Question

Previous 21 22 23 24 25 26 27 28 Next Question 21 of 28 (4 points) View problem in a pop-up 8.3 Section Exercise 19 A survey of 15 large U.S. cities finds that the average commute time one way is 25.4 minutes. A chamber of commerce executive feels that the commute time in his city is less and wants to publicize this. He randomly selects 29 commuters and finds the average is 24.6 minutes with a standard deviation of 5.3 minutes. Ata -0.05, is he correct? Assume the population is normally distributed, Gu Source: New York Times Almanac. Part I out of 5 Ask State the hypotheses and identify the claim with the correct hypothesis. Lin 125.4 ||not claim!▼ 25.4 ||claim is a twortailed - test. (select) - This h

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Answer 1

Answei Date: 17/04/2019 To test the hypothesis is that the mean commute time less than 25.40 minutes at 5% significance level The null and alternative hypothesis is Ho : μ 25 .40 Ha 25.40 The hypothesis is left-tailed test The t-test statistics is, 24.60 25.40 5.30/29 -0.813 The t-test statistics is -0.813.

The t-critical value is, The sample size is small and left-tailed test. Look in the column headed α = 0.05 and the row headed in the t distribution table by using degree of freedom is, d.f.-n-1 - 28 Thet-critical value is [-1.701] Decision The statistical decision is that the t-test statistics which is -0.813 is less than t-critical value which is-1.701, so the null hypothesis is fails to reject at 5% level of significance. There is insufficient evidence to indicate that the mean commute time less than 25.40 minutes. The result is not statistically significant.