Question

Suppose that the director of manufacturing at a clothing factory needs to determine whether a new machine is producing a particular type of cloth according to the manufacturer’s specifications, which indicate that the cloth should have a mean breaking strength of 70 pounds and a standard deviation of 3.5 pounds. A sample of 49 pieces reveals a sample mean of 69.1 pounds. The director is willing to tolerate a probability of .05 of rejecting the null hypothesis when it is true. (5) State the null and alternative hypothesis. H_0: H_A: (3) State the level of significance and the critical value. (5) State the decision rules, label the sketch to show rejection region. (5) Compute the test statistic and the p-value and interpret its meaning (2) Is there evidence that the machine is not meeting the manufacturer’s specifications in terms of the average breaking strength? (Use α = 0.05).

Answer 1

2 Tailed Z test, Single Mean

Given: = 70 lbs, = 69.1 lbs, = 3.5 lbs, n = 49, = 0.05

The Hypothesis:

H0: = 70 : The mean breaking strength is equal to 70 pounds..

Ha: 70: The mean breaking strength is not equal to 70 pounds..

This is a 2 tailed test

The Level of significance = 0.05

The Critical Value: Z critical values = +1.96 and -1.96

The Decision Rule:   Reject H0, If Zobserved is > 1.96 or if Z observed is < -1.96

Also if P value is < , Then Reject H0..

The Test Statistic: The test statistic is given by the equation:

Z observed = -1.8

The p Value: The p value for Z = -1.8, p value = 0.0718

The probability of obtaining a test statistic as extreme as or greater than the one obtained, assuming the null hypothesis is true is 0.0718.

The Decision: Since Zobserved (-1.8) is greater than -1.96, We Fail to Reject H0.

Also since P value (0.0718) is < (0.05) , We Fail to Reject H0.

The Conclusion: There isn't sufficient evidence at the 95% significance level to conclude that the machine is not meeting the manufacturers specifications.

The rejection region is as below