Statistics Assignment: MPG, Audit Time, and Confidence Intervals

This statistics assignment solution addresses two key problems. The first problem involves a company's fuel economy goal, requiring the calculation of a Z-score and P-value to determine if the goal is being met. The second problem focuses on audit fees and the time accountants spend on audits, involving the calculation of the sample mean, standard deviation, and a 99% confidence interval. The solution also includes finding critical values of t for both 99% and 95% confidence intervals with varying degrees of freedom. The document utilizes statistical formulas and concepts to provide comprehensive solutions to these problems, demonstrating a solid understanding of statistical analysis and hypothesis testing. The assignment references relevant statistical resources to support the findings.

Running head: STATISTICS ASSIGNMENT 1
Statistics Assignment
Student’s Name
Institutional Affiliation

STATISTICS ASSIGNMENT 2
STATISTICS ASSIGNMENT
QUESTION 1
A company with a large fleet of cars hopes to keep gasoline costs down and sets a goal of
attaining a fleet average of at least 29 miles per gallon. To see if the goal is being met, they
check the gasoline usage for 50 company trips chosen at random, finding a mean of 28.32 mpg
and a standard deviation of 5.25 mpg. Is this strong evidence they have failed to attain their fuel
economy goal?
Find the P-value.
This follows a normal distribution. The sample mean is referred by X bar and the standard
deviation as σ while the population mean is given by μ (King'oriah,2004).
To solve this we need to calculate the Z score which is given by the formula=Z= x−μ
σ / √n
In this question, the population mean μ=28.32 while the standard deviation σ =5.25 and X
bar=29.
Therefore Z= X−μ
σ = 29−28.32
5.25/ √(50)= 0.916
The company goal is to ensure that least 29 miles per gallon is attained hence, the probability
that at least 29 miles are attained is;
P (Z>0.916) =1-0.820=0.1799 which is 0.180(one-tailed) and 0.360 (two –tailed) at 95%
confidence level. The result is therefore not significant
QUESTION 2
While reviewing the 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

STATISTICS ASSIGNMENT 3
better understand the fee structure, the senior accountant requests a new sample that measures
the time the accountants spent on the audit. Last year, the average hours charged per client audit
was 2.84 hours. A new sample of 15 audit times shows the following times in hours. Complete
parts a and b below.
2.7 4.6 3.1 4.1 2.6
4.7 3.9 2.5 4.1 3.1
3.7 2.9 4.6 4.3 3.7
Solution
Average Hours
2.7
4.6
3.1
4.1
2.6
4.7
3.9
2.5
4.1
3.1
3.7
2.9
4.6
4.3
3.7
The Sample mean is given by X bar ( X ) = ∑ X
N = 2.7+4.6 +…+3.7
15 = 54.6
15 =3.64
The value of Standard deviation is given by formula √¿ ¿ =0.770
The Z-statistic formula can be expressed as; Z= x−μ
σ / √n = 3.64−2.84
0.770 / √15 =4.02
99% Confidence Interval For sample

STATISTICS ASSIGNMENT 4
Confidence interval is the value of a statistic such as sample mean added and subtracted by a
margin of error.
According to Gupta and Kapoor (2019), confidence interval (C.I) can be expressed by formula
X ∓ z
Given that () the standard error ( σ
√ n ) -population mean and the z value is the z score
The value of Z score at 99% confidence level is 2.58
From the above calculations; z= (2.58*0.770/√ 15¿=0.513
C.I= 3.64 ± 0.513
Upper Limit therefore is 3.64  (0.513) =4.153
Lower Limit is C.I  3.64 (0.513) = 3.127
The 99% confidence interval to estimate the mean audit time is from 3.13 hours to 4.15 hours.
For parts a and b, use technology to estimate the following.
a) The critical value of t for a 99% confidence interval with df= 8.
A critical value is the number of standard errors that are usually added or subtracted to get a
given margin of error.
To get the critical value in a t-distribution we need to check the student t distribution tables.
The critical value at 99% confidence level where the d.f = 8
The alpha value ∝ = 1 - .99 = .01 in a two tailed test, the critical value is
T = 3.355.
T=3.36 (2 decimal places)
b) The critical value of t for a 95% confidence interval with df=108.
The critical value at 95% confidence level where the d.f = 108
The alpha value ∝ = 1 - .95 = .05 in a two tailed test, the critical value is
T = 1.960

STATISTICS ASSIGNMENT 5
T=1.96 (2 decimal places)
References
Gupta, S.C. and Kapoor, V.K. (2019), Fundamentals of applied statistics. Sulthan Chand &
Sons.
King'oriah, G. K.,(2004). Fundamentals of applied statistics. Nairobi: The Jomo Kenyatta
Foundation.