4. Grace is a nail artist and the IRS is auditing her tax return. In her tax ret

Question

4. Grace is a nail artist and the IRS is auditing her tax return. In her tax return, Grace indicated that her mean tip last year was $4.75. The IRS claims that mean tip Grace received was more than S4.75. Grace sent the IRS a random sample of 15 credit card receipts showing the amount of the tip re s = $1.15. Do these receipts support the claim of the IRS that the mean tip Grace received was more than $4.75. Use = .05 level ofsignificance. What is the conclusion? ceived. For this sample, the IRS computed the mean, x = $5.25 and the standard deviation.

Explanation / Answer

Given that,
population mean(u)=4.75
sample mean, x =5.25
standard deviation, s =1.15
number (n)=15
null, Ho: =4.75
alternate, H1: >4.75
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.761
since our test is right-tailed
reject Ho, if to > 1.761
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =5.25-4.75/(1.15/sqrt(15))
to =1.6839
| to | =1.6839
critical value
the value of |t | with n-1 = 14 d.f is 1.761
we got |to| =1.6839 & | t | =1.761
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :right tail - Ha : ( p > 1.6839 ) = 0.05718
hence value of p0.05 < 0.05718,here we do not reject Ho
ANSWERS
---------------
null, Ho: =4.75
alternate, H1: >4.75
test statistic: 1.6839
critical value: 1.761
decision: do not reject Ho
p-value: 0.05718
mean tip grace received was more than $4.75

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