Question

It is reported in USA Today that the average flight cost nationwide is $507.93. You have never paid close to that amount and you want to perform a hypothesis test that the true average is actually greater than $507.93. The hypotheses for this situation are as follows: Null Hypothesis: μ ≤ 507.93, Alternative Hypothesis: μ > 507.93. A random sample of 27 flights shows an average cost of $501.435 with a standard deviation of $60.2994. What is the test statistic and p-value for this test?

Question 10 options:

	1) Test Statistic: 0.56, P-Value: 0.7098
	2) Test Statistic: 0.56, P-Value: 0.2902
	3) Test Statistic: -0.56, P-Value: 0.2902
	4) Test Statistic: -0.56, P-Value: 0.7098
	5) Test Statistic: -0.56, P-Value: 1.4196

Answer 1

Test Statistic :-
t = ( X̅ - µ ) / (S / √(n) )
t = ( 501.435 - 507.93 ) / ( 60.2994 / √(27) )
t = -0.56

Decision based on P value
P - value = P ( t > 0.5597 ) = 0.7098

Reject null hypothesis if P value < α = 0.01 level of significance
P - value = 0.7098 > 0.01 ,hence we fail to reject null hypothesis
Conclusion :- Fail to reject null hypothesis

4)	Test Statistic: -0.56, P-Value: 0.7098