Solution

Calculation:

To test the null hypothesis, follow the below mentioned steps in Minitab:

Step 1: Follow the Steps 1 to 5 performed in section 1 of part (a).

Step 2: Go to Stat→Basic statistics→Paired t.

Step 3: In the dialogue box that appears, select samples in columns.

Step 4: Enter “Operator 1” under the field marked as “First sample” and “Operator 2” under the field marked as “Second sample.”

Step 5: In the dialogue box that appears, Click on “Option.” Enter 95.0 in the field of Confidence interval, 0.0 in “Test mean,” and “not equal” in “Alternative.”

From Minitab results, the value of test statistic is −0.35 and the p- value is 0.739.

D.F = n –.

D.F = 8 –.

D.F =.

Conclusion: If the level of significance is 5% then as the p- value is greater than 0.05 the null hypothesis will not be rejected and it is concluded that the two means do not differ significantly at 95% confidence level.

(c)

To determine

To find: The confidence interval for difference.

(c)

Expert Solution

Answer to Problem 43E

Solution: The required confidence interval is (−0.01172, 0.00872)_.

Explanation of Solution

Calculation: To compute the confidence interval, follow the below mentioned steps in Minitab:

Step 1: Follow the steps 1 to 5 performed in section 1 of part (a).

Step 2: Go to Stat→Basic statistics→Paired t.

Step 3: In the dialogue box that appears, select samples in columns.

Step 4: Enter “Operator 1” under the field marked as “First sample” and “Operator 2” under the field marked as “Second sample.”

Step 5: In the dialogue box that appears, Click on “Option.” Enter 95.0 in the field of Confidence interval, 0.0 in “Test mean,” and “not equal” in “Alternative.”

From the Minitab results, the confidence interval is (−0.01172, 0.00872)_.

Interpretation: The confidence interval denotes that the differences recorded will lie in between −0.01172 and 0.00872 with a possible chance of 5% error.

(d)

To determine

The validity of the results if the samples are not random.

(d)

Expert Solution

Answer to Problem 43E

Solution: The testing results and the confidence interval may not be valid if the samples are non-random.

Explanation of Solution

The selection of the sample must be random to construct the confidence interval and to perform testing. The t tests assume the samples to simple random. But, in this case, the sample is not taken at random. So, the results may deviate from the actual ones.

Answer 6

Chapter 7.1, Problem 43E

(a)

Section 1:

To determine

To find: The difference between the TBBMC recorded for Operator 1 and the TBBMC for Operator 2.

(a)

Section 1:

Expert Solution

Answer to Problem 43E

Solution: The table for difference is as follows:

Operator 1	Operator 2	Difference
1.328	1.323	0.005
1.342	1.322	0.02
1.075	1.073	0.002
1.228	1.233	−0.005
0.939	0.934	0.005
1.004	1.019	−0.015
1.178	1.184	−0.006
1.286	1.304	−0.018

Explanation of Solution

Calculation:

To calculate the difference, follow the below mentioned steps in Minitab:

Step 1: Enter the variable name as “Operator 1” in column C1 and enter the data for “Operator 1” and enter the variable name as “Operator 2” in column C2 and enter the data for “Operator 2.”

Step 2: Enter variable name as “Difference” in column C3.

Step 3: Go to “Calc”→“Calculator”.

Step 4: In the dialog box that appears, select “Difference” in “Store result in variable.”

Step 5: Enter “Operator 1”-“Operator 2” in “Expression.” Click on “Ok” on the dialog box.

From Minitab result, the table for difference is as follows:

Operator 1	Operator 2	Difference
1.328	1.323	0.005
1.342	1.322	0.02
1.075	1.073	0.002
1.228	1.233	−0.005
0.939	0.934	0.005
1.004	1.019	−0.015
1.178	1.184	−0.006
1.286	1.304	−0.018

Interpretation: It is clear from the table that the range for difference is −0.018 to 0.02.

Answer 7

Chapter 7.1, Problem 43E

(a)

Section 1:

To determine

To find: The difference between the TBBMC recorded for Operator 1 and the TBBMC for Operator 2.

Answer 8

(a)

Section 1:

Expert Solution

Answer to Problem 43E

Solution: The table for difference is as follows:

Operator 1	Operator 2	Difference
1.328	1.323	0.005
1.342	1.322	0.02
1.075	1.073	0.002
1.228	1.233	−0.005
0.939	0.934	0.005
1.004	1.019	−0.015
1.178	1.184	−0.006
1.286	1.304	−0.018

Answer 9

(a)

Section 1:

Expert Solution

Answer 10

(a)

Section 1:

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Calculation:

To calculate the difference, follow the below mentioned steps in Minitab:

Step 1: Enter the variable name as “Operator 1” in column C1 and enter the data for “Operator 1” and enter the variable name as “Operator 2” in column C2 and enter the data for “Operator 2.”

Step 2: Enter variable name as “Difference” in column C3.

Step 3: Go to “Calc”→“Calculator”.

Step 4: In the dialog box that appears, select “Difference” in “Store result in variable.”

Step 5: Enter “Operator 1”-“Operator 2” in “Expression.” Click on “Ok” on the dialog box.

From Minitab result, the table for difference is as follows:

Operator 1	Operator 2	Difference
1.328	1.323	0.005
1.342	1.322	0.02
1.075	1.073	0.002
1.228	1.233	−0.005
0.939	0.934	0.005
1.004	1.019	−0.015
1.178	1.184	−0.006
1.286	1.304	−0.018

Interpretation: It is clear from the table that the range for difference is −0.018 to 0.02.

Answer 13

Section 2:

To determine

To test: Whether t- methods can be used or not.

Section 2:

Expert Solution

Answer to Problem 43E

Solution: The t- method can be used.

Explanation of Solution

Calculation:

To determine if the t methods can be used or not, a probability plot is drawn in Minitab by following the below mentioned steps:

Step 1: Follow steps 1 and 2 similar to that in Section 1.

Step 2: Go to “Graph” option and click on “Probability Plot.” In the dialog box that appears, select “Single” and click on OK.

Step 3: Then enter the name of the column containing the data of the differences in the “Graph variables” field and click on OK.

The following probability plot is obtained:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 7.1, Problem 43E

Conclusion: The probability plot shows that the data is approximately normal as all the data points fall within the range and, hence, it is appropriate to use t procedures.

Answer 14

Section 2:

To determine

To test: Whether t- methods can be used or not.

Answer 15

Section 2:

Expert Solution

Answer to Problem 43E

Solution: The t- method can be used.

Answer 16

Section 2:

Expert Solution

Answer 17

Section 2:

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Calculation:

To determine if the t methods can be used or not, a probability plot is drawn in Minitab by following the below mentioned steps:

Step 1: Follow steps 1 and 2 similar to that in Section 1.

Step 2: Go to “Graph” option and click on “Probability Plot.” In the dialog box that appears, select “Single” and click on OK.

Step 3: Then enter the name of the column containing the data of the differences in the “Graph variables” field and click on OK.

The following probability plot is obtained:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 7.1, Problem 43E

Conclusion: The probability plot shows that the data is approximately normal as all the data points fall within the range and, hence, it is appropriate to use t procedures.

Answer 20

(b)

To determine

To test: The significance of the difference between the means of two operators.

(b)

Expert Solution

Answer to Problem 43E

Solution: The test statistic value is −0.35 with degree of freedom 7_ and the p- value 0.739, and the means of the two operators do not differ significantly.

Explanation of Solution

Calculation:

To test the null hypothesis, follow the below mentioned steps in Minitab:

Step 1: Follow the Steps 1 to 5 performed in section 1 of part (a).

Step 2: Go to Stat→Basic statistics→Paired t.

Step 3: In the dialogue box that appears, select samples in columns.

Step 4: Enter “Operator 1” under the field marked as “First sample” and “Operator 2” under the field marked as “Second sample.”

Step 5: In the dialogue box that appears, Click on “Option.” Enter 95.0 in the field of Confidence interval, 0.0 in “Test mean,” and “not equal” in “Alternative.”

From Minitab results, the value of test statistic is −0.35 and the p- value is 0.739.

D.F = n –.

D.F = 8 –.

D.F =.

Conclusion: If the level of significance is 5% then as the p- value is greater than 0.05 the null hypothesis will not be rejected and it is concluded that the two means do not differ significantly at 95% confidence level.

Answer 21

(b)

To determine

To test: The significance of the difference between the means of two operators.

Answer 22

(b)

Expert Solution

Answer to Problem 43E

Solution: The test statistic value is −0.35 with degree of freedom 7_ and the p- value 0.739, and the means of the two operators do not differ significantly.

Answer 23

(b)

Expert Solution

Answer 24

(b)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Calculation:

To test the null hypothesis, follow the below mentioned steps in Minitab:

Step 1: Follow the Steps 1 to 5 performed in section 1 of part (a).

Step 2: Go to Stat→Basic statistics→Paired t.

Step 3: In the dialogue box that appears, select samples in columns.

Step 4: Enter “Operator 1” under the field marked as “First sample” and “Operator 2” under the field marked as “Second sample.”

Step 5: In the dialogue box that appears, Click on “Option.” Enter 95.0 in the field of Confidence interval, 0.0 in “Test mean,” and “not equal” in “Alternative.”

From Minitab results, the value of test statistic is −0.35 and the p- value is 0.739.

D.F = n –.

D.F = 8 –.

D.F =.

Conclusion: If the level of significance is 5% then as the p- value is greater than 0.05 the null hypothesis will not be rejected and it is concluded that the two means do not differ significantly at 95% confidence level.

Answer 27

(c)

To determine

To find: The confidence interval for difference.

(c)

Expert Solution

Answer to Problem 43E

Solution: The required confidence interval is (−0.01172, 0.00872)_.

Explanation of Solution

Calculation: To compute the confidence interval, follow the below mentioned steps in Minitab:

Step 1: Follow the steps 1 to 5 performed in section 1 of part (a).

Step 2: Go to Stat→Basic statistics→Paired t.

Step 3: In the dialogue box that appears, select samples in columns.

Step 4: Enter “Operator 1” under the field marked as “First sample” and “Operator 2” under the field marked as “Second sample.”

Step 5: In the dialogue box that appears, Click on “Option.” Enter 95.0 in the field of Confidence interval, 0.0 in “Test mean,” and “not equal” in “Alternative.”

From the Minitab results, the confidence interval is (−0.01172, 0.00872)_.

Interpretation: The confidence interval denotes that the differences recorded will lie in between −0.01172 and 0.00872 with a possible chance of 5% error.

Answer 28

(c)

To determine

To find: The confidence interval for difference.

Answer 29

(c)

Expert Solution

Answer to Problem 43E

Solution: The required confidence interval is (−0.01172, 0.00872)_.

Answer 30

(c)

Expert Solution

Answer 31

(c)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Calculation: To compute the confidence interval, follow the below mentioned steps in Minitab:

Step 1: Follow the steps 1 to 5 performed in section 1 of part (a).

Step 2: Go to Stat→Basic statistics→Paired t.

Step 3: In the dialogue box that appears, select samples in columns.

Step 4: Enter “Operator 1” under the field marked as “First sample” and “Operator 2” under the field marked as “Second sample.”

Step 5: In the dialogue box that appears, Click on “Option.” Enter 95.0 in the field of Confidence interval, 0.0 in “Test mean,” and “not equal” in “Alternative.”

From the Minitab results, the confidence interval is (−0.01172, 0.00872)_.

Interpretation: The confidence interval denotes that the differences recorded will lie in between −0.01172 and 0.00872 with a possible chance of 5% error.

Answer 34

(d)

To determine

The validity of the results if the samples are not random.

(d)

Expert Solution

Answer to Problem 43E

Solution: The testing results and the confidence interval may not be valid if the samples are non-random.

Explanation of Solution

The selection of the sample must be random to construct the confidence interval and to perform testing. The t tests assume the samples to simple random. But, in this case, the sample is not taken at random. So, the results may deviate from the actual ones.

Answer 35

(d)

To determine

The validity of the results if the samples are not random.

Answer 36

(d)

Expert Solution

Answer to Problem 43E

Solution: The testing results and the confidence interval may not be valid if the samples are non-random.

Answer 37

(d)

Expert Solution

Answer 38

(d)

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

The selection of the sample must be random to construct the confidence interval and to perform testing. The t tests assume the samples to simple random. But, in this case, the sample is not taken at random. So, the results may deviate from the actual ones.