At the time she was hired as a server at the Grumney Family Restaurant, Beth Bri
ID: 3154938 • Letter: A
Question
At the time she was hired as a server at the Grumney Family Restaurant, Beth Brigden was told, “You can average $71 a day in tips.” Assume the population of daily tips is normally distributed with a standard deviation of $4.48. Over the first 33 days she was employed at the restaurant, the mean daily amount of her tips was $73.33. At the 0.05 significance level, can Ms. Brigden conclude that her daily tips average more than $71?
At the time she was hired as a server at the Grumney Family Restaurant, Beth Brigden was told, “You can average $71 a day in tips.” Assume the population of daily tips is normally distributed with a standard deviation of $4.48. Over the first 33 days she was employed at the restaurant, the mean daily amount of her tips was $73.33. At the 0.05 significance level, can Ms. Brigden conclude that her daily tips average more than $71?
Explanation / Answer
a)
Formulating the null and alternative hypotheses,
Ho: u <= 71
Ha: u > 71 [ANSWER, A]
*****************************
b)
As we can see, this is a right tailed test.
Thus, getting the critical z, as alpha = 0.05 ,
alpha = 0.05
zcrit = + 1.644853627
hence,
OPTION D: Reject H0 if z > 1.65 [ANSWER]
*****************************
c)
Getting the test statistic, as
X = sample mean = 73.33
uo = hypothesized mean = 71
n = sample size = 33
s = standard deviation = 4.48
Thus, z = (X - uo) * sqrt(n) / s = 2.987685484 [ANSWER]
****************************
d)
As z > 1.65, we REJECT HO. [ANSWER]
***************************
e)
Also, the p value is, as this is right tailed, using table/technology,
p = 0.001405493 [ANSWER]
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.