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2.The Metropolitan Bus Company claims that the mean waiting time for a bus during rush hour is less than 5 minutes. A random sample of 20 waiting times has a mean of 3.7 minutes with a standard deviation of 2.1 minutes. At α = 0.01, test the bus company’s claim. Assume the distribution is normally distributed.

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In this problem, we know that the standard deviation is known and that the population is normally distributed.

Ha: population mean is less than 5 minutes

Ho: population mean is greater than or equal to 5 minutes

a = 0.01

This problem would be an example of a left-tailed test because the claim represents a mean that is less than 5 minutes; therefore our critical value would be a z-score corresponding to an area of a.

critical value = -2.29 so our rejection region would be the area of the graph to left of -2.29

standardized test statistic = z = 3.7-5/2.1/square root of 20 = -0.44 (corresponds to 0.33)

P is the area of the left tail; therefore, P = 0.33.

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We would fail to reject the null hypothesis in this situation because our test-statistic 0.33 falls outside of the rejection range; therefore at the 1% level of significance, there is not sufficient evidence to support the company’s claim that the mean waiting time is less than 5 minutes.

R Given that: x 3.7; s 2.1 and n 20 Here, the population standard deviation is unknown so we use t-distribution.Null hypothesis, population mean is greater than or equal to 5 minutesH0 :…

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