Question

A lumber company is making doors that are 2058.02058.0 millimeters tall. If the doors are too long they must be trimmed, and if the doors are too short they cannot be used. A sample of 77 is made, and it is found that they have a mean of 2046.02046.0 millimeters with a variance of 625.00625.00. A level of significance of 0.10.1 will be used to determine if the doors are either too long or too short. Assume the population distribution is approximately normal. Make the decision to reject or fail to reject the null hypothesis.

Answer 1

Given that, sample size (n) = 77, sample mean = 2046.0 millimetres and variance = 625.00

sample standard deviation (s) = √(625.00) = 25.0

The null and alternative hypotheses are,

H0 : μ = 2058.0

Ha : μ ≠ 2058.0

This hypothesis test is a right-tailed test.

Test statistic is,

=> Test statistic = -4.212

Degrees of freedom = 77 - 1 = 76

Using Excel we find the p-value as follows :

Excel Command : =TDIST (4.212, 76, 2) = 0.0001

=> p-value = 0.0001

Since, p-value is less than 0.10 level of significance,

The decision reached is Reject the null hypothesis.

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