Question 1
Using these results:
75
56
98
73
89
92
82
52
78
65
final calculus mark
52
34
75
28
47
57
64
21
43
39
achievement test
a) Calculate the coefficient of correlation for this data.
Solution:
data:
a8
x
=
460,
a8
x
2
=
23634
a8
y
=
760
a8
y
2
=
59816
a8
xy
=
36854
r
=
n
a8
xy
−
a8
x
a8
y
n
(
a8
x
2
)
−
(
a8
x
)
2
n
(
a8
y
2
)
−
(
a8
y
)
2
r
=
10
(
36854
) − (
460
)(
760
)
10
(
23634
)
−
(
460
)
2
10
(
59816
)
−
(
760
)
2
=
0.8398
l
0.84
b) Find the equation of the line of best fit for this data.
Solution:
b
0
= 40.78
b
1
= 0.766
y
= 40.78 + 0.766
x
c) Predict the final calculus grade for a student who scored 50 on the Achievement Test.
Solution:
y
= 40.78 + 0.766 (50) = 79.08 or 79%.
Question 2
a) For a random sample of 16 summer days in Montreal, the mean level of CO was found to be 4.9
ppm. With a standard deviation of 1.2. Based on this, construct a 95% confidence interval estimate
of the true mean level of CO in summer in Montreal.
Solution:
Since
n
= 16, this a small sample
t
distribution situation.
We know that
,
s
= 1.2, and
x
=
4.9
s
x
=
1.2
16
=
0.3
degrees of freedom
= 15,
so
t
a/2
, 15
= 2.131,
so our confidence limits are 4.9 ± 2.131(0.3) = 4.9 ± 0.639
the confidence interval for the mean level of CO is
[ 4.26,
5.54 ]
parts per million.
b) A preliminary study indicated that a new drug was effective in 75% of its trials. Accordingly, how
large a sample would be required to estimate the true effective rate of the drug to within 3% with
95% confidence.
Solution:
We’re finding sample size given a proportion, the size of the error and the level of
confidence. The formula for sample size under these conditions is:
n
=
p
(
1
−
p
)
z
a14
/2
E
2
95% confidence means
z
= 1.96,
we know
p
= 0.75
and
E
= 0.03
, so sample size is 801 people.
n
=
0.75
(
0.25
)
1.96
0.03
2
=
800.3333
COMM 215 MOCK FINAL EXAM SOLUTIONS
exam solutions page 1
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Question 3
A report said the average height of CEGEP basketball players is at least 180 cm. At a recent
tournament a random sample of 25 players yielded the following data
196
182
176
177
187
190
184
180
185
178
183
180
189
183
181
197
195
180
179
177
180
178
200
181
176
height in cm of
25 Cégep basketball players
Do these results support the report at the 0.05 Level of Significance?
Solution:
Since
n
= 25, we’ll use the
t
distribution for this upper tail test statistic.
a8
x
=
4594,
x
=
185.76,
s
=
6.98,
t
0.05, 24
=
1.711
Conclusion:
the test supports the claim that the average height of CEGEP basketball players is at least
180 cm..
Decision:
4.12 is WAY BIGGER than 1.711, we
must heave Ho.
Test Statistic:
t
=
x
−
a25
s
/
n
=
185.76
−
180
6.98/
25
=
4.12
Criteria:
if
t
> 1.711, heave Ho
Level of significance:
=
5 %
,
df
= 24
a14
Ha:
a25
> 180
Hypotheses:
Ho:
a25
[
180
Value of interest:
µ , mean height of players
Small Sample Hypothesis Test on a Mean
Question 4
In a study of airbag effectiveness, it was found that in 821 crashes of midsize cars equipped with
airbags, 46 of the crashes resulted in the hospitalization of the drivers.
Use a 0.01 level of significance to test the claim that the hospitalization rate for crashes of
midsize cars equipped with airbags is less than 7.8%.
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 Spring '08
 GHATRI
 Statistics, Statistical hypothesis testing

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