BASIC PRACTICE OF STATISTICS+LAUNCHPAD
BASIC PRACTICE OF STATISTICS+LAUNCHPAD
8th Edition
ISBN: 9781319053093
Author: Moore
Publisher: MAC HIGHER
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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
Check Mark

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:μ32,500

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 STATISTICS+LAUNCHPAD, 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-valueα(=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
Check Mark

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:μ31,500

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 STATISTICS+LAUNCHPAD, 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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