Question

A carpenter is making doors that are 2058.02058.0 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 88 doors is made, and it is found that they have a mean of 2070.02070.0 millimeters with a standard deviation of 29.029.0. Is there evidence at the 0.0250.025 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.

Step 5 of 5: Make the decision to reject or fail to reject the null hypothesis.

Answer 1

Solution:

1)The null and alternative hypothesis are

H0: = 2058.0

H1: > 2058.0

2)

n = 88

= 2070.0

s = 29.0

= 0.025

Since population SD is unknown,we use t test.

The test statistics t is given by ..

t =  

= (2070.0 - 2058.0)/(29.0/88)

= 3.882

Test statistic t = 3.882

3)

One tailed test

(> sign in H1 indicates that "Right tailed test"(One tailed right sided )

4)

Here , n = 88 d.f. = n - 1 = 87

Now, >  sign in H1 indicates that the Right tailed test.

So, the critical value is   i.e.  

The critical region : t >    

  =   _0.025,87 = 1.988 (using t table)

Decision Rule : Reject the null hypothesis H₀ if the test statistic t > 1.988

5)

Decision: Reject the null hypothesis

(because t = 3.882 > )