Question

The roundtrip airfare between two cities is normally distributed. A random sample of 16 flights has an average airfare of $422.10 with a standard deviation of $56.30. a. Find the standard error of the sample mean. b. What is the critical value for the 90% confidence interval for the mean airfare? c. What is the margin of error for a 90% confidence interval for the mean airfare? d. What is the 90% confidence interval for the mean airfare? e. Interpret the 90% confidence interval from part (d). f. Several years ago, the mean airfare was $403.80. Using the confidence interval from part (d), is there evidence that the mean airfare has changed? Explain. Please show work for all parts.

Answer 1

Solution Crven, D = 16 X = 422.10 S - 56.30 @ Standard error of the sample mean S.Ez = = 56.30 = 14.075 o Coticol Volve por 90%. C. I. for men. Sina ppulation SD is unknown, we use t distybution the critical qalue is to.nl n=16 7 d. f. an-1 =15 c = 90% = 0.go &=1-6 = 0.10 and of = 0.05 753 ... using t table ta n- TOO5,15 @ margin of the error E E = tas * S. E - 1.753 * 14.075 = 24.673475 & golf confidence interval for means x + margin of error Ł ŁE 422.10 + 24.673 475 .. (422.10-24.673475, 422.10 + 24.673475) ( 397.43, 446.77)

Interpret the intentas. If we chow several samples from the given population and construct the 90% confidena intervals using these samples, then go% of these intervals could contain the true populalation mean. Given that el= 403.80 claim el has changed Hypothesis axe Ho: el=403.80 vs Hill 403.80 (397-43, 446.77) Interval is the volue ... will fail 403.80 is lie in the interval to reject Hoi el=403.80. there is no evidence that the mean airfare has changed.