Lazurus Steel Corporation produces iron rods that are supposed to be 36 inches l
ID: 3312263 • Letter: L
Question
Lazurus 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 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 0.035 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, inches, against the alternative hypothesis, u does not equal 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.
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 a Type I error to be 0.025?
Use the normal distribution table. Round your answer to four decimal places.
Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: = 36
Alternative hypothesis: 36
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample mean is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.025. The test method is a one-sample t-test.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = s / sqrt(n)
S.E = 0.00783
DF = n - 1 = 20 - 1
D.F = 19
t = (x - ) / SE
t = 1.92
where s is the standard deviation of the sample, x is the sample mean, 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 t statistic having 19 degrees of freedom is less than -1.92 or greater than 1.92.
Thus, the P-value = 0.0699
Interpret results. Since the P-value (0.0699) is greater than the significance level (0.025), we cannot reject the null hypothesis.
From the above test we do have sufficient evidence in the favor of the claim that The machine that makes these rods does not produce each rod exactly 36 inches long.
z-test
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: = 36
Alternative hypothesis: 36
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample mean is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.025. 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 - ) / SE
z = 1.92
where s is the standard deviation of the sample, x is the sample mean, 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 having 19 degrees of freedom is less than -1.92 or greater than 1.92.
Thus, the P-value = 0.0548
Interpret results. Since the P-value (0.0548) is greater than the significance level (0.025), we cannot reject the null hypothesis.
From the above test we do have sufficient evidence in the favor of the claim that The machine that makes these rods does not produce each rod exactly 36 inches long.
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