At the time she was hired as a server at the Grumney Family Restaurant, Beth Bri

Question

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? a. State the null hypothesis and the alternate hypothesis. H_0: mu > 80: H_1: mu = 80 H_0: mu greaterthanorequalto 80: H_1: mu 80 H_0: mu = 80: H_1: mu notequalto 80 b. State the decision rule. Reject H_0 if z > 2.326 Reject H_1 if z > 2.326 Reject H_0 if z

Explanation / Answer

Given that,
population mean(u)=80
standard deviation, =9.95
sample mean, x =84.45
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 = 84.45-80/(9.95/sqrt(35)
zo = 2.64588
| zo | = 2.64588
critical value
the value of |z | at los 1% is 2.326
we got |zo| =2.64588 & | z | = 2.326
make decision
hence value of | zo | > | z | and here we reject Ho
p-value : right tail - ha : ( p > 2.64588 ) = 0.00407
hence value of p0.01 > 0.00407, here we reject Ho

ANSWERS
---------------
a.
null, Ho: =80
alternate, H1: >80

b.
decision: reject Ho if z>2.326
c.
test statistic: 2.64588
critical value: 2.326
p-value: 0.00407

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