Answer to Question #337535 in Statistics and Probability for ann

Question

The mean length of certain construction lumber is supposed to be 8.5 feet. A random

sample of 21 pieces of such lumbers gives a sample mean of 8.3 feet and a sample

standard deviation of 1.2 feet. A builder claims that the mean of the lumber is different

from 8.5 feet. Does the datasupport the builder's claim at a= 0.05?

Step:

1.State the null and alternative hypothesis concerning the population mean, $\mu$ and the type of test to be used.

2.Specify the level of significance $\alpha$

3.State the decision rule

4.Collect the data and perform calculations.

5. Make a statistical decision.

6.State the conclusion.

Answer 1

1. The following null and alternative hypotheses need to be tested:

$H_0:\mu=8.5$

$H_a:\mu\not=8.5$

This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

2. Based on the information provided, the significance level is $\alpha = 0.05,$ $df=n-1=20$ degrees of freedom, and the critical value for a two-tailed test is $t_c = 2.085963.$

3. The rejection region for this two-tailed test is $R = \{t: |t| >2.085963\}.$

4. The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{8.3-8.5}{1.2/\sqrt{21}}\approx-0.76376$$

5. Since it is observed that $|t| =0.76376< 2.085963=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for two-tailed, $df=20$ degrees of freedom, $t=-0.76376$ is $p=0.453927,$ and since $p= 0.453927>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

6.Therefore, there is not enough evidence to claim that the population mean $\mu$

is different than 8.5, at the $\alpha = 0.05$ significance level.