Question

The metropolitan bus company claims that the mean wait time for a bus during rush hour is less than 7 minutes. A random sample of 20 waiting times has a mean of 5.6 minutes with a standard deviation of 2.1 minutes. At a= 0.01, test the bus company’s claim. Assume the distribution is normally distributed.

Answer 1

Solution:-

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

Null hypothesis: u > 7
Alternative hypothesis: u < 7

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.01. 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 = 0.46957

z = (x - u) / SE

z = - 2.98

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.

The observed sample mean produced a z statistic test statistic of 2.981.

Thus the P-value in this analysis is 0.001

Interpret results. Since the P-value (0.001) is less than the significance level (0.01), we have to reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that the mean wait time for a bus during rush hour is less than 7 minutes.