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 $80 a day in tips.” Assume the population of daily tips is normally distributed with a standard deviation of $9.95. Over the first 35 days she was employed at the restaurant, the mean daily amount of her tips was $84.85. At the .01 significance level, can Ms. Brigden conclude that her daily tips average more than $80? Compute the value of the test statistic. (Round your answer to 2 decimal places.) What is the p-value? (Round your answer to 4 decimal places.)

Solution

a)

Formulating the null and alternative hypotheses,              
              
Ho:   u   <=   80  
Ha:    u   >   80  
              
As we can see, this is a    right   tailed test.      
              
Thus, getting the critical z, as alpha =    0.01   ,      
alpha =    0.01          
zcrit =    +   2.326347874      
              
Getting the test statistic, as              
              
X = sample mean =    84.85          
uo = hypothesized mean =    80          
n = sample size =    35          
s = standard deviation =    9.95          
              
Thus, z = (X - uo) * sqrt(n) / s =    2.883717281 = 2.88 [ANSWER, TEST STATISTIC]

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B)          
              
Also, the p value is, as this is right tailed,              
              
p =    0.0020 [ANSWER]

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As P < 0.01, we   REJECT THE NULL HYPOTHESIS.      

Thus, Ms. Brigden can conclude at 0.01 level that her daily tips average more than $80. [CONCLUSION]