Question

The management for a large restaurant claimed the waiters and waitresses average $75 a day in tips. The staff felt the figure was to high an collected the amount of tips for a period of 20 days. If the sample of 85 yielded an average of $73.50 per day with a standard deviation of $15.48, can they prove the $75 figure is too high? What is null hypothesis, alternative hypothesis and whether it is accepted or rejected and why?

Answer 1

To Test :-

H0 :- \mu = 75

H1 :- \mu > 75

Test Statistic :-
t = ( \bar{X} - \mu ) / (S /\sqrt{n})
t = ( 73.5 - 75 ) / ( 15.48 /\sqrt{ 85 })
t = -0.8934

Test Criteria :-
Reject null hypothesis if t \; > \;t_{\alpha, n-1}
t_{\alpha, n-1} = t_{0.05 , 85-1} = 1.663
t > t_{\alpha, n-1} = -0.8934 < 1.663
Result :- Fail to reject null hypothesis

Decision based on P value
P - value = P ( t > 0.8934 ) = 0.8129
Reject null hypothesis if P value < \alpha level of significance
P - value = 0.8129 > 0.05 ,hence we fail to reject null hypothesis
Conclusion :- Fail to reject null hypothesis

There is insufficient evidence to support the claim that the $75 figure is too high.