Statistics Assignment: Mean, Median, Mode, Hypothesis Testing - Solved

This document presents a detailed solution to a statistics assignment involving calculations of mean, median, and mode for multiple data sets, followed by an analysis of their agreement and appropriateness. It also covers the construction of a confidence interval for a proportion, hypothesis testing, and the identification of Type I and Type II errors. The solution provides explanations for each step, including the interpretation of regression statistics and the implications of different error types in a practical context. The assignment explores concepts like data symmetry, skewness, and the importance of sampling in quality control, offering a comprehensive overview of fundamental statistical principles. Desklib provides this document along with other past papers and solved assignments for students.

Statistics
Name:
Institution:
29th May 2018

Total Marks: This assignment is worth 40 marks (40 % of the total marks in the unit).
Instructions:
• Students are required to cover all stated requirements.
• Your answer must be both uploaded to Moodle in word file and handed over a printed copy.
• You need to support your answers with appropriate Harvard style references where necessary.
Each of the following question worth 10 marks, please answer all of them. If needed please use Harvard
referencing style. There is no word limit, but it is necessary that you provide answers with explanations.
Question 1:
Prof. Hardtack gave four Friday quizzes last semester in his 10-student senior tax accounting class.
Quiz 1: 60, 60, 60, 60, 71, 73, 74, 75, 88, 99
Quiz 2: 65, 65, 65, 65, 70, 74, 79, 79, 79, 79
Quiz 3: 66, 67, 70, 71, 72, 72, 74, 74, 95, 99
Quiz 4: 10, 49, 70, 80, 85, 88, 90, 93, 97, 98
(a) Find the mean, median, and mode for each quiz.
Solution
Mean for quiz 1
Mean , x=∑ xi
n = 60+60+ …+88+99
10 =72 0
10 =72 .0
Mean for quiz 2
Mean , x=∑ xi
n = 65+65+…+79+7 9
10 = 720
10 =72.0
Mean for quiz 3
Mean , x=∑ xi
n = 66+67+…+9 5+ 9 9
10 =760
10 =76.0
Mean for quiz 4
Mean , x=∑ xi
n = 10+49+…+97 +9 8
10 = 7 6 0
10 =76 .0
Median for quiz 1:
Median= 71+73
2 =144
2 =72
Median for quiz 2:
Median= 70+74
2 = 144
2 =72
Median for quiz 3:
Median= 72+72
2 =144
2 =72
Median for quiz 4:

Median= 85+88
2 = 173
2 =86.5
Mode for quiz 1:
Mode=60
Mode for quiz 2:
Mode=65
Mode for quiz 3:
Mode=72
Mode for quiz 4:
Mode=Null
(b) Do these measures of center agree? Explain.
Solution
No they don’t agree; this is because in some instances the median is greater than the mean by far (David, 2003)
(c) For each data set, note strengths or weaknesses of each statistic of center.
Solution
Where the mean and median disagree, one should look at the shape of the distribution to see which measure is
more appropriate. The mode is a reliable measure of central tendency for qualitative data (Merkle, 2005)
Where the mean and median agree, one should look at the shape of the distribution to see which measure is
more appropriate. The mode is an unreliable measure of central tendency for quantitative data.
(d) Are the data symmetric or skewed? If skewed, which direction?
Solution
Quiz 1: Symmetric
Quiz 2: Symmetric
Quiz 3: Skewed right
Quiz 4: Skewed left
(e) Briefly describe and compare student performance on each quiz.
Question 2:
In a sample of 100 Planter's Mixed Nuts, 19 were found to be almonds.
(a) Construct a 90 percent confidence interval for the true proportion of almonds.
Solution
C.I: p ± Zα / 2 SE
p= 19
100 =0.19
Zα/2 =1.645
Standard Error ( SE )= √ p (1− p )
n = √ 0.19(1−0.19)
100 = √0.001539=0.03923
C.I: p ± Zα/ 2 SE

: 0.19 ± 1.645∗0.03923
0.19 ± 0.06453335
Lower bound: 0.19−0.0645=0.1255
Upper bound: 0.19+0.0645=0.2545
Therefore the 90% confidence interval is between 0.1255 and 0.2545.
(b) May normality be assumed? Explain.
Solution
np=100∗0.19=19>10
Also
n ( 1− p ) =100∗0.81=81>10
Since both np > 10 and n(1-p) > 10; then normality can be assumed
(c) What sample size would be needed for 90 percent confidence and an error of ± 0.03?
Solution
ME=z √ ^p (1− ^p)
n
ME2=z2
[ ^p(1− ^p)
n ]
n= ^p( 1− ^p) z2
ME2 =0.19 (1−0.19)1.6452
0.032 =462.73 ≈ 473
Thus the sample size would be 473.
(d) Why would a quality control manager at Planter's need to understand sampling?
Solution
Sampling help produce accurate results and as such the control manager should understand sampling in order to
produce accurate results. A good sampling should also be able to be replicated and therefore a control manager
would want to have a good sample that he/she might replicate.
Question 3:
Consider the following Excel regression of perceived sound quality as a function of price for 27 stereo speakers.
(a) Is the coefficient of Price significantly different from zero at α = .05?
Solution
No the coefficient of Price is not significantly different from zero at α = .05. This is because the p-value of the
coefficient is 0.6019 (a value greater than 5% level of significance), which leads to failure to reject the null
hypothesis of coefficient being to zero.
(b) What does the R2 tell you?
Solution
The value of R2 is 0.01104; this tells us that only 1.104% of the variation in the dependent variable (perceived
sound quality) is explained by the price.
(c) Given these results, would you conclude that a higher price implies higher sound quality?
Solution
No. The results shows the coefficient of Price to be -0.00239; this means that higher price implies lower sound
quality.
Regression
Statistics
R Square 0.01104
Standard Error 4.02545
Observations 27

Statistic Coefficients Std Error t Stat P-value Lower 95% Upper 95%
Intercept 88.4902 1.67814 52.731 0.0000 85.0340 91.9464
Price −0.00239 0.00453 −0.528 0.6019 −0.01172 0.00693
Question 4：
Suppose a pickup and delivery company states that their packages arrive within two days or less on average. You want
to find out whether the actual average delivery time is longer than this. You conduct a hypothesis test.
(a) Set up the null and alternative hypotheses.
Solution
H0 : μ=2
H A : μ>2
(b) Suppose you conclude wrongly that the company's statement about average delivery time is within two days.
What type of error is being committed and what is the impact of that error?
Solution
Type II error; this is where we falsely retain the null hypothesis. The impact this will have is that people will go for
the company believing that the deliveries will be made in 2 days yet the deliveries will take longer than 2 days on
average.
(c) Suppose you conclude wrongly that the delivery company's average time to delivery is in fact longer than two
days. What type of error did you commit and what is the impact of this error?
Solution
Type I error; this is falsely rejecting the null hypothesis (Derrick, et al., 2016). The impact of this is that people
may tend to run away from the company believing that the delivery will take a longer period than 2 days yet it can
be done in 2 days.
(d) Which error is worse from the company's standpoint, a Type I or a Type II error? Why?
Solution
Type I is worse from the company’s standpoint. This is because people might have a wrong notion that they don’t
deliver within a shortest time and by this they might end up losing customers who would want their goods
delivered in 2 days or less.
(e) Which error is worse from a consumer standpoint, a Type I or a Type II error? Why?
Solution
Type II is worse from a customer standpoint. This is because the customers might use the company believing
that their deliveries will be done in 2 days but the deliveries would actually take longer than 2 days on average.

References
David, J. S., 2003. Handbook of Parametric and Nonparametric Statistical Procedures.
Derrick, B., Toher, D. & White, P., 2016. Why Welchs test is Type I error robust. The Quantitative Methods
for Psychology, 12(1), p. 30–38.
Merkle, M., 2005. Jensen's inequality for medians. Statistics & Probability Letters, 71(3), p. 277–281.