At the time she was hired as a server at the Grumney Family Restaurant, Beth Bri
ID: 3177547 • Letter: A
Question
At the time she was hired as a server at the Grumney Family Restaurant, Beth Brigden was told, "You can average more than $80 a day in tips." Assume that the population of daily tips is normally distributed with a standard deviation of $3.24. Over the first 35 days she was employed at the restaurant, the mean daily amount of her tips was $84.65. At the 0.01 significance level, can Ms. Brigden conclude that her daily tips average more than $80? a. State the null hypothesis and the alternate hypothesis. H_0: mu 80 H_1: mu 80 b. State the decision rule? (Round the final answer to 2 decimal places.) H_0 when z is. c. Compute the value of the test statistic. (Round the final answer to 2 decimal places.) Value of the test statistic d. What is your decision regarding H_0? H_0. e. What is the p-value?Explanation / Answer
Given that,
population mean(u)=80
standard deviation, =3.24
sample mean, x =84.65
number (n)=35
null, Ho: =80
alternate, H1: >80
level of significance, = 0.01
from standard normal table,right tailed z /2 =2.326
since our test is right-tailed
reject Ho, if zo > 2.326
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 84.65-80/(3.24/sqrt(35)
zo = 8.49067
| zo | = 8.49067
critical value
the value of |z | at los 1% is 2.326
we got |zo| =8.49067 & | z | = 2.326
make decision
hence value of | zo | > | z | and here we reject Ho
p-value : right tail - ha : ( p > 8.49067 ) = 0
hence value of p0.01 > 0, here we reject Ho
ANSWERS
---------------
null, Ho: <80
alternate, H1: >80
test statistic: 8.49067
critical value: 2.326
decision: reject Ho
p-value: 0
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