Suppose that you are testing the following hypotheses where the variance is known: Find the P -value for the following values of the test statistic. H 0 : μ = 100 H 1 : μ ≠ 100 (a) Z 0 = 2.75 (b) Z 0 = 1.86 (c) Z 0 = −2.05 (d) Z 0 = −1.86

Question

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Answer 1

Textbook Question

Chapter 4, Problem 1E

Suppose that you are testing the following hypotheses where the variance is known:

Find the P-value for the following values of the test statistic.

$H 0 : μ = 100 H 1 : μ ≠ 100$

(a) Z₀ = 2.75

(b) Z₀ = 1.86

(c) Z₀ = −2.05

(d) Z₀ = −1.86

a.

Expert Solution

To determine

Find P-value of the test statistic $Z0=2.75$ .

Answer to Problem 1E

The P-value is 0.0060.

Explanation of Solution

As the alternative hypothesis states that $μ≠100$ , the given test is a two-tailed test.

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

Choose Graph>Probability Distribution Plot, choose View Probability> OK.
From the Distribution drop down list box, choose Normal distribution.
Click the Shaded Area tab.
Choose the X Value and Both Tails for the region of the curve to be shaded.
Enter the X Value as 2.75.
Click OK.
The output obtained is as follows:

Introduction to Statistical Quality Control, Chapter 4, Problem 1E , additional homework tip 1

From the output, the P-value at each tail is obtained as 0.002980.

Since the alternative hypothesis (H₁) states that the mean is not equal to 100 (two-tail test), the P-value of the test statistic is calculated by adding the Probability values at both ends of the tail.

The P-value is calculated for two-tailed test is as follows:

$P-Value=0.002980+0.002980=0.00596≃0.0060$

Thus, the P-value is 0.0060.

b.

Expert Solution

To determine

Find P-value of the test statistic $Z0=1.86$ .

Answer to Problem 1E

The P-value is 0.0629.

Explanation of Solution

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

Choose Graph>Probability Distribution Plot, choose View Probability> OK.
From the Distribution drop down list box, choose Normal distribution.
Click the Shaded Area tab.
Choose the X Value and Two Tail for the region of the curve to be shaded.
Enter the X Value as 1.86.
Click OK.
The output obtained is as follows:

Introduction to Statistical Quality Control, Chapter 4, Problem 1E , additional homework tip 2

From the output, the P-value at each tail is obtained as 0.03144.

The P-value is calculated for two-tailed test is as follows:

$P-Value=0.03144+0.03144=0.06288≃0.0629$

Thus, the P-value is 0.0629.

c.

Expert Solution

To determine

Find P-value of the test statistic $Z0=−2.05$ .

Answer to Problem 1E

The P-value is 0.0404.

Explanation of Solution

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

Choose Graph>Probability Distribution Plot, choose View Probability> OK.
From the Distribution drop down list box, choose Normal distribution.
Click the Shaded Area tab.
Choose the X Value and Two Tail for the region of the curve to be shaded.
Enter the X Value as -2.05.
Click OK.
The output obtained is as follows:

Introduction to Statistical Quality Control, Chapter 4, Problem 1E , additional homework tip 3

From the output, the P-value at each tail is obtained as 0.02018.

The P-value is calculated for two-tailed test is as follows:

$P-Value=0.02018+0.02018=0.04036≃0.0404$

Thus, the P-value is 0.0404.

d.

Expert Solution

To determine

Find P-value of the test statistic $Z0=−1.86$ .

Answer to Problem 1E

The P-value is 0.0629.

Explanation of Solution

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

Choose Graph>Probability Distribution Plot, choose View Probability> OK.
From the Distribution drop down list box, choose Normal distribution.
Click the Shaded Area tab.
Choose the X Value and Two Tail for the region of the curve to be shaded.
Enter the X Value as –1.86.
Click OK.
The output obtained is as follows:

Introduction to Statistical Quality Control, Chapter 4, Problem 1E , additional homework tip 4

From the output, the P-value at each tail is obtained as 0.03144.

The P-value is calculated for two-tailed test is as follows:

$P-Value=0.03144+0.03144=0.06288≃0.0629$

Thus, the P-value is 0.0629.

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Answer 2

Textbook Question

Chapter 4, Problem 1E

Suppose that you are testing the following hypotheses where the variance is known:

Find the P-value for the following values of the test statistic.

$H 0 : μ = 100 H 1 : μ ≠ 100$

(a) Z₀ = 2.75

(b) Z₀ = 1.86

(c) Z₀ = −2.05

(d) Z₀ = −1.86

a.

Expert Solution

To determine

Find P-value of the test statistic $Z0=2.75$ .

Answer to Problem 1E

The P-value is 0.0060.

Explanation of Solution

As the alternative hypothesis states that $μ≠100$ , the given test is a two-tailed test.

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

Choose Graph>Probability Distribution Plot, choose View Probability> OK.
From the Distribution drop down list box, choose Normal distribution.
Click the Shaded Area tab.
Choose the X Value and Both Tails for the region of the curve to be shaded.
Enter the X Value as 2.75.
Click OK.
The output obtained is as follows:

Introduction to Statistical Quality Control, Chapter 4, Problem 1E , additional homework tip 1

From the output, the P-value at each tail is obtained as 0.002980.

Since the alternative hypothesis (H₁) states that the mean is not equal to 100 (two-tail test), the P-value of the test statistic is calculated by adding the Probability values at both ends of the tail.

The P-value is calculated for two-tailed test is as follows:

$P-Value=0.002980+0.002980=0.00596≃0.0060$

Thus, the P-value is 0.0060.

b.

Expert Solution

To determine

Find P-value of the test statistic $Z0=1.86$ .

Answer to Problem 1E

The P-value is 0.0629.

Explanation of Solution

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

Choose Graph>Probability Distribution Plot, choose View Probability> OK.
From the Distribution drop down list box, choose Normal distribution.
Click the Shaded Area tab.
Choose the X Value and Two Tail for the region of the curve to be shaded.
Enter the X Value as 1.86.
Click OK.
The output obtained is as follows:

Introduction to Statistical Quality Control, Chapter 4, Problem 1E , additional homework tip 2

From the output, the P-value at each tail is obtained as 0.03144.

The P-value is calculated for two-tailed test is as follows:

$P-Value=0.03144+0.03144=0.06288≃0.0629$

Thus, the P-value is 0.0629.

c.

Expert Solution

To determine

Find P-value of the test statistic $Z0=−2.05$ .

Answer to Problem 1E

The P-value is 0.0404.

Explanation of Solution

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

Choose Graph>Probability Distribution Plot, choose View Probability> OK.
From the Distribution drop down list box, choose Normal distribution.
Click the Shaded Area tab.
Choose the X Value and Two Tail for the region of the curve to be shaded.
Enter the X Value as -2.05.
Click OK.
The output obtained is as follows:

Introduction to Statistical Quality Control, Chapter 4, Problem 1E , additional homework tip 3

From the output, the P-value at each tail is obtained as 0.02018.

The P-value is calculated for two-tailed test is as follows:

$P-Value=0.02018+0.02018=0.04036≃0.0404$

Thus, the P-value is 0.0404.

d.

Expert Solution

To determine

Find P-value of the test statistic $Z0=−1.86$ .

Answer to Problem 1E

The P-value is 0.0629.

Explanation of Solution

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

Choose Graph>Probability Distribution Plot, choose View Probability> OK.
From the Distribution drop down list box, choose Normal distribution.
Click the Shaded Area tab.
Choose the X Value and Two Tail for the region of the curve to be shaded.
Enter the X Value as –1.86.
Click OK.
The output obtained is as follows:

Introduction to Statistical Quality Control, Chapter 4, Problem 1E , additional homework tip 4

From the output, the P-value at each tail is obtained as 0.03144.

The P-value is calculated for two-tailed test is as follows:

$P-Value=0.03144+0.03144=0.06288≃0.0629$

Thus, the P-value is 0.0629.

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Answer 3

Textbook Question

Chapter 4, Problem 1E

Suppose that you are testing the following hypotheses where the variance is known:

Find the P-value for the following values of the test statistic.

$H 0 : μ = 100 H 1 : μ ≠ 100$

(a) Z₀ = 2.75

(b) Z₀ = 1.86

(c) Z₀ = −2.05

(d) Z₀ = −1.86

a.

Expert Solution

To determine

Find P-value of the test statistic $Z0=2.75$ .

Answer to Problem 1E

The P-value is 0.0060.

Explanation of Solution

As the alternative hypothesis states that $μ≠100$ , the given test is a two-tailed test.

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

Choose Graph>Probability Distribution Plot, choose View Probability> OK.
From the Distribution drop down list box, choose Normal distribution.
Click the Shaded Area tab.
Choose the X Value and Both Tails for the region of the curve to be shaded.
Enter the X Value as 2.75.
Click OK.
The output obtained is as follows:

Introduction to Statistical Quality Control, Chapter 4, Problem 1E , additional homework tip 1

From the output, the P-value at each tail is obtained as 0.002980.

Since the alternative hypothesis (H₁) states that the mean is not equal to 100 (two-tail test), the P-value of the test statistic is calculated by adding the Probability values at both ends of the tail.

The P-value is calculated for two-tailed test is as follows:

$P-Value=0.002980+0.002980=0.00596≃0.0060$

Thus, the P-value is 0.0060.

b.

Expert Solution

To determine

Find P-value of the test statistic $Z0=1.86$ .

Answer to Problem 1E

The P-value is 0.0629.

Explanation of Solution

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

Choose Graph>Probability Distribution Plot, choose View Probability> OK.
From the Distribution drop down list box, choose Normal distribution.
Click the Shaded Area tab.
Choose the X Value and Two Tail for the region of the curve to be shaded.
Enter the X Value as 1.86.
Click OK.
The output obtained is as follows:

Introduction to Statistical Quality Control, Chapter 4, Problem 1E , additional homework tip 2

From the output, the P-value at each tail is obtained as 0.03144.

The P-value is calculated for two-tailed test is as follows:

$P-Value=0.03144+0.03144=0.06288≃0.0629$

Thus, the P-value is 0.0629.

c.

Expert Solution

To determine

Find P-value of the test statistic $Z0=−2.05$ .

Answer to Problem 1E

The P-value is 0.0404.

Explanation of Solution

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

Choose Graph>Probability Distribution Plot, choose View Probability> OK.
From the Distribution drop down list box, choose Normal distribution.
Click the Shaded Area tab.
Choose the X Value and Two Tail for the region of the curve to be shaded.
Enter the X Value as -2.05.
Click OK.
The output obtained is as follows:

Introduction to Statistical Quality Control, Chapter 4, Problem 1E , additional homework tip 3

From the output, the P-value at each tail is obtained as 0.02018.

The P-value is calculated for two-tailed test is as follows:

$P-Value=0.02018+0.02018=0.04036≃0.0404$

Thus, the P-value is 0.0404.

d.

Expert Solution

To determine

Find P-value of the test statistic $Z0=−1.86$ .

Answer to Problem 1E

The P-value is 0.0629.

Explanation of Solution

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

Choose Graph>Probability Distribution Plot, choose View Probability> OK.
From the Distribution drop down list box, choose Normal distribution.
Click the Shaded Area tab.
Choose the X Value and Two Tail for the region of the curve to be shaded.
Enter the X Value as –1.86.
Click OK.
The output obtained is as follows:

Introduction to Statistical Quality Control, Chapter 4, Problem 1E , additional homework tip 4

From the output, the P-value at each tail is obtained as 0.03144.

The P-value is calculated for two-tailed test is as follows:

$P-Value=0.03144+0.03144=0.06288≃0.0629$

Thus, the P-value is 0.0629.

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Answer 4

Textbook Question

Chapter 4, Problem 1E

Suppose that you are testing the following hypotheses where the variance is known:

Find the P-value for the following values of the test statistic.

$H 0 : μ = 100 H 1 : μ ≠ 100$

(a) Z₀ = 2.75

(b) Z₀ = 1.86

(c) Z₀ = −2.05

(d) Z₀ = −1.86

a.

Expert Solution

To determine

Find P-value of the test statistic $Z0=2.75$ .

Answer to Problem 1E

The P-value is 0.0060.

Explanation of Solution

As the alternative hypothesis states that $μ≠100$ , the given test is a two-tailed test.

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

Choose Graph>Probability Distribution Plot, choose View Probability> OK.
From the Distribution drop down list box, choose Normal distribution.
Click the Shaded Area tab.
Choose the X Value and Both Tails for the region of the curve to be shaded.
Enter the X Value as 2.75.
Click OK.
The output obtained is as follows:

Introduction to Statistical Quality Control, Chapter 4, Problem 1E , additional homework tip 1

From the output, the P-value at each tail is obtained as 0.002980.

Since the alternative hypothesis (H₁) states that the mean is not equal to 100 (two-tail test), the P-value of the test statistic is calculated by adding the Probability values at both ends of the tail.

The P-value is calculated for two-tailed test is as follows:

$P-Value=0.002980+0.002980=0.00596≃0.0060$

Thus, the P-value is 0.0060.

b.

Expert Solution

To determine

Find P-value of the test statistic $Z0=1.86$ .

Answer to Problem 1E

The P-value is 0.0629.

Explanation of Solution

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

Choose Graph>Probability Distribution Plot, choose View Probability> OK.
From the Distribution drop down list box, choose Normal distribution.
Click the Shaded Area tab.
Choose the X Value and Two Tail for the region of the curve to be shaded.
Enter the X Value as 1.86.
Click OK.
The output obtained is as follows:

Introduction to Statistical Quality Control, Chapter 4, Problem 1E , additional homework tip 2

From the output, the P-value at each tail is obtained as 0.03144.

The P-value is calculated for two-tailed test is as follows:

$P-Value=0.03144+0.03144=0.06288≃0.0629$

Thus, the P-value is 0.0629.

c.

Expert Solution

To determine

Find P-value of the test statistic $Z0=−2.05$ .

Answer to Problem 1E

The P-value is 0.0404.

Explanation of Solution

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

Choose Graph>Probability Distribution Plot, choose View Probability> OK.
From the Distribution drop down list box, choose Normal distribution.
Click the Shaded Area tab.
Choose the X Value and Two Tail for the region of the curve to be shaded.
Enter the X Value as -2.05.
Click OK.
The output obtained is as follows:

Introduction to Statistical Quality Control, Chapter 4, Problem 1E , additional homework tip 3

From the output, the P-value at each tail is obtained as 0.02018.

The P-value is calculated for two-tailed test is as follows:

$P-Value=0.02018+0.02018=0.04036≃0.0404$

Thus, the P-value is 0.0404.

d.

Expert Solution

To determine

Find P-value of the test statistic $Z0=−1.86$ .

Answer to Problem 1E

The P-value is 0.0629.

Explanation of Solution

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

Choose Graph>Probability Distribution Plot, choose View Probability> OK.
From the Distribution drop down list box, choose Normal distribution.
Click the Shaded Area tab.
Choose the X Value and Two Tail for the region of the curve to be shaded.
Enter the X Value as –1.86.
Click OK.
The output obtained is as follows:

Introduction to Statistical Quality Control, Chapter 4, Problem 1E , additional homework tip 4

From the output, the P-value at each tail is obtained as 0.03144.

The P-value is calculated for two-tailed test is as follows:

$P-Value=0.03144+0.03144=0.06288≃0.0629$

Thus, the P-value is 0.0629.

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

Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

a.

Expert Solution

To determine

Find P-value of the test statistic $Z0=2.75$ .

Answer to Problem 1E

The P-value is 0.0060.

Explanation of Solution

As the alternative hypothesis states that $μ≠100$ , the given test is a two-tailed test.

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

Choose Graph>Probability Distribution Plot, choose View Probability> OK.
From the Distribution drop down list box, choose Normal distribution.
Click the Shaded Area tab.
Choose the X Value and Both Tails for the region of the curve to be shaded.
Enter the X Value as 2.75.
Click OK.
The output obtained is as follows:

Introduction to Statistical Quality Control, Chapter 4, Problem 1E , additional homework tip 1

From the output, the P-value at each tail is obtained as 0.002980.

Since the alternative hypothesis (H₁) states that the mean is not equal to 100 (two-tail test), the P-value of the test statistic is calculated by adding the Probability values at both ends of the tail.

The P-value is calculated for two-tailed test is as follows:

$P-Value=0.002980+0.002980=0.00596≃0.0060$

Thus, the P-value is 0.0060.

b.

Expert Solution

To determine

Find P-value of the test statistic $Z0=1.86$ .

Answer to Problem 1E

The P-value is 0.0629.

Explanation of Solution

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

Choose Graph>Probability Distribution Plot, choose View Probability> OK.
From the Distribution drop down list box, choose Normal distribution.
Click the Shaded Area tab.
Choose the X Value and Two Tail for the region of the curve to be shaded.
Enter the X Value as 1.86.
Click OK.
The output obtained is as follows:

Introduction to Statistical Quality Control, Chapter 4, Problem 1E , additional homework tip 2

From the output, the P-value at each tail is obtained as 0.03144.

The P-value is calculated for two-tailed test is as follows:

$P-Value=0.03144+0.03144=0.06288≃0.0629$

Thus, the P-value is 0.0629.

c.

Expert Solution

To determine

Find P-value of the test statistic $Z0=−2.05$ .

Answer to Problem 1E

The P-value is 0.0404.

Explanation of Solution

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

Choose Graph>Probability Distribution Plot, choose View Probability> OK.
From the Distribution drop down list box, choose Normal distribution.
Click the Shaded Area tab.
Choose the X Value and Two Tail for the region of the curve to be shaded.
Enter the X Value as -2.05.
Click OK.
The output obtained is as follows:

Introduction to Statistical Quality Control, Chapter 4, Problem 1E , additional homework tip 3

From the output, the P-value at each tail is obtained as 0.02018.

The P-value is calculated for two-tailed test is as follows:

$P-Value=0.02018+0.02018=0.04036≃0.0404$

Thus, the P-value is 0.0404.

d.

Expert Solution

To determine

Find P-value of the test statistic $Z0=−1.86$ .

Answer to Problem 1E

The P-value is 0.0629.

Explanation of Solution

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

Choose Graph>Probability Distribution Plot, choose View Probability> OK.
From the Distribution drop down list box, choose Normal distribution.
Click the Shaded Area tab.
Choose the X Value and Two Tail for the region of the curve to be shaded.
Enter the X Value as –1.86.
Click OK.
The output obtained is as follows:

Introduction to Statistical Quality Control, Chapter 4, Problem 1E , additional homework tip 4

From the output, the P-value at each tail is obtained as 0.03144.

The P-value is calculated for two-tailed test is as follows:

$P-Value=0.03144+0.03144=0.06288≃0.0629$

Thus, the P-value is 0.0629.

Answer 8

a.

Expert Solution

To determine

Find P-value of the test statistic $Z0=2.75$ .

Answer to Problem 1E

The P-value is 0.0060.

Explanation of Solution

As the alternative hypothesis states that $μ≠100$ , the given test is a two-tailed test.

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

Choose Graph>Probability Distribution Plot, choose View Probability> OK.
From the Distribution drop down list box, choose Normal distribution.
Click the Shaded Area tab.
Choose the X Value and Both Tails for the region of the curve to be shaded.
Enter the X Value as 2.75.
Click OK.
The output obtained is as follows:

Introduction to Statistical Quality Control, Chapter 4, Problem 1E , additional homework tip 1

From the output, the P-value at each tail is obtained as 0.002980.

Since the alternative hypothesis (H₁) states that the mean is not equal to 100 (two-tail test), the P-value of the test statistic is calculated by adding the Probability values at both ends of the tail.

The P-value is calculated for two-tailed test is as follows:

$P-Value=0.002980+0.002980=0.00596≃0.0060$

Thus, the P-value is 0.0060.

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

Find P-value of the test statistic $Z0=2.75$ .

Answer 13

Answer to Problem 1E

The P-value is 0.0060.

Answer 14

Explanation of Solution

As the alternative hypothesis states that $μ≠100$ , the given test is a two-tailed test.

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

Choose Graph>Probability Distribution Plot, choose View Probability> OK.
From the Distribution drop down list box, choose Normal distribution.
Click the Shaded Area tab.
Choose the X Value and Both Tails for the region of the curve to be shaded.
Enter the X Value as 2.75.
Click OK.
The output obtained is as follows:

Introduction to Statistical Quality Control, Chapter 4, Problem 1E , additional homework tip 1

From the output, the P-value at each tail is obtained as 0.002980.

Since the alternative hypothesis (H₁) states that the mean is not equal to 100 (two-tail test), the P-value of the test statistic is calculated by adding the Probability values at both ends of the tail.

The P-value is calculated for two-tailed test is as follows:

$P-Value=0.002980+0.002980=0.00596≃0.0060$

Thus, the P-value is 0.0060.

Answer 15

b.

Expert Solution

To determine

Find P-value of the test statistic $Z0=1.86$ .

Answer to Problem 1E

The P-value is 0.0629.

Explanation of Solution

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

Choose Graph>Probability Distribution Plot, choose View Probability> OK.
From the Distribution drop down list box, choose Normal distribution.
Click the Shaded Area tab.
Choose the X Value and Two Tail for the region of the curve to be shaded.
Enter the X Value as 1.86.
Click OK.
The output obtained is as follows:

Introduction to Statistical Quality Control, Chapter 4, Problem 1E , additional homework tip 2

From the output, the P-value at each tail is obtained as 0.03144.

The P-value is calculated for two-tailed test is as follows:

$P-Value=0.03144+0.03144=0.06288≃0.0629$

Thus, the P-value is 0.0629.

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

Find P-value of the test statistic $Z0=1.86$ .

Answer 20

Answer to Problem 1E

The P-value is 0.0629.

Answer 21

Explanation of Solution

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

Choose Graph>Probability Distribution Plot, choose View Probability> OK.
From the Distribution drop down list box, choose Normal distribution.
Click the Shaded Area tab.
Choose the X Value and Two Tail for the region of the curve to be shaded.
Enter the X Value as 1.86.
Click OK.
The output obtained is as follows:

Introduction to Statistical Quality Control, Chapter 4, Problem 1E , additional homework tip 2

From the output, the P-value at each tail is obtained as 0.03144.

The P-value is calculated for two-tailed test is as follows:

$P-Value=0.03144+0.03144=0.06288≃0.0629$

Thus, the P-value is 0.0629.

Answer 22

c.

Expert Solution

To determine

Find P-value of the test statistic $Z0=−2.05$ .

Answer to Problem 1E

The P-value is 0.0404.

Explanation of Solution

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

Choose Graph>Probability Distribution Plot, choose View Probability> OK.
From the Distribution drop down list box, choose Normal distribution.
Click the Shaded Area tab.
Choose the X Value and Two Tail for the region of the curve to be shaded.
Enter the X Value as -2.05.
Click OK.
The output obtained is as follows:

Introduction to Statistical Quality Control, Chapter 4, Problem 1E , additional homework tip 3

From the output, the P-value at each tail is obtained as 0.02018.

The P-value is calculated for two-tailed test is as follows:

$P-Value=0.02018+0.02018=0.04036≃0.0404$

Thus, the P-value is 0.0404.

Answer 23

c.

Expert Solution

Answer 24

c.

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

Find P-value of the test statistic $Z0=−2.05$ .

Answer 27

Answer to Problem 1E

The P-value is 0.0404.

Answer 28

Explanation of Solution

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

Choose Graph>Probability Distribution Plot, choose View Probability> OK.
From the Distribution drop down list box, choose Normal distribution.
Click the Shaded Area tab.
Choose the X Value and Two Tail for the region of the curve to be shaded.
Enter the X Value as -2.05.
Click OK.
The output obtained is as follows:

Introduction to Statistical Quality Control, Chapter 4, Problem 1E , additional homework tip 3

From the output, the P-value at each tail is obtained as 0.02018.

The P-value is calculated for two-tailed test is as follows:

$P-Value=0.02018+0.02018=0.04036≃0.0404$

Thus, the P-value is 0.0404.

Answer 29

d.

Expert Solution

To determine

Find P-value of the test statistic $Z0=−1.86$ .

Answer to Problem 1E

The P-value is 0.0629.

Explanation of Solution

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

Choose Graph>Probability Distribution Plot, choose View Probability> OK.
From the Distribution drop down list box, choose Normal distribution.
Click the Shaded Area tab.
Choose the X Value and Two Tail for the region of the curve to be shaded.
Enter the X Value as –1.86.
Click OK.
The output obtained is as follows:

Introduction to Statistical Quality Control, Chapter 4, Problem 1E , additional homework tip 4

From the output, the P-value at each tail is obtained as 0.03144.

The P-value is calculated for two-tailed test is as follows:

$P-Value=0.03144+0.03144=0.06288≃0.0629$

Thus, the P-value is 0.0629.

Answer 30

d.

Expert Solution

Answer 31

d.

Expert Solution

Answer 32

Expert Solution

Answer 33

To determine

Find P-value of the test statistic $Z0=−1.86$ .

Answer 34

Answer to Problem 1E

The P-value is 0.0629.

Answer 35

Explanation of Solution

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

Choose Graph>Probability Distribution Plot, choose View Probability> OK.
From the Distribution drop down list box, choose Normal distribution.
Click the Shaded Area tab.
Choose the X Value and Two Tail for the region of the curve to be shaded.
Enter the X Value as –1.86.
Click OK.
The output obtained is as follows: