Question

While reviewing a sample of audit fees, a senior accountant for the firm notes that the fee charged by the firm's accountants depends on the complexity of the return. A comparison of actual charges therefore might not provide the information needed to set next year's fees. To understand the fee structure, the senior accountant requests a new sample that measures the time the accountant spent on the audit. Last year, the average number of hours spent per client was 3.25. A new sample of 20 audit times shows the following times in terms of hours:

4.2	4.5	3.4	2.7	3.9
3.7	4.2	4.2	3.3	5.2
2.9	4.1	4.4	4.3	3.6
3.1	5	3.5	4	3.2

We plan to perform a hypothesis test to determine if the true mean audit time for this year's audits is statistically significantly higher from last year's average which was 3.25 hours.

1. What is the alternative hypothesis in this problem?

2. What is the appropriate distribution of the test statistic, assuming the null hypothesis is true?

a. A standard normal distribution

b. A t-distribution with n - 1 degrees of freedom

c. An F-distribution with 1 and n degrees of freedom for the numerator and denominator, respectively

d. An approximate t-distribution, where the degrees of freedom is approximated by Satterthwaite's approximation

e. A t-distribution with n1+n2−2n1+n2−2 degrees of freedom  

3. Set this problem up in excel to find the p-value

3.a What is the P-Value (using excel)

a. 0.0030

b. 4.16

c. 0.0008

d. 0.0003

e. -4.16

f. 0.3085

Answer 1

Given that,
population mean(u)=3.25
sample mean, x =3.87
standard deviation, s =0.649
number (n)=20
null, Ho: μ=3.25
alternate, H1: μ>3.25
level of significance, α = 0.05
from standard normal table,right tailed t α/2 =1.729
since our test is right-tailed
reject Ho, if to > 1.729
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =3.87-3.25/(0.649/sqrt(20))
to =4.272
| to | =4.272
critical value
the value of |t α| with n-1 = 19 d.f is 1.729
we got |to| =4.272 & | t α | =1.729
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :right tail - Ha : ( p > 4.2723 ) = 0.00021
hence value of p0.05 > 0.00021,here we reject Ho
ANSWERS
---------------
1.
null, Ho: μ=3.25
alternate, H1: μ>3.25
2.
a. A standard normal distribution
test statistic: 4.272
critical value: 1.729
decision: reject Ho
3.
p-value: 0.00021
option:D
we have enough evidence to support the claim that the true mean audit time for this year's audits is statistically significantly higher from last year's average which was 3.25 hours.