Answer to Question #342353 in Statistics and Probability for Mac

Question #342353

In a study made on time and motion, it was found out that a certain manual work can be finished at an average time of 40 minutes with a standard deviation of 18 minutes. A group of 16 workers is given a special training and then found to average only 35 minutes. Can we conclude that the special training can speed up the work using a 0.05 level? 

1
Expert's answer
2022-05-19T09:38:20-0400

The following null and alternative hypotheses need to be tested:

"H_0:\\mu\\ge40"

"H_1:\\mu<40"

This corresponds to a left-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

Based on the information provided, the significance level is "\\alpha = 0.05," and the critical value for a left-tailed test is "z_c = -1.6449."

The rejection region for this left-tailed test is "R = \\{z:z<-1.6449\\}."

The z-statistic is computed as follows:


"z=\\dfrac{\\bar{x}-\\mu}{\\sigma\/\\sqrt{n}}=\\dfrac{35-40}{18\/\\sqrt{16}}\\approx-1.111111"

Since it is observed that "z=-1.111111>-1.6449=z_c," it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is "p=P(z<-1.111111)=0.13326," and since "p=0.13326>0.05=\\alpha," it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean "\\mu"

is less than 40, at the "\\alpha = 0.05" significance level.



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