Chapter 17, Problem 17.38E

(a)

To determine

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

Answer to Problem 17.38E

There is statistically significant difference at $\alpha =0.10$ 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 $\sigma =3,000$. Moreover, a sample of 20 pieces is selected from the population. Is there statistically significant evidence at the $\alpha =0.10$ level against the hypothesis that the mean is 32,500 pounds for the two-sided alternative?

PLAN:

Parameter:

Define the parameter $\mu$ 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:

$H_0:\mu =32,500$

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

Alternative hypothesis:

$H_a:\mu \ne 32,500$

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 $\sigma =3,000$.

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:

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 $P\text{-value}\le \alpha \left(=0.10\right)$, then reject the null hypothesis $\left(H_0\right)$.

If $P\text{-value}>\alpha \left(=0.10\right)$, then fail to reject the null hypothesis $\left(H_0\right)$.

CONCLUDE:

Use a significance level, $\alpha =0.10$.

Here, P-value is 0.013, which is lesser than the value of $\alpha =0.10$.

That is, $P\text{-value}\left(=0.013\right)<\alpha \left(=0.10\right)$.

Therefore, the null hypothesis is rejected.

Thus, there is statistically significant at $\alpha =0.10$ 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 $\alpha =0.10$ level against the hypothesis that the mean is 31,500 pounds or not for the two-sided alternative.

Answer to Problem 17.38E

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

Explanation of Solution

Calculation:

STATE:

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

PLAN:

Parameter:

Define the parameter $\mu$ 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:

$H_0:\mu =31,500$

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

Alternative hypothesis:

$H_a:\mu \ne 31,500$

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 $\sigma =3,000$.

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:

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

CONCLUDE:

Use a significance level, $\alpha =0.10$.

Here, P-value is 0.326, which is greater than the value of $\alpha =0.10$.

That is, $P\text{-value}\left(=0.326\right)>\alpha \left(=0.10\right)$.

Therefore, the null hypothesis is not rejected.

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