Unformatted text preview: COMM 215 MOCK FINAL EXAM SOLUTIONS
exam solutions page 1
Question 1
Using these results:
achievement test 39 43 21 64 57 47 75 34 52 89 final calculus mark 65
78
52
82
92
a) Calculate the coefficient of correlation for this data. 28
73 98 56 75 ¨ x = 460, ¨ x 2 = 23634 ¨ y = 760 ¨ y 2 = 59816 ¨ xy = 36854
n ¨ xy − ¨ x ¨ y
r=
n(¨ x 2 ) − (¨ x ) 2 n(¨ y 2 ) − (¨ y ) 2 Solution: data: 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: b0 = 40.78 b1 = 0.766
y = 40.78 + 0.766x
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 tdistribution situation.
We know that x = 4.9 , s = 1.2, and degrees of freedom = 15, so s x = 1.2 = 0.3
16 ta/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 / 2
E 2 95% confidence means z = 1.96, we know p = 0.75 and E = 0.03 n = 0.75(0.25) 1.96
0.03 2 = 800.3333 , so sample size is 801 people. Solutions will be posted at www.prep101.com/solutions COMM 215 MOCK FINAL EXAM SOLUTIONS
exam solutions page 2
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
height in cm of 25 Cégep basketball players
176 181 200 178 180 177 179 180 197 181 183 189 180 183 178 185 180 184 190 187 177 176 182 195 196 Do these results support the report at the 0.05 Level of Significance?
Solution: Since n = 25, we’ll use the tdistribution for this upper tail test statistic. ¨ x = 4594, x = 185.76, s = 6.98, t 0.05, 24 = 1.711 Small Sample Hypothesis Test on a Mean
Value of interest: , mean height of players
Ha: % > 180
Hypotheses: Ho: % [ 180
Criteria: if t > 1.711, heave Ho
Level of significance: = 5 %, df = 24
Test Statistic: t = x − % 185.76 − 180
=
= 4.12
s/ n
6.98/ 25 Decision: 4.12 is WAY BIGGER than 1.711, we
must heave Ho. Conclusion: the test supports the claim that the average height of CEGEP basketball players is at least
180 cm..
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%.
Solution: lower tail hypothesis test on a proportion.
First we find the sample statistics: p = 46 = 0.056 1 =
p
821 p(1 − p)
= 0.0094
n Large Sample Hypothesis Test on a Proportion
Value of interest: p , proportion of hospitalizations from crashes with air bags.
Ha: p < 0.078
Hypotheses: Ho: p m 0.078
Criteria: if z < – 2.33, heave Ho
Level of significance: = 1 %,
Test Statistic: z = p−p
= 0.056 − 0.078 = −2.34
0.0094
p(1 − p)
n Decision: since – 2.34 is just a teeny bit less than
– 2.33, we might not heave Ho. We might run
another test. Conclusion: the test barely supports the claim that the hospitalization rate for crashes of midsize cars
equipped with airbags is less than 7.8%. Solutions will be posted at www.prep101.com/solutions COMM 215 MOCK FINAL EXAM SOLUTIONS
exam solutions page 3
Question 5
Suppose that we want to investigate whether there is a relationship between the test scores of
persons who have gone through a certain job training program and their subsequent performance
on the job. A random sample of 400 cases yields these results.
job performance
test score
Poor
Fair
Good
Total
Below average
67
64
25
156
Average
42
76
56
174
Above average
10
23
37
70
Total
119
163
118
400
Test at the 0.01 level of significance whether on the job performance of workers who took the
training program is independent of their test score.
Solution: this is an r × c table and a chisquare test for independence. First, we find the expected values
for the 9 data cells. Degrees of freedom = (r – 1)(c – 1) = 4 e= (row total) % (column total)
, so, e 11 = 156 % 119 = 46.41 etc.
400
grand total observed and expected values for test scores vs. job performance
o11 = 67 o12 = 64 o13 = 25 e11 = 46.41 e12 = 63.57 e13 = 46.02 o21 = 42
e21 =
51.77 o22 = 76 o23 = 56 o31 = 10 o32 = 23 o33 = 37 e22 = 70.91 e23 = 51.33 e31 = 20.83 e32 = 28.53 e33 = 20.65 Ho: the test scores are independent of workers’ job performance.
Ha: the test scores are not independent of workers’ job performance.
alpha = 1% 72 =¨ dof = 4 test criteria: if 7 2 > 13.277, heave Ho. (o − e ) 2 (67 − 46.41 ) 2 (64 − 63.57 ) 2
(37 − 20.65 ) 2
=
+
+ ..... +
= 41.02
e
46.41
63.57
20.65 Decision: since 41.02 > 13.277, we must reject Ho
Conclusion: the test indicates that after taking a training program the test scores and job performance are
not independent of each other. Solutions will be posted at www.prep101.com/solutions COMM 215 MOCK FINAL EXAM SOLUTIONS
exam solutions page 4
Question 6
These are the average weekly losses of manhours due to accidents in 10 industrial plants before and after
a certain safety program was put into operation. Use the 0.05 level of significance to test whether the
safety program is effective.
Solution:
We do a small sample, 2tail tdistribution test on whether the average difference = 0.
before
45
57
73
83
46
34
124
26
33
17
after
36
51
60
77
44
29
119
24
35
11
difference
9
6
7
6
2
5
5
2
–2
6
statistics
¨d
2
¨d
% d = n i = 4.6, ¨ x 2 = 300, (¨ x ) = 2116
% d = n i = 4.6 sd = n ¨ x 2 − (¨ x) 2
n(n − 1) Ho: % d = 0
Level of significance: = 5 % df = 9
Statistic: t = sd = 10(300) − 2116
= 3.13
90 Ha: % d ! 0
Criteria: if  t  > 2.262, reject Ho %d
4.6
=
= 4.65
1d/ n
3.13/ 10 Decision: 4.65 > 2.262 therefore, we must reject the null hypothesis Ho.
Conclusion: The result indicates that the change in the number of man hours lost due to accidents is
significant.
Question 7
A random sample of 120 passengers arriving at Dorval airport took an average of 24.15 minutes
with s = 3.29 minutes to retrieve their luggage and get through customs.
a) construct a 95% confidence interval for the average time it takes for passengers at Dorval to
retrieve their luggage and get through customs.
Solution:
95% confidence interval makes z = ± 1.96 x = 24.15 ! 1.96 3.29
120
the confidence interval about = 24.15 ! 0.59
is: 23.56 < < 24.74 b) By how much is the maximum error increased if we use a 99% level of confidence?
Solution:
99% confidence interval makes z = ± 2.575 E = ! 2.575 3.29
120 = 0.77 minutes 0.77  0.59 = 0.18 minute increase in the maximum error.
c) A random sample of 40 cans of pineapple slices has an estimated mean weight of 15.85 ounces
and a standard deviation of 0.23 ounces. With what confidence can we assert that this estimate
is "off" by at most 0.06 ounces?
Solution: z= E = 0.06 = 1.65 .
s
0.23
n
40 Confidence level is 2(.4505) = 90.1% confidence. Solutions will be posted at www.prep101.com/solutions COMM 215 MOCK FINAL EXAM SOLUTIONS
exam solutions page 5
Question 8
Thirteen percent (13%) of the manuscripts submitted to The New England Journal of Medicine
are accepted for publication. Harvard Medical School students and profs have submitted 50
manuscripts and they claim they have a higher than average acceptance rate.
a) What are Ho and Ha, the null and alternative hypotheses to test their claim.
Solution:
Ho: p = 0.13 Ha: p > 0.13 b) State the decision rule for a 5% significance test of their claim.
Solution:
It’s an upper tail test with alpha = 5% If z > 1.65, reject Ho. c) Test their claim at a 5% level of significance if 10 of their 50 papers are accepted.
Solution:
The sample proportion p = 10 = 0.2
50 p−p
z = 1 p = 0.20 − 0.13 = 1.47
0.048 the standard deviation 1 p = .13(.87)
= 0.048
50 since 1.47 < 1.65, we cannot reject Ho. d) Find the pvalue and use it to state at what level of significance we can reject their claim.
Solution: z = 1.47, therefore the pvalue = 0.5000 – 0.4292 = 0.0708, which is P(z > 1.47).
We could reject their claim at 7.1% of significance.
Question 9
The table lists data on 6 different destinations of all international travellers from Canada in
percent and the number who travelled to each destination in the last month from a sample of 200
customers of a local travel agency.
Run a Chisquare goodness of fit test to see if the local travellers’ choice of destination differs
from the national average. Use the 5% level of significance.
destination Europe Far East South Am. Middle East S.E. Asia others national % 42 % 20 % 16 % 6% 12 % 4% o = 80
o = 44
o = 34
o = 16
o = 20
e = 84
e = 40
e = 32
e = 12
e = 24
Ho: The local and national proportions are the same for the 6 destinations.
Ha: The local and national proportions are NOT the same for the 6 destinations. o=6
e=8 # from 200 dof = 5 test criteria: if 7 2 > 11.070, heave Ho. alpha = 5%
72 =¨ (o − e ) 2 (80 − 84 ) 2 (44 − 40 ) 2
(6 − 8 ) 2
=
+
+ ..... +
= 3.215
e
84
40
8 Decision: since 3.215 < 11.070, we cannot reject Ho
Conclusion: the local travellers’ choice of destination is not different from the national average. Solutions will be posted at www.prep101.com/solutions COMM 215 MOCK FINAL EXAM SOLUTIONS
exam solutions page 6
Question 10
A jobtraining program offers training in advanced computer programming. We suspect that the
probability that an individual who completed the program will be able to find a programming job
within six months, depends on previous related work experience.
The data for a random sample of 80 individuals trained in the program, are in the table:
Years of Experience
0
1
more than 1 # in Program
52
16
12 # who Found a Job
20
8
10 a) At a 5% level of significance, is there enough evidence to support the hypothesis that the
probability of finding a job (within six months) depends on prior related working experience?
Solution:
Ho: Finding a job is independent of work experience.
Ha: Finding a job is NOT independent of work experience.
alpha = 5% dof = 2 test criteria: if 7 2 > 5.99, heave Ho. The pooled proportion of the # who found a job = (20 + 8 + 10) ÷ 80 = 0.475
The expected value in each cell is 0.475 × the # in the program.
0.475 × 52 = 24.7
72 =¨ 0.475 × 16 = 7.6 and 0.475 × 12 = 5.7. (o − e ) 2 (20 − 24.7 ) 2 (8 − 7.6 ) 2 (10 − 5.7 ) 2
=
+
+
= 4.159
e
5.7
24.7
7.6 Decision: since 4.159 < 5.99, we cannot reject Ho
Conclusion: The test indicates that finding a job is independent of work experience.
b) At a 10% level of significance, is there any evidence against the hypothesis that the percentage of
the 80 individuals in the program with “zero”, “one year” and “more than one year” of
experience are respectively 70%, 15%, and 15%?
Solution:
Ho: p 1 = .70; p 2 = .15; p 3 = .15.
Ha: at least one of the proportions is different from the stated values.
alpha = 10% dof = 2 test criteria: if 7 2 > 4.60517, heave Ho. the expected values are: (.70)(80) = 56;
72 =¨ (.15)(80) = 12 ; (.15)(80) = 12 (o − e ) 2 (52 − 56 ) 2 (16 − 12 ) 2 (12 − 12 ) 2
=
+
+
= 1.6193
e
56
12
12 Decision: since 1.6193 < 4.60517, we cannot reject Ho
Conclusion: there is not enough evidence to support the claim that the proportions are different from
those suggested. Solutions will be posted at www.prep101.com/solutions COMM 215 MOCK FINAL EXAM SOLUTIONS
exam solutions page 7
Question 11
An insurance company is analyzing the profiles of its individual policyholders as they relate to
the amount of life insurance coverage carried. Three independent variables are bein g considered:
policyholders annual income (X 1), policyholders number of children under 21 (X 2) and
policyholders age (X 3).
The equation for the line of regression is
^ y = 2.43 + 0.32X 1 + 8.89X 2 + 0.14X 3 where the variables are measured in thousands of dollars.
a) How much life insurance would a 40 year old woman be expected to carry if she has an annual
income of $50,000 and 3 children?
Solution: We set X1 = 50 and X2 = 3 and X3 = 40 to get
^ y = 2.43 + 0.32(50) + 8.89(3) + 0.14(40) = 50.7 She should carry $50 700 coverage.
b) How much does the amount of coverage increase with each child?
Solution: since the coefficient of X 2 = 8.89, each child increases the coverage by $8 890.
c) If two 40yearold policyholders each have five children, but one makes $40,000 per year and the
other makes $50,000 per year, how much difference would you expect there to be in their
respective insurance coverages?
Solution: Since the only thing different is the annual income, the difference will be
50(0.32) – 40 (0.32) = 10(0.32) = 3.2 which is $3 200. Solutions will be posted at www.prep101.com/solutions COMM 215 MOCK FINAL EXAM SOLUTIONS
exam solutions page 8
Question 12:
This is the Excel Regression output for data on the annual income of 5 executives, related to the
executive’s age and the number of years of college he or she has completed.
SUMMARY OUTPUT
Regression Statistics
Multiple R
R Square
Adjusted R Square
Standard Error
Observations 0.9998
0.9995
0.9990
237.1248
5 ANOVA
df SS
233,239,544
112,456
233,352,000 MS
116,619,772
56,228 Coefficients
Standard Error
16740
1629.26
1961
35.29
5976
93.88 t Stat
10.27
55.56
63.65 Regression
Residual
Total 2
2
4 Intercept
age (years)
years of college F
2074 Significance F
0.0005 Pvalue
0.0093
0.0003
0.0002 a) What is the equation for the line of multiple regression? (use x 1 for age, x 2 for college years)
Solution:
From the column marked Coeffiecients we find b0 the constant, b 1 the coefficient of the “age”
variable, and b 2, the coefficient of the “years of college completed” variable.
^ y = −16740 + 1961 x 1 + 5976 x 2 b) Use the equation from part (a) to estimate the annual income of a 39yearold executive who
has completed 4 years of college.
Solution:
Set x 1 = 39 and x 2 = 4 to get
^ y = −16740 + 1961 (39) + 5976(4) = 83, 643 his annual income would be $83,600 (rounded to the nearest hundred).
c) By how much will a 36yearold executive’s income increase once he completes one more
year of college?
Solution:
Since b 2 = 5976, his income will increase by $5976 or about $6000.
d) Find the coefficient of determination and interpret its meaning.
Solution:
According to the Excel sheet R² = .9995 which means that 99.95% of the variations in an
executive’s annual income are due to variations in age and number of years of college completed. Solutions will be posted at www.prep101.com/solutions ...
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This note was uploaded on 09/17/2010 for the course COMM 215 taught by Professor Ghatri during the Spring '08 term at Concordia Canada.
 Spring '08
 GHATRI

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