Question

Given in class on 12/11/2019 4. A tourist agency in Hawaii claims the mean daily cost of meals and lodging for a family of 4 travelling in Hawaii is at most $6.50; you work for the consumer protection advocate and want to test this claim. In $6.57 with a standard deviation of $40. At a = 0 O wan, u have enough evidence to reject the tourist agency's claim? Let 0= trus mean da ly Cost (10) meols f laspey for a lam ly t4 had %3D

Answer 1

Given: x= 6.57, n = 45, s = 40 and u = 6.5 Ho:u= 6.5 Hu> 6.5 For a= 0.05 and df=n-1 = 45-1 =44 Critical value t0.05,44 = 1.68 (right tailed) Test statistic = "= 40°* = 0.01174 P-value (At df= 44,t> 0.01174)=0.4954 P-value=0.4954 Level of significance a = 0.05 Since P-value (0.4954)> 0 (0.05), don't reject Ho At 5% level of significance, failed to reject H, and there is insufficient evidence to reject the tourist agency's claim.

5) Given: x = 48, n=25, s = 5.4 and u = 45 Null hypothesis: Ho:u= 45 Alternate hypothesis: H : u> 45 For a=0.1 and df=n-1 = 25-1 =24 Critical value t0.1,24 = 1.318 (right tailed) Test statistic = u in 48-450 =2.7778 P-value (At degree of freedom = 24)= 0.0052 Level of significance a = 0.10 Since P-value (0.0052)<a (0.10), Reject the null hypothesis (HO) At 10% level of significance, we reject null hypothesis and conclude of having enough evidence to support the county's claim