Stat 141 R1 - Lecture #36
Announcements:
1. Office Hours: Fri 2-4, Tues 2-4
2. Do as we did in class …
3. You are responsible for all chapters …
Review:
1)
The statistics marks at a university have an unknown distribution
with a mean of 65 and a standard deviation of 10. In a sample of 50
randomly selected Stat XXX students, what is the probability that
the average mark in the sample is less than 60?
Sol.: 𝑃(𝑦̅ < 60) = 𝑃 (𝑧 <
2)
60−65
10
√50
) = 𝑃(𝑧 < −3.54) = 0.0002
A producer of combo-packs consisting of 3 containers of flavoured
coffee is advertising that the total weight of the pack is
approximately 900 grams. If the weights of the three containers are
independent with distributions N(260,10), N(300,15), and
N(350,12) then what is the probability that a randomly selected
combo-pack weighs less than 900 grams?
Sol.: Let the weights of the three containers in the pack be u, v, and w
=> u+v+w has a normal distribution N(  ) where:

  102  152  122  469  21.65
P(u+v+w < 900) = P(z <
900  910
)
21.65
= P(z < - 0.46 ) = 0.3228
3)
A company wishes to predict next month’s trade with its customers
in the US. 30 months were randomly selected and the trade with
the US recorded. The mean trade per month amounted to $90.000
with a standard deviation of $10.000.
a) Determine the 90% confidence interval for the mean monthly
trade numbers.
b) Determine the sample size required for the next sample to lead
to a 90% confidence interval with an approximate margin of error
of $100.
Sol.:
s
10000
 90000  1.699 
 (86898,93102)
n
30
a) x  t0.05, n 1
b) 𝑛 = (
4)
𝑡𝑠
𝑀𝐸
2
) >(
𝑧𝑠 2
𝑀𝐸
1.645∗10000 2
) =(
100
) = 27060.25 => 27061
A random sample of 100 observations drawn from a normal
population with unknown mean and variance 25 was found to
have sample mean of 121. In testing Ho: =120 against H1:  
120, what will the p-value of the test be? What would the
conclusion be if =0.05 ?
Sol.: Ho: =120 vs. H1:   120
test-statistic : z 0 
x  0
/ n

121  120
5 / 100
 2.00
p-value = 2P(z>2.00) = 2 0.0228  0.0456
Since the p-value < =0.05: reject Ho in favor of H1:   120.
5)
A new diet pill tested on four people gave the following data on
weigh before and after the diet:
#1
#2
#3
#4
before
180
160
165
120
after
178
159
165
115
Test if there is sufficient evidence that the diet pill works, i.e. on
the average the weight after is less than before. Use the pvalue to
make your decision.
Sol.: Consider the paired differences d = “before-after”:
nd = 4, x d  2 , sd  2.16
Ho:d = 0 vs. H1: d > 0
test-statistic: t 0 
xd   0
sd / n

20
2.16 / 4
 1.85
where df = n – 1 = 3
P(t  2.353)  0.05, P(t  1.638)  0.10 =>
pvalue = P(t > 1.85) lies between 0.05 and 0.10
=> moderate to suggestive, but inconclusive evidence against Ho in
favour of H1.
6)
Research psychologists asked students to indicate what emotions
they associate with the word “math” and “meth”. The response
and gender of the students were summarized in the following table.
Is there a strong evidence to conclude that the proportion of math
respondents saying “addictive” is greater than the proportion of
meth respondents? Use a 1% level of significance.
math
meth
addictive
34
27
non-addictive
38
39
Sol.: Let “math” = #1 , “meth” = #2
34
27
 0.472 , n2  66 , pˆ 2 
 0.409 ,
n1  72 , pˆ 1 
72
Ho:
p1  p2
66
vs. H1:
teststatistic:
z0 
p1  p2
pˆ 1  pˆ 2
1
1
pˆ (1  pˆ )(  )
n1 n2

pˆ 
34  27
 0.442
72  66
0.472  0.409
1
1
0.442(1  0.442)(  )
72 66
 0.74
p-value = P(z > 0.74) = 0.2297
conclusion: since p-value >   0.01 , do not reject Ho.
There is insufficient evidence to conclude that H1: p1  p2
7)
Luigi Bonelli is an Italian restaurant owner who is interested
whether or not his customers would prefer a redesigned menu over
the current one. Bonelli randomly samples 60 customers and then
randomly separates them into two groups. The customers in the
first group are to rate the old menu on a scale of 1 to 10 and the
customers in the second group are to rate the new menu on a scale
of 1 to 10.
The data obtained is summarized in the following table:
group
𝑦̅
s
n
#1
6.6
1.9
28
#2
7.3
0.9
32
Use a hypothesis test to determine if there is enough statistical
evidence to conclude that Bonelli should change his menu with
  = 0.05. Assume that the population variances of the responses
were approximately equal?
Sol.: assuming 𝜎1 2 = 𝜎2 2
Ho: 2 = 0
HA: 2 < 0
Test statistic:
t
y1  y 2  0
6.6  7.3  0

 1.87
1
1
1
1
sp

1.45

n1 n2
28 32
with df = 28 + 32 – 2 = 58 => round down to 50
where sp =
(n1  1) s12  (n2  1) s22
n1  n2  2
=
27 * 1.9 2  31 * 0.9 2
 1.45
28  32  2
Pvalue = P(t <-1.87) = P(t > 1.87) = between 0.025 and 0.05
Conclusion: Reject Ho since pvalue < = 0.05
There is sufficient evidence to conclude that Bonelli should change
his menu with = 0.05.
8)
The following data pertain to the numbers of hours jet aircraft
engines have been used and the number of hours for repair.
number of hours (hundreds)
1
2
3
4
5
a)
b)
c)
repair time (hours)
10
40
30
80
90
Write the complete model for linear regression in general
terms.
Construct the 95% confidence limits for  1 .
Test the null hypothesis  =15 against  >15 at the 5% level
of significance.
Summary: 𝑥̅ = 3, 𝑦̅ = 50, 𝑏0 = −10, 𝑏1 = 20, sx = 1.58, SSE = 600
Sol.: 𝑠𝑒2 =
𝑆𝑆𝐸
𝑛−2
= 200, 𝑆𝐸 (𝑏1 ) =
a)
yˆ   0  1 x  
b)
𝑏1 ± 𝑡 ∗ 𝑆𝐸(𝑏1 ) =
c)
Ho:  =15 vs H1 : >15
test-statistic: 𝑡 =
𝑠𝑒
𝑠𝑥 √𝑛−1
𝑏1 −𝛽1
𝑆𝐸(𝑏1 )
= 4.47
where df = n-2 = 3
pvalue > 0.05
do NOT reject Ho: =15 in favor of H1 :>15.
9)
The race times of five formula one drivers in 10 randomly selected
rounds on a test track are recorded. The results are recorded in the
following partial ANOVA table:
DF
Racer
Error
Total
SS
600
MS
F
Pvalue
0.005
1000
Complete the ANOVA table, and decide if there is sufficient
evidence to conclude that the average performances of these three
racers are different?
Sol.: The completed ANOVA table is as follows:
Racer
Error
Total
DF
4
45
49
SS
600
400
1000
MS
150
8.9
F
16.9
Pvalue
0.005
Pvalue <0.01
=> strong evidence against H0: the average race times are all equal.