Looking for a great airfare? Consumer Reports has several hints about how to minimize your costs, which include being flexible about travel dates and times, checking multiple websites, and knowing when to book your flight. One suggestion involved checking fares at “secondary” airports—airports that might be slightly farther from your home, but where fares are lower. For example, suppose the average of all domestic ticket prices at one airport, x1, was quoted as $347 compared to an average price of $286 at a nearby airport, x2. Suppose that these estimates were based on random samples of 1000 domestic tickets at each airport and that the standard deviation of the prices at both airports was $130. (a) Is there sufficient evidence to indicate that the mean ticket prices differ for these two airports (μ1 − μ2) at the α = 0.05 level of significance? Use the large-sample z-test.
(a) Null and alternative hypotheses: H0: (μ1 − μ2) ≠ 0 versus Ha: (μ1 − μ2) = 0 H0: (μ1 − μ2) = 0 versus Ha: (μ1 − μ2) ≠ 0 H0: (μ1 − μ2) = 0 versus Ha: (μ1 − μ2) > 0 H0: (μ1 − μ2) < 0 versus Ha: (μ1 − μ2) > 0 H0: (μ1 − μ2) = 0 versus Ha: (μ1 − μ2) < 0
What is the p-value for this test? (Round your answer to four decimal places.) p-value = Conclusion: H0 is not rejected. There is sufficient evidence to indicate a difference in the mean ticket prices. H0 is rejected. There is insufficient evidence to indicate a difference in the mean ticket prices. H0 is rejected. There is sufficient evidence to indicate a difference in the mean ticket prices. H0 is not rejected. There is insufficient evidence to indicate a difference in the mean ticket prices.
(b) Construct a 95% confidence interval for (μ1 − μ2). (Round your answers to two decimal places.) $ to $ Does this interval confirm your conclusions in part (a)? Since (μ1 − μ2) = 0 does not fall in the confidence interval, it is likely that the means are the same. This contradicts the conclusion from part (a). Since (μ1 − μ2) = 0 does not fall in the confidence interval, it is likely that the means are the same. This confirms the conclusion from part (a). Since (μ1 − μ2) = 0 does not fall in the confidence interval, it is likely that the means are different. This contradicts the conclusion from part (a). Since (μ1 − μ2) = 0 falls in the confidence interval, it is likely that the means are different. This confirms the conclusion from part (a). Since (μ1 − μ2) = 0 does not fall in the confidence interval, it is likely that the means are different. This confirms the conclusion from part (a).
