Question

A company announced that the mean breaking strength of the cables it produces is 2140 lb with a variance of 32895 . A consumer protection agency, skeptical of the consistency among the company's several factories, claims that the actual variance,o^2 , is higher. The agency takes a random sample of 16 cables produced by the company. The cables have a sample mean breaking strength of 1644, with a sample variance of 66919. Is there enough evidence to conclude, at the .10 level of significance, that the true variance is higher than 32895?

Answer 1

To test whether the true variance of the breaking strength of the cables is higher than 32895, a one-tailed F-test can be used. The calculated F-value is greater than the critical F-value, providing enough evidence to support the agency's claim.

Step-by-step explanation:

To test whether the true variance of the breaking strength of the cables is higher than 32895, we can use a one-tailed F-test. The formula for the F-test is:

F = (sample variance)/(population variance)

In this case, the sample variance is 66919 and the population variance is 32895. Plugging these values into the formula gives:

F = 66919/32895 = 2.035

We can compare this F-value to the critical F-value at the .10 level of significance. Looking up the critical F-value in a table, we find that the critical F-value is 1.34.

Since the calculated F-value of 2.035 is greater than the critical F-value of 1.34, we have enough evidence to conclude, at the .10 level of significance, that the true variance is higher than 32895. Therefore, the consumer protection agency's claim is supported.