This document contains 5 questions and their answers. Question 1 analyzes survey data to determine if more than 61% of people sleep 7 or more hours per night on weekends. Question 2 calculates a p-value for a hypothesis test comparing the means of two employment tests. Question 3 performs a hypothesis test to examine if a sample's mean score differs from the expected population mean. Question 4 uses a chi-squared test to determine if there is a preference for certain class times. Question 5 provides commute data and asks to calculate the line of best fit, confidence intervals, and determine if distance can indicate travel time.

1. 1
Contents
Question 1 ...................................................................................................................................1
Answer number 1 ..................................................................................................................... 1
Question 2: ..................................................................................................................................2
Answer number 2 ..................................................................................................................... 2
Question 3 ...................................................................................................................................3
Answer number 3 ..................................................................................................................... 3
Question 4: ..................................................................................................................................4
Answer Number 4..................................................................................................................... 4
Question 5 ...................................................................................................................................6
Answer Number 5..................................................................................................................... 7
Question 1
1. Many people sleeplate onthe weekendstomake upfor “shortnights”duringthe work week.The
Better Sleep council reports that 61% of us get more than 7 hours of sleep per night on the
weekend.A randomsample of 350 adults found that 235 had more than 7 hours of sleep each night
lastweekend.Atthe 0.05 level of significance,does this evidence show that more than 61% sleep 7
hours or more per night on the weekend?
Answer number 1
1. p is the proportion of adults who get more than 7 hours of sleep each night on weekends.
Hypothesis :
Ho: p = P(7+ hours of sleep) ≤ 0.61 (no more than 61% or equal to 61%)
Ha: p > 0.61 (more than 61%)
The n =350 which is greater than 20 so we use the standard normal z.
np and nq is also greater than 5 . np=(350)(0.61) = 213.5 ; nq = (350)(0.39) = 136.5.
p is expected to be normally distributed.
p’=
𝑥
𝑛
=
235
350
= 0.671
z’ =
𝑝′
−𝑝
√
𝑝𝑞
𝑛
=
0.671 −0.61
√
(0.61)(0.39)
350
=
0.061
√0.0006797
=
0.061
0.0261
= 2.34
By using the standard normal probability table, the pvalue is 1 - .9904 = 0.0096
Or, by using calculator, we get the p value = 0.0096

2. 2
So the Pvalue is smaller than The level of signifiance is  = 0.05 .
Conclusion :
We reject Ho . There is sufficient reason to conclude that the proportion of adults in the
sampled population who are getting more than 7 hours of sleep nightly on weekends is
significantly higher than 61% at the 0.05 level of significance
Question 2:
For years, many large companies in Melbourne have used a reputable employment agency for
testing perspective employees. The employment selection test used has historically resulted in scores
of normally distributed about a mean of 82 and a standard deviation of 8. A competitor has developed
a new psychoanalytic test that it claims to be quicker and easier to administer and therefore less
expensive. The competitor claims that its test results are the same as those obtained by the reputable
agency test. Many of the large companies, facing the financial crunch, are considering a change to cut
costs. However, they are unwilling to change if the competitor firm’s test results have a different
mean value. An independent testing firm tested 36 prospective employees with the Brown test. A
sample mean of 70 resulted. Determine the p value associated with this hypothesis test. (Assume
population standard deviation to be 8).
Answer number 2
The population mean . The mean of all test scores using the Brown Agency test.
Hypothesis :
Ho:  = 82 (Test results have the same mean)
Ha:   82 (Test results have different mean)
𝞼 is known. If the brown test scores are distributed the same as the kelley test scores, they
will be normally distributed and the sampling distribution will be normal for all sample sizes.
The standard normal probability distribution and the test statistic :
z’ =
𝑋̅− 
𝑠/√ 𝑛
will be used with 𝞼 = 8
Level of significance is not used because the question asks for Pvalue only, so there will
be no decision making for this case.
n= 36 ; 𝑋̅ = 79
z’ =
𝑋̅− 
𝑠/√ 𝑛
; z’ =
79−82
8/√36
=
−3
1.3333
= -2.25
The alternative hypothesis is two tailed test, we must find the probability associated with
both tails.
z’ = -2.25, the value of | 𝑧′| = 2.25.

3. 3
The Pvalue = 2 * P (z >2.25) = 2* (0.5000 – 0.4878) = 2(0.0122) = 0.0244
The Pvalue for the hypothesis test is 0.0255. Each individual company now will decide
whether to continue to use the Kelley Agency’s Servies or change to Brown Agency. We can
decide based on the hypothesis it if the case have the level of significance.
Question 3
On a popularself-image test that results in normally distributed scores, the mean score for public-
assistance recipientsisexpectedtobe 65. A randomsample of 28 publicassistance recipientsinNTG
is given the test. They achieve a mean score of 62.1, and their scores have a standard deviation of
5.83. Do the NTG publicassistance recipients test differently on average, than what is expected, at
the 0.02 level of significance?
Answer number 3
 is the mean self image test score for all NTG public assistance recipients.
The hypothesis for the null hypothesis and alternative hypothesis
Ho:  = 65 (mean is 65)
Ha:   65 (mean is different from 65)
For this case,the assumption of normal distribution has been satisfied based on the case. The level of
significance is  = 0.02. So we use t’ for the statistical test.
n= 28, 𝑋̅= 62.1 and s = 5.83 .
The value of the test statistics :
t’ =
𝑋̅− 
𝑠/√ 𝑛
t’ =
62.1− 65
5.83/√28
=
−2.9
1.1018
= -2.632
=-2.63
Because the alternative hypothesis express the difference of mean, so we use both tail to calculate the
pvalue for the test statistic.
P = P(t –2.63) + P(t 2.63) = 2  P(t 2.63), with df = 27.
with t=2.63 and df = 27,
The Pvalue is P = 0.0139.
Conclusion:

4. 4
The Pvalue is smaller than the level of significance (P = 0.0139 <  = 0.02). So we reject Ho . At the
level of significance 2% or0.02,we have sufficient evidence to conclude that the NTG recipients tests
significantly different on average fromthe expected 65
Question 4:
College students have regularly insisted on freedom of choice when they register for tutorial classes.
In one semester there were seven tutorials offered at different times in the Decision making unit.
Table below provides the number of students who selected each of the seven tutorial classes:
Tutorial 1 2 3 4 5 6 7 Total
No. of
Students
18 12 25 23 8 19 14 119
Does the data indicate that students have a preference for certain tutorials or do they indicate that each
tutorial was equally likely to be chosen? Explain your answer fully. Show all your workings clearly.
Answer Number 4 :
If there are no preference for certain tutorials for the students, then we expect that 119 students in total
are equally distributed for each tutorial. We test hypothesis using 95% confidence level.
Parameter of interest : Preference each tutorials.
Hypothesis :
H0 : There was no preference shown
H1 : There was a preference shown.
119 students in total is a random sample of population of all student who register for the tutorial.
We are using the chi-square distribution for the statistical test with df = 6, α=0.05
𝜒2 = ∑
(𝑂−𝐸)2
𝐸
𝜒2 = ∑
(18−17)2
17
+
(18−17)2
17
+
(12−17)2
17
+
(25−17)2
17
+
(23−17)2
17
+
(8−17)2
17
+
(19−17)2
17
+
(14−17)2
17
𝜒2 =
12+(−5)2+(82)+(62)+(−92)+(2)2+(−32)
17
𝜒2 =
1+25+64+36+81+4+9
17
𝜒2 =
220
17
= 𝟏𝟐. 𝟗𝟒
Pvalue = P(𝜒2 > 12.94 | 𝑑𝑓 = 6) . By using calculator, the pvalue = 0.044

5. 5
Decision : For this case we Reject H0 because pvalue is smaller than the level of significance of 5%
or 0.05. The decision shows us that there was a preference shown, but we cannot decide yet what
preference is relevance for this occasion, it could be the time preference or teacher preference.

6. 6
Question 5
Suppose you move to another city after graduation to take up a new job. You will, of course, be
concernedaboutthe problemsyouwill face commuting to and from work. You would like to know,
for example,howlongyouwill ittake youtodrive to workeach morning.Using“one waydistance to
work” as a measure of where you live, your employer has collected the following data based on
fifteen (15) of the employees in the company. Fifteen of your potential coworkers have given the
following data of one-way travel time and distances to work.
Data on Commute DistancesandTime
Coworker Distance in Kilometers Time to travel in minutes
1 3 7
2 5 20
3 7 20
4 8 15
5 10 25
6 11 17
7 12 20
8 12 35
9 13 26
10 15 25
11 15 35
12 16 32
13 18 44
14 19 37
15 20 45
`
(a) Findthe line of bestfitand the variance of Y about the line of bestfit.Explaininyourwordswhatdo
these calculationsindicate.
(b) Findthe 95% confidence interval forthe populationslope.
(c) Is the line of bestfituseful inmakingadecisionthatone-waydistance issuitable inindicatingone-
waytravel time?
(d) Constructa 95% confidence interval forthe meantravel time forthe coworkerswhotravel 7
Kilometerstowork.

7. 7
Answer Number 5
Coworker
Distance in
Kilometers
Time to travel
in minutes
X2 XY Y2
1 3 7 9 21 49
2 5 20 25 100 400
3 7 20 49 140 400
4 8 15 64 120 225
5 10 25 100 250 625
6 11 17 121 187 289
7 12 20 144 240 400
8 12 35 144 420 1225
9 13 26 169 338 676
10 15 25 225 375 625
11 15 35 225 525 1225
12 16 32 256 512 1024
13 18 44 324 792 1936
14 19 37 361 703 1369
15 20 45 400 900 2025
Total 184 403 2616 5623 12493
A.
Line of best fit can find the association between two variables, in this case are
Distance in Kilometers and Time to Travel in minutes. To estimate the line of best fit, we
must obtain the Sum Square of Distance in Kilometer(x) and Time to travel in minutes (y).
The above table helps us to calculate the Sum Square of both variables
SS(x) = ∑x2=∑x2 / n
= 2616- (184)2/15
=358.9333
So, the sum square of both variables
SS(xy) = ∑xy -
∑x .∑y
𝑛
= SS(xy) =5623 –
(184)(403)
15
= 679.5333
We calculate b1 for the slope,
b1 =
SS(xy)
𝑆𝑆(𝑥)
:
679 .5333
358 .9333
= 1.893202 = 1.89
the y intercept is b0 ,
b0 =
∑y−(b1.∑x)
𝑛
=
403−(1.893202)(184)
15
= 3.64
The equation of the line of best fit is,
𝑦̂ = 3.64 + 1.89𝑥

8. 8
The variance of y about the regression line,
𝑠 𝑒
2
=
∑y2
−(b0)(∑y)−(b1)(∑xy)
𝑛−2
=
(12,493)− (3.643387)(403)− (1.893202 )(5623 )
15 −2
379.2402
13
= 29.17
=29.17
𝑠 𝑒
2
𝑖𝑠 𝑡ℎ𝑒 𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒 𝑜𝑓 𝑦
B. 1 is the parameter of interest of the line of best fit for the population. Since x and
y from random sample so we assume that the values of minutes and miles are normally
distributed with 95% level of confidence.n = 15, b1 = 1.89 , 𝑠 𝑏1
2
= 0.0813.
t(df,α/2) = t(13,0.025) = 2.16.
The maximum error of estimate :E = t(n-2,α/2) . 𝑠 𝑏1
E = (2.16) . √0.0813 = 0.6159
So the lower and upper confidence limits are :
b1 – E to b1 + E ; 1.89 – 0.62 to 1.89 + 0.62 ; 1.27 to 2.51
1.27 to 2.51 is the 95% confidence interval for 1
We believe that the slope of line of best fit of the population from which the sample
was drawn is between the 95% confidence interval of 1.27 to 2.51. The interval indicates that
we are 95% confident that for every extra mile will take 1.27 minutes( 1 minute 16 seconds)
and 2.5 minutes(2 minutes, 31 seconds) to make the commute.
C. We have to consturct the hypothesis to determine whether the line of best fit can be
useful to predict y and so we use α = 0.05.
H0 : 1 = 0 (x cannot predict y)
H1 : 1 > 0 ( y increases as the distance of x increases)
We assume that y and x have normal distribution because the minutes and miles from
a random sample.

9. 9
df = n-2=13
To test the hypothesis, we are using the t test statistic with the level of significance α
= 0.05.
n = 15 , b1 = 1.89 𝑠 𝑏1
2
= 0.0813
Statistic Test :
t =
𝑏1−1
𝑠 𝑏1
t =
1.89−0
√0.0813
= 6.63
So the t is 6.63
P = P(t>6.63, df=13)
So the Pvalue : P <0.005. So we decide to reject H0 which means that there is a linear
relationship and that one-way distance(x) can be used to predict the travel time to work(y)
D. Travel time for one co-worker travels 7 miles to work. Assume that y values minutes and x
miles have normal distribution with 95% level of confidence.
𝑠 𝑒
2 = 29.17
𝑆𝑒 = 5.40
𝑦̂ = 3.64 + 1.89𝑥 = 3.64 + 1.89(7) = 16.87
Using the table, we obrain t(13,0.025) = 2.16
Maximum error of estimate :
E= t(n-2,α/2).Se . √1
𝑛
+
(𝑥0−𝑥)̂2
358.933
E= (2.16(5.40) . √
1
15
+
(7−12.272
358.933
=12.48
Lower and Upper confidence Limits
𝑦̂ − 𝐸 𝑡𝑜 𝑦̂ + 𝐸
16.87 – 12.48 to 16.87 + 12.48

10. 10
So we have 4.39 to 29.35 with 95% confidence interval. This indicates that the co-worker
travel times for commuters who travel 7 miles is between 4.39 minutes (4 minutes, 23 seconds) and
29.35 minutes(29 minutes,21 seconds)