Question

A carpenter is making doors that are 20582058 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 3333 doors is taken, and it is found that they have a mean of 20472047 millimeters. Assume a population standard deviation of 2323. Is there evidence at the 0.020.02 level that the doors are too short and unusable?

1.)Find the value of the test statistic. Round your answer to two decimal places.

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

3.)Find the P-value of the test statistic. Round your answer to four decimal places.

4.)Identify the level of significance for the hypothesis test.

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

Answer 1

Given a carpenter is making doors that are = 2058 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 = 33 doors is taken, and it is found that they have a mean of = 2047 millimeters. Assuming a population standard deviation of = 23.

The claim is that, is there evidence at the = 0.02 level that the doors are too short and unusable?

Thus based on the claim the hypotheses are:

Since the sample size is greater than 30 and the population standard deviation is known hence Z-statistic is applicable for hypothesis testing.

1) Test statistic:

Thus Z = -2.75

2) Type of test

Now based on the hypothesis it will be one-tailed test.

3) P-value:

The P-value is computed using the excel formual for normal distribution which is =NORM.S.DIST(-2.747, TRUE), thus the P-value is computed as:

P-value = 0.0030.

Rejection region:

Reject Ho if P-value is less than 0.02.

4) Level of significance is given as = 0.02.

5) Conclusion:

Since the P-value = 0.0030 which is less than = 0.02 hence we reject the null hypothesis and conclude that there is sufficient evidence to support the claim that the doors are too short and unusable.