Question

A carpenter is making doors that are 2058.0 millimeters tall. If the doors are too long they must be trimmed, and if they are too short they cannot be used. A sample of 14 doors is made, and it is found that they have a mean of 2071.0 millimeters with a standard deviation of 32.0. Is there evidence at the 0.05 level that the doors are too long and need to be trimmed? Assume the population distribution is approximately normal.

Step 1 of 5:

State the null and alternative hypotheses.

Step 2 of 5:

Find the value of the test statistic. Round your answer to three decimal places.

Step 3 of 5:

Specify if the test is one-tailed or two-tailed.

Step 4 of 5:

Determine the decision rule for rejecting the null hypothesis. Round your answer to three decimal places.

Answer

Step 5 of 5:

Make the decision to reject or fail to reject the null hypothesis.

Answer 1

Given that A carpenter is making doors that are = 2058.0 millimeters tall. If the doors are too long they must be trimmed, and if they are too short they cannot be used. A sample of n = 14 doors is made, and it is found that they have a mean of = 2071.0 millimeters with a standard deviation of s = 32.0.

Given that the distribution is normal hence at given 0.05 level of significance the following hypothesis is tested using the t-distribution.

Based on the claim the hypotheses are:

Step -1

Step-2

Test Statistic:

t = 1.520

Step 3:

Based on the hypothesis it will be a right tailed test.

Step - 4

Rejection region:

Based on the type of hypothesis the critical value for rejection region is calculated using the excel formula for t-distribution which is =T.INV(1-0.05, 13), thus the critical value is computed as 1.771.

So, reject the Ho if t > tc

Step-5

Conclusion:

Since the test statistic is less than critical score hence we do not reject the null hypothesis and conclude that there is insufficient evidence to support the claim that at the 0.05 level that the doors are too long and need to be trimmed.