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In: Statistics and Probability

Exercise 11-22 (LO11-1) Clark Heter is an industrial engineer at Lyons Products. He would like to...

Exercise 11-22 (LO11-1) Clark Heter is an industrial engineer at Lyons Products. He would like to determine whether there are more units produced on the night shift than on the day shift. The mean number of units produced by a sample of 58 day-shift workers was 343. The mean number of units produced by a sample of 64 night-shift workers was 353. Assume the population standard deviation of the number of units produced on the day shift is 24 and 33 on the night shift. At the 0.02 significance level, is the number of units produced on the night shift larger? Is this a one-tailed or a two-tailed test? One-tailed test Two-tailed test State the decision rule. (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.) Compute the value of the test statistic. (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.) What is your decision regarding H0? Reject H0. Do not reject H0.

Solutions

Expert Solution

Solution:-

This is one tailed test.

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

Null hypothesis: uDay> uNight
Alternative hypothesis: uDay < uNight

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.02. Using sample data, we will conduct a two-sample z-test of the null hypothesis.

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

SE = sqrt[(s12/n1) + (s22/n2)]
SE = 5.19102
z = [ (x1 - x2) - d ] / SE

z = - 1.926

where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is thesize of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between population means, and SE is the standard error.

The observed difference in sample means produced a z statistic of - 1.926.

Therefore, the P-value in this analysis is 0.027.

Interpret results. Since the P-value (0.027) is greater than the significance level (0.02), we failed to reject the null hypothesis.

Do not reject the H0.

At the 0.02 significance level, the number of units produced on the night shift larger.


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