Question

A bolt manufacturer is trying to calibrate a new machine that is to produce 0.23cm0.23⁢cm bolts. The manufacturer takes a random sample of 2929 bolts produced by the new machine and finds that the standard deviation is 0.13350.1335. If the machine is operating properly, the variance of the diameters of the bolts should be 0.0150.015. Does the manufacturer have evidence at the α=0.01α=0.01 level that the variance of the bolt diameters is more than required? Assume the population is normally distributed.

Step 1 of 5:

State the null and alternative hypotheses. Round to four decimal places when necessary.

Step 2 of 5:

Determine the critical value(s) of the test statistic. If the test is two-tailed, separate the values with a comma. Round your answer to three decimal places.

Step 3 of 5:

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

Step 4 of 5:

Make the decision.

Step 5 of 5:

What is the conclusion?

Answer 1

Step 1 of 5 : The null and alternative hypothesis is ,

Step 2 of 5 : Now , df=degrees of freedom=n-1=29-1=28

The critical value is ,

; The Excel function is ,=CHIINV(0.01,28)

Step 3 of 5 : The value of the test statistic is ,

Step 4 of 5 : Decision : Here , the value of the test statistic does not lies in the rejection region

Therefore , fail to reject Ho.

Step 5 of 5 : Conclusion : Hence , there is not enough evidence to support the claim that the variance of the bolt diameters is more than required.