To test: Whether there is any statistically significant evidence at the α = 0.10 level against the hypothesis that the mean is 32,500 pounds or not for the two-sided alternative.

Question

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Answer 1

Question

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.

Chapter 17, Problem 17.38E

(a)

To determine

To test: Whether there is any statistically significant evidence at the α=0.10 level against the hypothesis that the mean is 32,500 pounds or not for the two-sided alternative.

(a)

Expert Solution

Answer to Problem 17.38E

There is statistically significant difference at α=0.10 level against the hypothesis that the mean is 32,500 pounds.

Explanation of Solution

Given info:

The data represents the sample of strength of pieces of wood and standard deviation 3,000 pounds.

Calculation:

STATE:

The strength of pieces of wood follows normal distribution with standard deviation σ=3,000 . Moreover, a sample of 20 pieces is selected from the population. Is there statistically significant evidence at the α=0.10 level against the hypothesis that the mean is 32,500 pounds for the two-sided alternative?

PLAN:

Parameter:

Define the parameter μ as the mean strength of pieces of wood.

The hypotheses are given below:

The claim of the problem is the mean strength is different from 32,500.

Null Hypothesis:

H0:μ=32,500

That is, the mean strength is equal to 32,500.

Alternative hypothesis:

Ha:μ≠32,500

That is, the mean strength is not equal to 32,500.

SOLVE:

Conditions for valid test:

A sample of 20 pieces of wood is randomly selected and strength of pieces of wood follows normal distribution with standard deviation σ=3,000 .

Test statistic and P-value:

Software procedure:

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

Choose Stat > Basic Statistics > 1-Sample Z.
In Samples in Column, enter the column of Strength of pieces.
In Standard deviation, enter 3,000.
In Perform hypothesis test, enter the test mean as 32,500.
Check Options, enter Confidence level as 90.
Choose not equal in alternative.
Click OK in all dialogue boxes.

Output using the MINITAB software is given below:

BASIC PRACTICE OF STATS-LL W/SAPLINGPLU, Chapter 17, Problem 17.38E , additional homework tip 1

From the MINITAB output, the test statistic is –2.47 and the P-value is 0.013.

Decision criteria for the P-value method:

If P-value≤α(=0.10) , then reject the null hypothesis (H0) .

If P-value>α(=0.10) , then fail to reject the null hypothesis (H0) .

CONCLUDE:

Use a significance level, α=0.10 .

Here, P-value is 0.013, which is lesser than the value of α=0.10 .

That is, P-value(=0.013)<α(=0.10) .

Therefore, the null hypothesis is rejected.

Thus, there is statistically significant at α=0.10 level against the hypothesis that the mean is 32,500 pounds.

(b)

To determine

To test: Whether there is any statistically significant evidence at the α=0.10 level against the hypothesis that the mean is 31,500 pounds or not for the two-sided alternative.

(b)

Expert Solution

Answer to Problem 17.38E

There is no statistically significant difference at α=0.10 level against the hypothesis that the mean is 31,500 pounds.

Explanation of Solution

Calculation:

STATE:

Is there statistically significant evidence at the α=0.10 level against the hypothesis that the mean is 31,500 pounds for the two-sided alternative?

PLAN:

Parameter:

Define the parameter μ as the mean strength of pieces of wood.

The hypotheses are given below:

The claim of the problem is the mean strength is different from 31,500.

Null Hypothesis:

H0:μ=31,500

That is, the mean strength is equal to 31,500.

Alternative hypothesis:

Ha:μ≠31,500

That is, the mean strength is not equal to 31,500.

SOLVE:

Conditions for valid test:

A sample of 20 pieces of wood is randomly selected and strength of pieces of wood follows normal distribution with standard deviation σ=3,000 .

Test statistic and P-value:

Software procedure:

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

Choose Stat > Basic Statistics > 1-Sample Z.
In Samples in Column, enter the column of Strength of pieces.
In Standard deviation, enter 3,000.
In Perform hypothesis test, enter the test mean as 31,500.
Check Options, enter Confidence level as 90.
Choose not equal in alternative.
Click OK in all dialogue boxes.

Output using the MINITAB software is given below:

BASIC PRACTICE OF STATS-LL W/SAPLINGPLU, Chapter 17, Problem 17.38E , additional homework tip 2

From the MINITAB output, the test statistic is –0.98 and the P-value is 0.326.

CONCLUDE:

Use a significance level, α=0.10 .

Here, P-value is 0.326, which is greater than the value of α=0.10 .

That is, P-value(=0.326)>α(=0.10) .

Therefore, the null hypothesis is not rejected.

Thus, there is no statistically significant difference at α=0.10 level against the hypothesis that the mean is 31,500 pounds.

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Answer 2

Question

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.

Chapter 17, Problem 17.38E

(a)

To determine

To test: Whether there is any statistically significant evidence at the α=0.10 level against the hypothesis that the mean is 32,500 pounds or not for the two-sided alternative.

(a)

Expert Solution

Answer to Problem 17.38E

There is statistically significant difference at α=0.10 level against the hypothesis that the mean is 32,500 pounds.

Explanation of Solution

Given info:

The data represents the sample of strength of pieces of wood and standard deviation 3,000 pounds.

Calculation:

STATE:

The strength of pieces of wood follows normal distribution with standard deviation σ=3,000 . Moreover, a sample of 20 pieces is selected from the population. Is there statistically significant evidence at the α=0.10 level against the hypothesis that the mean is 32,500 pounds for the two-sided alternative?

PLAN:

Parameter:

Define the parameter μ as the mean strength of pieces of wood.

The hypotheses are given below:

The claim of the problem is the mean strength is different from 32,500.

Null Hypothesis:

H0:μ=32,500

That is, the mean strength is equal to 32,500.

Alternative hypothesis:

Ha:μ≠32,500

That is, the mean strength is not equal to 32,500.

SOLVE:

Conditions for valid test:

A sample of 20 pieces of wood is randomly selected and strength of pieces of wood follows normal distribution with standard deviation σ=3,000 .

Test statistic and P-value:

Software procedure:

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

Choose Stat > Basic Statistics > 1-Sample Z.
In Samples in Column, enter the column of Strength of pieces.
In Standard deviation, enter 3,000.
In Perform hypothesis test, enter the test mean as 32,500.
Check Options, enter Confidence level as 90.
Choose not equal in alternative.
Click OK in all dialogue boxes.

Output using the MINITAB software is given below:

BASIC PRACTICE OF STATS-LL W/SAPLINGPLU, Chapter 17, Problem 17.38E , additional homework tip 1

From the MINITAB output, the test statistic is –2.47 and the P-value is 0.013.

Decision criteria for the P-value method:

If P-value≤α(=0.10) , then reject the null hypothesis (H0) .

If P-value>α(=0.10) , then fail to reject the null hypothesis (H0) .

CONCLUDE:

Use a significance level, α=0.10 .

Here, P-value is 0.013, which is lesser than the value of α=0.10 .

That is, P-value(=0.013)<α(=0.10) .

Therefore, the null hypothesis is rejected.

Thus, there is statistically significant at α=0.10 level against the hypothesis that the mean is 32,500 pounds.

(b)

To determine

To test: Whether there is any statistically significant evidence at the α=0.10 level against the hypothesis that the mean is 31,500 pounds or not for the two-sided alternative.

(b)

Expert Solution

Answer to Problem 17.38E

There is no statistically significant difference at α=0.10 level against the hypothesis that the mean is 31,500 pounds.

Explanation of Solution

Calculation:

STATE:

Is there statistically significant evidence at the α=0.10 level against the hypothesis that the mean is 31,500 pounds for the two-sided alternative?

PLAN:

Parameter:

Define the parameter μ as the mean strength of pieces of wood.

The hypotheses are given below:

The claim of the problem is the mean strength is different from 31,500.

Null Hypothesis:

H0:μ=31,500

That is, the mean strength is equal to 31,500.

Alternative hypothesis:

Ha:μ≠31,500

That is, the mean strength is not equal to 31,500.

SOLVE:

Conditions for valid test:

A sample of 20 pieces of wood is randomly selected and strength of pieces of wood follows normal distribution with standard deviation σ=3,000 .

Test statistic and P-value:

Software procedure:

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

Choose Stat > Basic Statistics > 1-Sample Z.
In Samples in Column, enter the column of Strength of pieces.
In Standard deviation, enter 3,000.
In Perform hypothesis test, enter the test mean as 31,500.
Check Options, enter Confidence level as 90.
Choose not equal in alternative.
Click OK in all dialogue boxes.

Output using the MINITAB software is given below:

BASIC PRACTICE OF STATS-LL W/SAPLINGPLU, Chapter 17, Problem 17.38E , additional homework tip 2

From the MINITAB output, the test statistic is –0.98 and the P-value is 0.326.

CONCLUDE:

Use a significance level, α=0.10 .

Here, P-value is 0.326, which is greater than the value of α=0.10 .

That is, P-value(=0.326)>α(=0.10) .

Therefore, the null hypothesis is not rejected.

Thus, there is no statistically significant difference at α=0.10 level against the hypothesis that the mean is 31,500 pounds.

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

Answer 3

Question

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.

Chapter 17, Problem 17.38E

(a)

To determine

To test: Whether there is any statistically significant evidence at the α=0.10 level against the hypothesis that the mean is 32,500 pounds or not for the two-sided alternative.

(a)

Expert Solution

Answer to Problem 17.38E

There is statistically significant difference at α=0.10 level against the hypothesis that the mean is 32,500 pounds.

Explanation of Solution

Given info:

The data represents the sample of strength of pieces of wood and standard deviation 3,000 pounds.

Calculation:

STATE:

The strength of pieces of wood follows normal distribution with standard deviation σ=3,000 . Moreover, a sample of 20 pieces is selected from the population. Is there statistically significant evidence at the α=0.10 level against the hypothesis that the mean is 32,500 pounds for the two-sided alternative?

PLAN:

Parameter:

Define the parameter μ as the mean strength of pieces of wood.

The hypotheses are given below:

The claim of the problem is the mean strength is different from 32,500.

Null Hypothesis:

H0:μ=32,500

That is, the mean strength is equal to 32,500.

Alternative hypothesis:

Ha:μ≠32,500

That is, the mean strength is not equal to 32,500.

SOLVE:

Conditions for valid test:

A sample of 20 pieces of wood is randomly selected and strength of pieces of wood follows normal distribution with standard deviation σ=3,000 .

Test statistic and P-value:

Software procedure:

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

Choose Stat > Basic Statistics > 1-Sample Z.
In Samples in Column, enter the column of Strength of pieces.
In Standard deviation, enter 3,000.
In Perform hypothesis test, enter the test mean as 32,500.
Check Options, enter Confidence level as 90.
Choose not equal in alternative.
Click OK in all dialogue boxes.

Output using the MINITAB software is given below:

BASIC PRACTICE OF STATS-LL W/SAPLINGPLU, Chapter 17, Problem 17.38E , additional homework tip 1

From the MINITAB output, the test statistic is –2.47 and the P-value is 0.013.

Decision criteria for the P-value method:

If P-value≤α(=0.10) , then reject the null hypothesis (H0) .

If P-value>α(=0.10) , then fail to reject the null hypothesis (H0) .

CONCLUDE:

Use a significance level, α=0.10 .

Here, P-value is 0.013, which is lesser than the value of α=0.10 .

That is, P-value(=0.013)<α(=0.10) .

Therefore, the null hypothesis is rejected.

Thus, there is statistically significant at α=0.10 level against the hypothesis that the mean is 32,500 pounds.

(b)

To determine

To test: Whether there is any statistically significant evidence at the α=0.10 level against the hypothesis that the mean is 31,500 pounds or not for the two-sided alternative.

(b)

Expert Solution

Answer to Problem 17.38E

There is no statistically significant difference at α=0.10 level against the hypothesis that the mean is 31,500 pounds.

Explanation of Solution

Calculation:

STATE:

Is there statistically significant evidence at the α=0.10 level against the hypothesis that the mean is 31,500 pounds for the two-sided alternative?

PLAN:

Parameter:

Define the parameter μ as the mean strength of pieces of wood.

The hypotheses are given below:

The claim of the problem is the mean strength is different from 31,500.

Null Hypothesis:

H0:μ=31,500

That is, the mean strength is equal to 31,500.

Alternative hypothesis:

Ha:μ≠31,500

That is, the mean strength is not equal to 31,500.

SOLVE:

Conditions for valid test:

A sample of 20 pieces of wood is randomly selected and strength of pieces of wood follows normal distribution with standard deviation σ=3,000 .

Test statistic and P-value:

Software procedure:

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

Choose Stat > Basic Statistics > 1-Sample Z.
In Samples in Column, enter the column of Strength of pieces.
In Standard deviation, enter 3,000.
In Perform hypothesis test, enter the test mean as 31,500.
Check Options, enter Confidence level as 90.
Choose not equal in alternative.
Click OK in all dialogue boxes.

Output using the MINITAB software is given below:

BASIC PRACTICE OF STATS-LL W/SAPLINGPLU, Chapter 17, Problem 17.38E , additional homework tip 2

From the MINITAB output, the test statistic is –0.98 and the P-value is 0.326.

CONCLUDE:

Use a significance level, α=0.10 .

Here, P-value is 0.326, which is greater than the value of α=0.10 .

That is, P-value(=0.326)>α(=0.10) .

Therefore, the null hypothesis is not rejected.

Thus, there is no statistically significant difference at α=0.10 level against the hypothesis that the mean is 31,500 pounds.

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

Answer 4

Question

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.

Chapter 17, Problem 17.38E

(a)

To determine

To test: Whether there is any statistically significant evidence at the α=0.10 level against the hypothesis that the mean is 32,500 pounds or not for the two-sided alternative.

(a)

Expert Solution

Answer to Problem 17.38E

There is statistically significant difference at α=0.10 level against the hypothesis that the mean is 32,500 pounds.

Explanation of Solution

Given info:

The data represents the sample of strength of pieces of wood and standard deviation 3,000 pounds.

Calculation:

STATE:

The strength of pieces of wood follows normal distribution with standard deviation σ=3,000 . Moreover, a sample of 20 pieces is selected from the population. Is there statistically significant evidence at the α=0.10 level against the hypothesis that the mean is 32,500 pounds for the two-sided alternative?

PLAN:

Parameter:

Define the parameter μ as the mean strength of pieces of wood.

The hypotheses are given below:

The claim of the problem is the mean strength is different from 32,500.

Null Hypothesis:

H0:μ=32,500

That is, the mean strength is equal to 32,500.

Alternative hypothesis:

Ha:μ≠32,500

That is, the mean strength is not equal to 32,500.

SOLVE:

Conditions for valid test:

A sample of 20 pieces of wood is randomly selected and strength of pieces of wood follows normal distribution with standard deviation σ=3,000 .

Test statistic and P-value:

Software procedure:

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

Choose Stat > Basic Statistics > 1-Sample Z.
In Samples in Column, enter the column of Strength of pieces.
In Standard deviation, enter 3,000.
In Perform hypothesis test, enter the test mean as 32,500.
Check Options, enter Confidence level as 90.
Choose not equal in alternative.
Click OK in all dialogue boxes.

Output using the MINITAB software is given below:

BASIC PRACTICE OF STATS-LL W/SAPLINGPLU, Chapter 17, Problem 17.38E , additional homework tip 1

From the MINITAB output, the test statistic is –2.47 and the P-value is 0.013.

Decision criteria for the P-value method:

If P-value≤α(=0.10) , then reject the null hypothesis (H0) .

If P-value>α(=0.10) , then fail to reject the null hypothesis (H0) .

CONCLUDE:

Use a significance level, α=0.10 .

Here, P-value is 0.013, which is lesser than the value of α=0.10 .

That is, P-value(=0.013)<α(=0.10) .

Therefore, the null hypothesis is rejected.

Thus, there is statistically significant at α=0.10 level against the hypothesis that the mean is 32,500 pounds.

(b)

To determine

To test: Whether there is any statistically significant evidence at the α=0.10 level against the hypothesis that the mean is 31,500 pounds or not for the two-sided alternative.

(b)

Expert Solution

Answer to Problem 17.38E

There is no statistically significant difference at α=0.10 level against the hypothesis that the mean is 31,500 pounds.

Explanation of Solution

Calculation:

STATE:

Is there statistically significant evidence at the α=0.10 level against the hypothesis that the mean is 31,500 pounds for the two-sided alternative?

PLAN:

Parameter:

Define the parameter μ as the mean strength of pieces of wood.

The hypotheses are given below:

The claim of the problem is the mean strength is different from 31,500.

Null Hypothesis:

H0:μ=31,500

That is, the mean strength is equal to 31,500.

Alternative hypothesis:

Ha:μ≠31,500

That is, the mean strength is not equal to 31,500.

SOLVE:

Conditions for valid test:

A sample of 20 pieces of wood is randomly selected and strength of pieces of wood follows normal distribution with standard deviation σ=3,000 .

Test statistic and P-value:

Software procedure:

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

Choose Stat > Basic Statistics > 1-Sample Z.
In Samples in Column, enter the column of Strength of pieces.
In Standard deviation, enter 3,000.
In Perform hypothesis test, enter the test mean as 31,500.
Check Options, enter Confidence level as 90.
Choose not equal in alternative.
Click OK in all dialogue boxes.

Output using the MINITAB software is given below:

BASIC PRACTICE OF STATS-LL W/SAPLINGPLU, Chapter 17, Problem 17.38E , additional homework tip 2

From the MINITAB output, the test statistic is –0.98 and the P-value is 0.326.

CONCLUDE:

Use a significance level, α=0.10 .

Here, P-value is 0.326, which is greater than the value of α=0.10 .

That is, P-value(=0.326)>α(=0.10) .

Therefore, the null hypothesis is not rejected.

Thus, there is no statistically significant difference at α=0.10 level against the hypothesis that the mean is 31,500 pounds.

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 17, Problem 17.38E

(a)

To determine

To test: Whether there is any statistically significant evidence at the α=0.10 level against the hypothesis that the mean is 32,500 pounds or not for the two-sided alternative.

(a)

Expert Solution

Answer to Problem 17.38E

There is statistically significant difference at α=0.10 level against the hypothesis that the mean is 32,500 pounds.

Explanation of Solution

Given info:

The data represents the sample of strength of pieces of wood and standard deviation 3,000 pounds.

Calculation:

STATE:

The strength of pieces of wood follows normal distribution with standard deviation σ=3,000 . Moreover, a sample of 20 pieces is selected from the population. Is there statistically significant evidence at the α=0.10 level against the hypothesis that the mean is 32,500 pounds for the two-sided alternative?

PLAN:

Parameter:

Define the parameter μ as the mean strength of pieces of wood.

The hypotheses are given below:

The claim of the problem is the mean strength is different from 32,500.

Null Hypothesis:

H0:μ=32,500

That is, the mean strength is equal to 32,500.

Alternative hypothesis:

Ha:μ≠32,500

That is, the mean strength is not equal to 32,500.

SOLVE:

Conditions for valid test:

A sample of 20 pieces of wood is randomly selected and strength of pieces of wood follows normal distribution with standard deviation σ=3,000 .

Test statistic and P-value:

Software procedure:

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

Choose Stat > Basic Statistics > 1-Sample Z.
In Samples in Column, enter the column of Strength of pieces.
In Standard deviation, enter 3,000.
In Perform hypothesis test, enter the test mean as 32,500.
Check Options, enter Confidence level as 90.
Choose not equal in alternative.
Click OK in all dialogue boxes.

Output using the MINITAB software is given below:

BASIC PRACTICE OF STATS-LL W/SAPLINGPLU, Chapter 17, Problem 17.38E , additional homework tip 1

From the MINITAB output, the test statistic is –2.47 and the P-value is 0.013.

Decision criteria for the P-value method:

If P-value≤α(=0.10) , then reject the null hypothesis (H0) .

If P-value>α(=0.10) , then fail to reject the null hypothesis (H0) .

CONCLUDE:

Use a significance level, α=0.10 .

Here, P-value is 0.013, which is lesser than the value of α=0.10 .

That is, P-value(=0.013)<α(=0.10) .

Therefore, the null hypothesis is rejected.

Thus, there is statistically significant at α=0.10 level against the hypothesis that the mean is 32,500 pounds.

(b)

To determine

To test: Whether there is any statistically significant evidence at the α=0.10 level against the hypothesis that the mean is 31,500 pounds or not for the two-sided alternative.

(b)

Expert Solution

Answer to Problem 17.38E

There is no statistically significant difference at α=0.10 level against the hypothesis that the mean is 31,500 pounds.

Explanation of Solution

Calculation:

STATE:

Is there statistically significant evidence at the α=0.10 level against the hypothesis that the mean is 31,500 pounds for the two-sided alternative?

PLAN:

Parameter:

Define the parameter μ as the mean strength of pieces of wood.

The hypotheses are given below:

The claim of the problem is the mean strength is different from 31,500.

Null Hypothesis:

H0:μ=31,500

That is, the mean strength is equal to 31,500.

Alternative hypothesis:

Ha:μ≠31,500

That is, the mean strength is not equal to 31,500.

SOLVE:

Conditions for valid test:

A sample of 20 pieces of wood is randomly selected and strength of pieces of wood follows normal distribution with standard deviation σ=3,000 .

Test statistic and P-value:

Software procedure:

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

Choose Stat > Basic Statistics > 1-Sample Z.
In Samples in Column, enter the column of Strength of pieces.
In Standard deviation, enter 3,000.
In Perform hypothesis test, enter the test mean as 31,500.
Check Options, enter Confidence level as 90.
Choose not equal in alternative.
Click OK in all dialogue boxes.

Output using the MINITAB software is given below:

BASIC PRACTICE OF STATS-LL W/SAPLINGPLU, Chapter 17, Problem 17.38E , additional homework tip 2

From the MINITAB output, the test statistic is –0.98 and the P-value is 0.326.

CONCLUDE:

Use a significance level, α=0.10 .

Here, P-value is 0.326, which is greater than the value of α=0.10 .

That is, P-value(=0.326)>α(=0.10) .

Therefore, the null hypothesis is not rejected.

Thus, there is no statistically significant difference at α=0.10 level against the hypothesis that the mean is 31,500 pounds.

Answer 6

Chapter 17, Problem 17.38E

(a)

To determine

To test: Whether there is any statistically significant evidence at the α=0.10 level against the hypothesis that the mean is 32,500 pounds or not for the two-sided alternative.

(a)

Expert Solution

Answer to Problem 17.38E

There is statistically significant difference at α=0.10 level against the hypothesis that the mean is 32,500 pounds.

Explanation of Solution

Given info:

The data represents the sample of strength of pieces of wood and standard deviation 3,000 pounds.

Calculation:

STATE:

The strength of pieces of wood follows normal distribution with standard deviation σ=3,000 . Moreover, a sample of 20 pieces is selected from the population. Is there statistically significant evidence at the α=0.10 level against the hypothesis that the mean is 32,500 pounds for the two-sided alternative?

PLAN:

Parameter:

Define the parameter μ as the mean strength of pieces of wood.

The hypotheses are given below:

The claim of the problem is the mean strength is different from 32,500.

Null Hypothesis:

H0:μ=32,500

That is, the mean strength is equal to 32,500.

Alternative hypothesis:

Ha:μ≠32,500

That is, the mean strength is not equal to 32,500.

SOLVE:

Conditions for valid test:

A sample of 20 pieces of wood is randomly selected and strength of pieces of wood follows normal distribution with standard deviation σ=3,000 .

Test statistic and P-value:

Software procedure:

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

Choose Stat > Basic Statistics > 1-Sample Z.
In Samples in Column, enter the column of Strength of pieces.
In Standard deviation, enter 3,000.
In Perform hypothesis test, enter the test mean as 32,500.
Check Options, enter Confidence level as 90.
Choose not equal in alternative.
Click OK in all dialogue boxes.

Output using the MINITAB software is given below:

BASIC PRACTICE OF STATS-LL W/SAPLINGPLU, Chapter 17, Problem 17.38E , additional homework tip 1

From the MINITAB output, the test statistic is –2.47 and the P-value is 0.013.

Decision criteria for the P-value method:

If P-value≤α(=0.10) , then reject the null hypothesis (H0) .

If P-value>α(=0.10) , then fail to reject the null hypothesis (H0) .

CONCLUDE:

Use a significance level, α=0.10 .

Here, P-value is 0.013, which is lesser than the value of α=0.10 .

That is, P-value(=0.013)<α(=0.10) .

Therefore, the null hypothesis is rejected.

Thus, there is statistically significant at α=0.10 level against the hypothesis that the mean is 32,500 pounds.

Answer 7

Chapter 17, Problem 17.38E

(a)

To determine

To test: Whether there is any statistically significant evidence at the α=0.10 level against the hypothesis that the mean is 32,500 pounds or not for the two-sided alternative.

Answer 8

(a)

Expert Solution

Answer to Problem 17.38E

There is statistically significant difference at α=0.10 level against the hypothesis that the mean is 32,500 pounds.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given info:

The data represents the sample of strength of pieces of wood and standard deviation 3,000 pounds.

Calculation:

STATE:

The strength of pieces of wood follows normal distribution with standard deviation σ=3,000 . Moreover, a sample of 20 pieces is selected from the population. Is there statistically significant evidence at the α=0.10 level against the hypothesis that the mean is 32,500 pounds for the two-sided alternative?

PLAN:

Parameter:

Define the parameter μ as the mean strength of pieces of wood.

The hypotheses are given below:

The claim of the problem is the mean strength is different from 32,500.

Null Hypothesis:

H0:μ=32,500

That is, the mean strength is equal to 32,500.

Alternative hypothesis:

Ha:μ≠32,500

That is, the mean strength is not equal to 32,500.

SOLVE:

Conditions for valid test:

A sample of 20 pieces of wood is randomly selected and strength of pieces of wood follows normal distribution with standard deviation σ=3,000 .

Test statistic and P-value:

Software procedure:

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

Choose Stat > Basic Statistics > 1-Sample Z.
In Samples in Column, enter the column of Strength of pieces.
In Standard deviation, enter 3,000.
In Perform hypothesis test, enter the test mean as 32,500.
Check Options, enter Confidence level as 90.
Choose not equal in alternative.
Click OK in all dialogue boxes.

Output using the MINITAB software is given below:

BASIC PRACTICE OF STATS-LL W/SAPLINGPLU, Chapter 17, Problem 17.38E , additional homework tip 1

From the MINITAB output, the test statistic is –2.47 and the P-value is 0.013.

Decision criteria for the P-value method:

If P-value≤α(=0.10) , then reject the null hypothesis (H0) .

If P-value>α(=0.10) , then fail to reject the null hypothesis (H0) .

CONCLUDE:

Use a significance level, α=0.10 .

Here, P-value is 0.013, which is lesser than the value of α=0.10 .

That is, P-value(=0.013)<α(=0.10) .

Therefore, the null hypothesis is rejected.

Thus, there is statistically significant at α=0.10 level against the hypothesis that the mean is 32,500 pounds.

Answer 13

(b)

To determine

To test: Whether there is any statistically significant evidence at the α=0.10 level against the hypothesis that the mean is 31,500 pounds or not for the two-sided alternative.

(b)

Expert Solution

Answer to Problem 17.38E

There is no statistically significant difference at α=0.10 level against the hypothesis that the mean is 31,500 pounds.

Explanation of Solution

Calculation:

STATE:

Is there statistically significant evidence at the α=0.10 level against the hypothesis that the mean is 31,500 pounds for the two-sided alternative?

PLAN:

Parameter:

Define the parameter μ as the mean strength of pieces of wood.

The hypotheses are given below:

The claim of the problem is the mean strength is different from 31,500.

Null Hypothesis:

H0:μ=31,500

That is, the mean strength is equal to 31,500.

Alternative hypothesis:

Ha:μ≠31,500

That is, the mean strength is not equal to 31,500.

SOLVE:

Conditions for valid test:

A sample of 20 pieces of wood is randomly selected and strength of pieces of wood follows normal distribution with standard deviation σ=3,000 .

Test statistic and P-value:

Software procedure:

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

Choose Stat > Basic Statistics > 1-Sample Z.
In Samples in Column, enter the column of Strength of pieces.
In Standard deviation, enter 3,000.
In Perform hypothesis test, enter the test mean as 31,500.
Check Options, enter Confidence level as 90.
Choose not equal in alternative.
Click OK in all dialogue boxes.

Output using the MINITAB software is given below:

BASIC PRACTICE OF STATS-LL W/SAPLINGPLU, Chapter 17, Problem 17.38E , additional homework tip 2

From the MINITAB output, the test statistic is –0.98 and the P-value is 0.326.

CONCLUDE:

Use a significance level, α=0.10 .

Here, P-value is 0.326, which is greater than the value of α=0.10 .

That is, P-value(=0.326)>α(=0.10) .

Therefore, the null hypothesis is not rejected.

Thus, there is no statistically significant difference at α=0.10 level against the hypothesis that the mean is 31,500 pounds.

Answer 14

(b)

To determine

To test: Whether there is any statistically significant evidence at the α=0.10 level against the hypothesis that the mean is 31,500 pounds or not for the two-sided alternative.

Answer 15

(b)

Expert Solution

Answer to Problem 17.38E

There is no statistically significant difference at α=0.10 level against the hypothesis that the mean is 31,500 pounds.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Calculation:

STATE:

Is there statistically significant evidence at the α=0.10 level against the hypothesis that the mean is 31,500 pounds for the two-sided alternative?

PLAN:

Parameter:

Define the parameter μ as the mean strength of pieces of wood.

The hypotheses are given below:

The claim of the problem is the mean strength is different from 31,500.

Null Hypothesis:

H0:μ=31,500

That is, the mean strength is equal to 31,500.

Alternative hypothesis:

Ha:μ≠31,500

That is, the mean strength is not equal to 31,500.

SOLVE:

Conditions for valid test:

A sample of 20 pieces of wood is randomly selected and strength of pieces of wood follows normal distribution with standard deviation σ=3,000 .

Test statistic and P-value:

Software procedure:

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

Choose Stat > Basic Statistics > 1-Sample Z.
In Samples in Column, enter the column of Strength of pieces.
In Standard deviation, enter 3,000.
In Perform hypothesis test, enter the test mean as 31,500.
Check Options, enter Confidence level as 90.
Choose not equal in alternative.
Click OK in all dialogue boxes.

Output using the MINITAB software is given below:

BASIC PRACTICE OF STATS-LL W/SAPLINGPLU, Chapter 17, Problem 17.38E , additional homework tip 2

From the MINITAB output, the test statistic is –0.98 and the P-value is 0.326.

CONCLUDE:

Use a significance level, α=0.10 .

Here, P-value is 0.326, which is greater than the value of α=0.10 .

That is, P-value(=0.326)>α(=0.10) .

Therefore, the null hypothesis is not rejected.

Thus, there is no statistically significant difference at α=0.10 level against the hypothesis that the mean is 31,500 pounds.

Answer 20

Ch. 17.1 - Prob. 1AYK Ch. 17.1 - Prob. 2AYK Ch. 17.2 - Prob. 3AYK Ch. 17.2 - Prob. 4AYK Ch. 17.2 - Prob. 5AYK Ch. 17.2 - Prob. 6AYK Ch. 17.2 - Prob. 7AYK Ch. 17.3 - Prob. 8AYK Ch. 17.3 - Prob. 9AYK Ch. 17.3 - Prob. 10AYK

To test: Whether there is any statistically significant evidence at the α = 0.10 level against the hypothesis that the mean is 32,500 pounds or not for the two-sided alternative.

Videos

To test: Whether there is any statistically significant evidence at the α=0.10 level against the hypothesis that the mean is 32,500 pounds or not for the two-sided alternative.

Answer to Problem 17.38E

Explanation of Solution

To test: Whether there is any statistically significant evidence at the α=0.10 level against the hypothesis that the mean is 31,500 pounds or not for the two-sided alternative.

Answer to Problem 17.38E

Explanation of Solution

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