Question

Lazarus Steel Corporation produces iron rods that are supposed to be 36 inches long. The machine that makes these rods does not produce each rod exactly 36 inches long. The lengths of the rods are normally distributed, and they vary slightly. It is known that when the machine is working properly, the mean length of the rods is 36 inches. The standard deviation of the lengths of all rods produced on this machine is always equal to .035 inch.

The quality control department at the company takes a sample of 20 such rods every week, calculates the mean length of these rods, and tests the null hypothesis, µ = 36 inches, against the alternative hypothesis, µ ? 36 inches. If the null hypothesis is rejected, the machine is stopped and adjusted. A recent sample of 20 rods produced a mean length of 36.015 inches.

a. Calculate the p-value for this test of hypothesis. Based on this p-value, will the quality control inspector decide to stop the machine and adjust it if he chooses the maximum probability of Type I error to be .02? What if the maximum probability of a Type I error is .10?

b. Test the hypothesis of part a using the critical-value approach and ? = 0.1. Does the machine need to be adjusted? What if ? = 0.5

Answer 1

Solution:-

a)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u = 36
Alternative hypothesis: u $\neq$ 36

Note that these hypotheses constitute a two-tailed test.  

Formulate an analysis plan. For this analysis, the significance level is 0.02. The test method is a one-sample z-test.

Analyze sample data. Using sample data, we compute the standard error (SE), z statistic test statistic (z).

SE = s / sqrt(n)

S.E = 0.00783

z = (x - u) / SE

z = 1.916

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.

Since we have a two-tailed test, the P-value is the probability that the z statistic less than -1.916 or greater than 1.916.

Thus, the P-value = 0.055

Interpret results. Since the P-value (0.055) is greater than the significance level (0.02), we cannot reject the null hypothesis.

Thus, the P-value = 0.055

Interpret results. Since the P-value (0.055) is less than the significance level (0.10), we have to reject the null hypothesis.

b)

z = 1.916

z_critical = + 1.645

Interpret results. Since the z-value (1.916) is greater than the z-critical value, hence we have to reject the null hypothesis.