Question

an individuals present route to work results in, on average, 40 minutes of travel time per trip. an alternate route has been suggested that will reduce the travel time. suppose the new route was tried on 10 randomly chosen occasions and the average travel time was 38.17 minutes and the standard deviation was 2.97 minutes. do these data establish the claim that the new route is shorter, at the 5% level significance (Assume the distribution is normal)

Answer 1

Solution:-

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

Null hypothesis: u > 40
Alternative hypothesis: u < 40

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 0.05. The test method is a one-sample z-test.

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

SE = s / sqrt(n)

S.E = 0.9392

z = (x - u) / SE

z = - 1.95

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 - 1.95.

Thus the P-value in this analysis is 0.026.

Interpret results. Since the P-value (0.026) is less than the significance level (0.05), we have to reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that new route is shorter, at the 5% level significance.

answered by: anonymous