Question

1. Expedia would like to test the hypothesis that the average roundtrip airfare between Philadelphia and Paris is higher for a flight originating in Philadelphia when compared to a flight originating in Paris. The following data summarizes the sample statistics for roundtrip flights originating in both cities. Assume that the population variances are equal.
   

   	Originating City
   	Philadelphia	Paris
   Sample mean	$1,240	$1,060
   Sample size	15	19
   Sample standard deviation	$270	$240

   
   If Population 1 is defined as flights originating in Philadelphia and Population 2 is defined as flights originating in Paris, and using α =  0.05, the conclusion for this hypothesis test would be that because the test statistic is

   		less than the critical value, Expedia can conclude that the average roundtrip airfare between Philadelphia and Paris is higher for a flight originating in Philadelphia when compared to a flight originating in Paris.
   		more than the critical value, Expedia cannot conclude that the average roundtrip airfare between Philadelphia and Paris is higher for a flight originating in Philadelphia when compared to a flight originating in Paris.
   		less than the critical value, Expedia can conclude that the average roundtrip airfare between Philadelphia and Paris is higher for a flight originating in Paris when compared to a flight originating in Philadelphia.
   		more than the critical value, Expedia can conclude that the average roundtrip airfare between Philadelphia and Paris is higher for a flight originating in Philadelphia when compared to a flight originating in Paris.

If Population 1 is defined as flights originating in Philadelphia and Population 2 is defined as flights originating in Paris, the correct hypothesis statement for this hypothesis test would be

		H0: μ1 - μ2 = 0; H1: μ1 - μ2 ≠ 0
		H0: μ1 - μ2 = 0; H1: μ1 - μ2 < 0
		H0: μ1 - μ2 = 0; H1: μ1 - μ2 > 0
		H0: 2 - 1 = 0; H1: 2 - 1 > 0.

Answer 1

Hypotheses : To test null hypothesis Hopi - M2 = 0 against alternative hypothesis H:41-42 > 0 This is a one tailed test as the alternative hypothesis contains '>' sign. Here, Ij = 1240.00, $1 = 270.00, n1 = 15, T2 = 1060.00, S2 = 240.00, n2 = 19 The test statistic can be written as: (21-22 - 0 t= SpV TH + 1 122 which under H, follows a t distribution with 32 df. where pooled variance (nı - 1)s +(n2 – 1)s ni + n2 - 2 64293.750000 and standard error for the difference between means 1 1 1 SEN SD * + = V64293.750000* 1 V 15 + = 22206.790002 ni n2 19 Decision rule / Rejection region : We reject Ho at 0.05 level of significance if P-value < 0.05 or if tstat > t0.05,761 +62-2 = t0.05,32 = 1.693889 0.4- 20.3 20.2- 0.1 0.0- t Now. Value of the test statistic : The value of the test statistic is (1240.00 – 1060.00) - 0 tstat = = 2.0552802.055 ✓ 64293.750000* 1 V 15 + 19 associated degrees of freedom = n1 + n2 - 2 = 15 + 19 - 2 = 32 The critical value tcritical = t0.05,df = t0.05,32 = 1.693889 1.694 P-value = P(tn, t12-2 > tstat) = 1 - P(tne +12-2 <tstat) = 1 - P(t32 <2.055280) = 1 -0.975955 = 0.024045 0.0240 Conclusion: Since p-value < 0.05 and tstat > t critical = 1.694, so we reject Ho at 0.05 level of significance

The test statistic value is more than the critical value, Expedia can conclude that the average roundtrip airfare between Philadelphia and Paris is higher for a flight originating in Philadelphia when compared to a flight originating in Paris.