Question

A prospective employee was told she could average more than $80 a day in tips. Assume the standard deviation of the population is $3.24. Over the first 35 days the employee worked the mean daily amount of her tips was $84.85. At the .01 significance level what will be the value of the critical value if we are testing to see if the true average daily amount of tips is greater than $80?   

Answer 1

Solution :

Given that,

Population mean = = 80

Sample mean = = 84.35

Population standard deviation = = 3.24

Sample size = n = 35

Level of significance = = 0.01

This a right (One) tailed test.

The null and alternative hypothesis is,  

Ho: 80

Ha: 80

The test statistics,

Z =( - )/ (/n)

= ( 84.35 - 80 ) / ( 3.24 / 35 )

= 7.94

Critical value of  the significance level is α = 0.01, and the critical value for a right-tailed test is

= 2.33

Since it is observed that z = 7.94 > = 2.33 , it is then concluded that the null hypothesis is rejected.

P-value = P(Z > z )

= 1 - P( Z < 7.94 )

= 1 - 1

= 0.000

The p-value is p = 0, and since p = 0 < 0.01, it is concluded that the null hypothesis is rejected.

Conclusion :

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the true average daily amount of tips is greater than $80, at the 0.01 significance level.