Question

2 Super CURAB Huert S UWOYU on __Suport CDM. 2. A tourist agency in California claims the mean daily cost of meals and lodging for a family of four traveling in California is at most $268. In an annual survey performed by the consumer protection advocacy you work for, a random sample of 45 families or four traveling in California had a mean daily cost of meals and lodging of $277. Assume the population standard deviation is $40. At a = 0.05, do you have enough evidence to reject the agency's claim? (277)-(268)= 9 40545 = 1.51 H: M > 269 Ha: 268 a=0.05 Test Statistic: Left, right or two-tailed? P-Value: Conclusion: (2 part)

Answer 1

Here we have 277, o 40, n=45 Hypotheses are: Но : и «-268 Н:и >268 a Level of significance: aa 0.05 est is one tailed (right tailed). Test Statistics: (277-268)/(40/sqrt(45)) Z = aln 1.51 Critical value: 1.645 Rejection Region: If z 1.645, Reject H0 Decision: Since test statistics does not lie in rejection region so we fail to reject the null hypothesis P-value 0.0655 P-value: Decision: Since p-value is not less than level of significance so we fail to reject the null hypothesis Excel function for critical value: =NORMSINV(1-0.05) Excel function for p-value: -1-NORMSDIST(1.51)

Conclusion: There is not enough evidence to reject the agency's claim.