Question

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 34 doors is taken, and it is found that they have a mean of 2048millimeters. Assume a population variance of 441. Is there evidence at the 0.1 level that the doors are too short and unusable?

Step 4 of 6:

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

Answer 1

P-value = 0.0027

Explanation:

The null and alternative hypotheses for this test are given as below:

Null hypothesis: H0: The average height of doors is 2058.

Alternative hypothesis: Ha: The average height of doors is less than 2058.

H0: µ = 2058 versus Ha: µ < 2058

This is one tailed (lower tailed) test.

We are given

Xbar = 2048

σ^2 = 441

σ = sqrt(441) = 21

µ = 2058

n = 34

Test statistic formula is given as below:

Z = (Xbar - µ)/[σ/sqrt(n)]

Z = (2048 - 2058)/[21/sqrt(34)]

Z = -10/ 3.60147

Z = -2.77664

P-value = 0.0027

(by using z-table or excel)

P-value = 0.0027