Question

For a certain group of individuals, the average cost of a trip to the Super Bowl was $75. The standard deviation of the population was $50. This year, 49 fans who scheduled the trip paid an average of $89 for the three-day trip. Test the claim that the average cost is greater than last year’s cost at 1% significance level?

Answer 1

Solution :

= 89

=75

=50

n = 49

This is the right tailed test .

The null and alternative hypothesis is ,

H0 :    = 89

Ha : > 89

Test statistic = z

= ( - ) / / n

= (75 - 89) / 50 / 49

= -1.96

Test statistic = z = -1.96

P(z >-1.96 ) = 1 - P(z < -1.96 ) = 1 - 0.0250

P-value =0.9750

= 0.01

P-value >

0.9750 > 0.01

Fail to reject the null hypothesis .

There is not sufficient evidence to suggest that