At the time she was hired as a server at the Grumney Family

At the time she was hired as a server at the Grumney Family Restaurant, Beth Brigden was told, “You can average $88 a day in tips.” Assume the population of daily tips is normally distributed with a standard deviation of $2.50. Over the first 50 days she was employed at the restaurant, the mean daily amount of her tips was $89.93. At the .10 significance level, can Ms. Brigden conclude that her daily tips average more than $88?

Solution

Set Up Hypothesis
Null, her daily tips average is equal or lessar $88 H0: U<=88
Alternate, her daily tips average more than $88 H1: U>88
Test Statistic
Population Mean(U)=88
Given That X(Mean)=89.93
Standard Deviation(S.D)=2.5
Number (n)=50
we use Test Statistic (Z) = x-U/(s.d/Sqrt(n))
Zo=89.93-88/(2.5/Sqrt(50)
Zo =5.4589
| Zo | =5.4589
Critical Value
The Value of |Z a| at LOS 0.1% is 1.28
We got |Zo| =5.4589 & | Z a | =1.28
Make Decision
Hence Value of | Zo | > | Z a| and Here we Reject Ho
P-Value : Right Tail - Ha : ( P > 5.4589 ) = 0
Hence Value of P0.1 > 0, Here we Reject Ho

ANS:
H0:µ = 88 ; H1: µ > 88
Reject H0 if z > 1.28
Zo =5.46
Reject Ho
( P > 5.4589 ) = 0