The diploma to statistics

Diploma statistics, final sample, due New York time 1.15 at 8:00 p.m

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STUDENT NUMBER:

MACQUARIE UNIVERSITY INTERNATIONAL COLLEGE
FORMAL EXAMINATION PERIOD: TERM 2, 2018

Unit Code: WSTA150

Unit Name: Business Statistic

s

Duration of Exam
(including reading time if applicable): 2 (two) hours plus 10 (ten) minutes reading time

Total No. of Questions: 5 (Five) questions

Total No. of Pages 15 (fifteen) pages(including this cover sheet):

GENERAL INSTRUCTIONS TO STUDENTS:
• Students are required to follow directions given by the Final Examination Supervisor and must refrain from communicating in any way with another student once they have enterec

the final examination venue.
• Students may not write or mark the exam materials in any way during reading time.
• Students may only access authorised materials during this examination. A list of authorised material is available on this cover sheet.
• All watches must be removed and placed at the top of the exam desk and must remain there for the duration of the exam. All alarms, notifications and alerts must be switched off.
• Students are not permitted to leave the exam room during the first hour (excluding reading time) and during the last 15 minutes of the examination.
• If it is alleged you have breached these rules at any time during the examination, the matter may be reported to a University Discipline Committee for determination.

EXAMINATION INSTRUCTIONS:
1. Write your Student ID number, Name and Surname and table number in the spaces provided at the top of

this page.
2. Place your Macquarie University Campus Card in the top right hand corner of your desk, with your

photograph facing upwards.
3. You must follow all instructions provided by the ExaminationInvigilators.
4. DO NOT open this examination paper or write anything until the Examination Supervisor instructs you to

do so.
5. This paper consists of the following questions:

Question Type of Question Content Tested Marks Allocated Instructions
1 20 marks Answers to short

answer/calculation
questions must be
written in the spaces
provided after each
question.

2 15 marks
Questions may be on any
content covered in this unit.

Short answer/calculation
questions3 10 marks

4 20 marks
5 10 marks

Exam TOTAL 5 questions 75 marks
6. Attempt ALL questions in this paper as per instructions provided.
7. Write your answers using a blue or black pen (answers in pencil will not be marked).
8. Additional writing space is available on the last two pages of the paper. If you use this space you must

indicate which question you are answering.
9. No marks will be awarded for unlabelled or illegible work.
10. If you have a question raise your hand and the Examination Invigilator will approachyou.

AIDS AND MATERIALS PERMITTED/NOT PERMITTED:
No dictionaries permitted.
Non-programmable calculators without text retrieval abilities, and calculators without “run”, “exe” or
“calc” keys are permitted.
A page of formulae and relevant Excel output are included in the final examination. Students are
permitted to take one A4 sheet (any colour), handwritten on both sides (using pens and/or pencils)
into the final examination. This sheet must be submitted with the final exam paper at the conclusion of
the exam.

Candidates must supply their own materials as these will not be supplied at the examination venue. Materials permitted in
the examination may not be shared among students.

Dictionaries:
Calculators:

Other:

Page 1 of 15

Question 1 (20 marks)

a. Suppose the length of time it takes one variety of plant seeds to germinate is a normally distributed
random variable with a mean of 15 days and a standard deviation of 4 days. Using the appropriate
Excel code, as provided below, answer the following questions:

Code Result Code Result

NORM.DIST(12, 15 ,4, TRUE) 0.2266 NORM.DIST(1,0,1, TRUE) 0.8413
NORM.DIST(19, 15, 4, TRUE) 0.8413 NORM.DIST(2,0,1, TRUE) 0.977

2

NORM.DIST(18, 15, 4, TRUE) 0.7733 NORM.DIST(3,0,1, TRUE) 0.9987
NORM.DIST(9,15,4, TRUE) 0.0668 NORM.INV(0.25, 15, 4) 12.3020
NORM.INV(0.90, 15, 4) 20.1262 NORM.INV(0.75, 15, 4) 17.6979

i) Find the probability that the seed should germinate within 19 days and shade the area of interest
on the bell curve. (3 marks)

ii) Find the probability that the seed should germinate in 12 or more days and shade the area of
interest on the bell curve. (3 marks)

iii) Find the probability that the seed should germinate between will be between 12 and 18 days and
shade the area of interest on the bell curve. (4 marks)

Page 2 of 15

iv) By what day should three-fourths of the seeds have germinated and shade the area of interest
on the bell curve. (3 marks)

v) Based on previous experience by the 15th day 60% of the seeds germinate. To test this claim a
researcher selected a random sample of 50 seeds and found that 33 seeds germinated by the
15th day. Carry out an appropriate hypothesis test on the research question below.

Use the following Excel code for this part: NORM.DIST(0.8660,0,1,TRUE)=0.8068
(7 marks)

Research Question: Was the proportion of seeds germinated by the 15th day equal to 60%?

Page 3 of 15

Question 2 (15 marks)

Research Question: Was the average waiting time for cancer surgery equal to 16 days, on
average?

The administrators, at hospitals across Australia, claim that the average waiting time for cancer
surgery is equal to 16 days, on average. Based on discussions you have had with friends, who
have visited hospital for cancer surgery recently, you dispute the administrators’ claim.

An independent expert is appointed to test the claim made by the administrators. Over the course
of a few weekends the expert records the waiting time for 36 randomly selected patients. The
average waiting time for these 36 patients is 15.42 days with a standard deviation of 6 days.

The following chart and Excel output were produced using the sample patients’ data gathered for
the study:

Summary statistics

Count Standard deviationMea

n

36 15.42

6

Histogram for cancer surgery data

12 –

i

10 –

& 8 –
T

ELL
4 –

2 –

0 –

1

r T T T T T 1

0 5 10 15 20 25 30

waiting time (days)

Excel Command Result

t.dist.2t(0.58,36) 0.5655

t.dist.2t(0.58,35) 0.5656

norm.s.dist(0.58,TRUE) 0.7190

t.inv.2t(0.05,35) 2.0301

Page 4 of 15

(a) Perform an appropriate hypothesis test to validate the claim by the administrators at 5% level of
significance. (10 marks)

(b) Calculate a 95% confidence interval (C.I.) for the average waiting time (days) for cancer surgery and
interpret the result. (5 marks)

Page 5 of 15

Question 3 (10 marks)

Research Question: Is there a difference between the average waiting time (AWT) for cancer surgery
(CS) in New South Wales and Victoria?

An independent expert is appointed to test the claim made by administrators from hospitals across
Australia in Question 2. Over the course of a few weekends the expert records the waiting time
for 36 randomly selected patients from each state. Use the following output to answer the research
question:

Histogram tor cancer surgery (VIC)
Histogram tor cancer surgery (NSW)Box plot for cancer surgery waiting time (days)

12 1

15 “I25 – 10 –

I J e – I 10I- !
I

s -K 10 –

2 -5 –

0 J0 J
5 10 15 30 25i ; ■ 10 15 20 25 30

waiting time (days)waiting time (days)

Test 1

t-Test: Two-Sample Assuming Equal Variances

AWT (NSW) AWT (VIC)
Mean
Variance
Observations
Pooled Variance
Hypothesized Mean Difference

15.4444
43.7397

13.8056
19.9325

36 36
31.8361

0
df 70
t Stat
P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail t Critical two-tail

1.2323
0.1110
1.6669
0.2220
1.9944

Test 2

t-Test: Paired Two Sample for Means

AWT (NSW) AWT (VIC)
Mean
Variance
Observations
Pearson Correlation
Hypothesized Mean Difference

15.4444
43.7397
13.8056
19.9325

36 36
-0.1625

0
df 35
t Stat
P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail t Critical two-tail

1.1488
0.1292
1.6896
0.2584
2.0301

Page 6 of 15

(a) There are two different hypothesis tests, which is the most appropriate test (Test 1 or Test 2) for
answering the research question above? Explain your choice. (2 marks)

(b) Carry out an appropriate hypothesis test to address the research question above.
(8 marks)

Page 7 of 15

Question 4 (20 marks)
A study was undertaken by MUIC to investigate the relationship between the class size (number of
students) and the academic performance of students in the unit WSTA150 Business Statistics. A random
sample of 15 classes was selected, which included the class size (number of students) and the average
marks per class (out of 100). The following chart and Excel output were produced using the sample data.
Use this information to answer the questions which follow:

Scatter Plot Class Size Residual Plot

Histogram for Residua

ls

7

6

5>u
4

3CTu 3
LL.

2
1

0
-1.8464 -0.7761 0.2941 More

Bin

Table 1:

Regression Statistics

Multiple R 0.9574
R Square 0.9167
Adjusted R Square 0.9103
Standard Error 2.0237
Observations 15

Coefficients Standard Error t Stat P-value
Intercept 91.1918 1.8261 49.9375 0.0001
Class Size -1.0568 0.0884 ***** 0.0002

Page 8 of 15

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(a) Examine the scatter plot and correctly assess the strength and the nature of the relationship
between the class size and average marks. (2 marks)

(b) Comment on the model assumption, in relation to the normality of the residuals, by examining
the appropriate plot. (2 marks)

(c) Has the assumption of constant spread of residuals been met? State which plot you need to
examine for testing this assumption and the reason for your conclusion. (2 marks)

(d) Calculate the correlation coefficient for the regression equation, ensure you include the
appropriate sign. Interpret the value of this coefficient. (2 marks)

(e) What is the value of the coefficient of determination for the relationship between the class size
and the average marks obtained by the students. Interpret this value accurately and in a
meaningful way. (2 marks)

Page 9 of 15

(f) Write down the least squares regression equation and correctly interpret the equation. (2 marks)

(g) By using the regression equation from part (f) predict the average marks in WSTA150 for a class
size of 20. (2 marks)

(h) Using the regression equation predict the average marks for a class size of 3. (1 marks)

(i) Undertake an appropriate test for determining whether the average marks and the class size are
significantly related (a = 0.05). When answering this question, you may assume that all assumptions
in relation to the least square regression have been met. (5 marks)

Page 10 of 15

Question 5 (10 marks)
Prior to the 2018 winter Olympics in Pyeongchang, South Korea, various broadcasters and sports
agencies used qualifying times to predict the number of Gold medals that each country might be
expected to win. One agency predicted that the Norway would win 20% of the gold medals,
Germany would win 9% of the gold medals, Canada and USA would each win 6% of the gold
medals and Netherland would win 5% of the gold medals, with other countries winning the
remainder of the medals. The actual number of medals won is shown in the Table below:

Number of gold medals won by each country in the 2018 winter Olympics
Country Norway Germany Canada USA Netherland Other

countries
Total

Gold
medals
won

14 14 11 9 8 47 103

Use a suitable hypothesis test to answer the following research question. Note that the p-value
for this test is =CHISQ.DIST.RT(12.487,5)=0.029.

Research Question: Among the countries competing at the 2018 winter Olympics did the Norway
win 20% of the gold medals, Germany win 9% of the gold medals, Canada and USA each win 6%
of the gold medals and Netherland win 5% of the gold medals?

Page 11 of 15

This space is for the answer to Question 5.

Page 12 of 15

ADDITIONAL WRITING SPACE

If you use this space you must label your work.

Page 13 of 15

ADDITIONAL WRITING SPACE
If you use this space you must label your work.

END OF EXAMINATION PAPER

Page 14 of 15

m MACQUARIE
University

WSTA 150 Statistical Formulas

y-py-p (for individuals) (for means)
a/Vn

P-P (for proportions) y = za + pz J?

a s
standard error (se) of y =

n
estimated se of y =

n

pqse of p = Jv estimated se of p = p

95% CI: p ± 1.96p-pz =
J?

„ = y-p o 95% CI: y±1.96ff

y-p o 95% CI: y±tv-1, (df: v = n-1)t =
s/Vn

yg-Po
95% CI: ^±^7=, (df: v = nd-1)t = sg/T^g

95% CI: (y! – y2) ± tvspJIlp=, (df: v =n± + n2-2)
SpJn1+n2

1 1
+t =

ni n-2

(ni – 1)s^ + (n2 – 1)s^
sp = n1+n2 — 2

\
bi (df: v = n-2) 95% CI: b1 ± tn-2se(b1)t =

se(bi)’

c (ObSj – Exp j )2
XI 2 = Z (df: v = c-1)

j=i Exp j

c (ObSj – Exp jj )2
x2 =S-=i z (df: v = (r-1)(c-1) )

j=i Expij

Page 15 of 15