DS-203
Fall 2004
Final Exam
Name: ___________________
Show All your Work
1.
An engine manufacturer tests eight engines for pollution (measured in milligrams of
particulate matter per cubic yard) with the following results:
72,
74,
75,
79,
81,
70,
77,
85
70
72
74
75
77
79
81
85
a)
Calculate the sample mean.
x 613
x

= 76.625
n
8
b)
c)
What is the sample Median?
75  77 152

= 76
2
2
Calculate the sample Standard deviation.
(x) 2
n 
n 1
x 2 i 
S=
d)
7
(613) 2
8
 24.268 = 4.93
25% of the data points are less than___73___. Fill in the blanks.
1 
e)
47141 
72  74 146

= 73
2
2
Calculate the 96% confidence interval for  (average pollution produced by these
type of engines)?
t* = t .02, 7 = 2.517
xt*
s
n
76.625  2.517
4.93
8
76.625  4.39 => (72.24, 81.02)
f)
How would your result in part (e) be affected if the sample size had been 25?
Would the interval be larger or smaller? Calculate the 96% confidence interval
with the new sample size.
s
4.93
Smaller interval x  t *
=> 76.625  2.517
= 76.625  2.48 = (74.14, 79.11)
n
25
1
2.
Foreign vessels fishing for haddock within 200 miles of the American coast
are allowed set limits of chinook salmon incidental to the catch. American
observers aboard the foreign ships monitor the number of chinook salmon
caught to make sure the limit is not exceeded. The observer counts the
Chinook in a sample of 50 trawls. If the mean number of Chinook salmon per
trawl is greater than 10 the vessel will be asked to leave the waters or risk
heavy fines and /or confiscation of the ship. A sample of 50 trawls from the
Japanese vessel Takahahi Maru No. 14 yielded a mean of 11.0 chinook
salmon with a standard deviation of 5.4 chinook salmon.
a) State the hypotheses the observer is testing with this sample.
H o :  =10
H a :  >0
b) What is the test statistic value?
Z=
x
s/ n
11  10

5.4 / 50
= 1.31
or
t=
x
s/ n

11  10
5.4 / 50
= 1.31
c) What is the rejection region? Use  = .05.
z
1.645
5%Reject H o
Do not reject H o
1.68
1.645
Do not reject H o
Reject H o
t (49)
d) What is your conclusion?
Do not reject H o since Z = 1.31 < 1.645
The Japanese Vessel did not exceed the limit.
2
3.
Different kinds of companies compensate their key employees in different
ways. Established companies may pay higher salaries, while new companies may
offer stock options that will be valuable if the company succeeds. Do high-tech
companies tend to offer stock options more often than other companies? One
study looked at a random sample of 200 companies. The result is summarized in
the following table:
High tech.
Non-high-tech.
n
91
109
X
73
75
Treat these two groups as SRSs of high-tech and non-high-tech companies.
a)
Give a 90% confidence interval for the difference in the proportions of the
two types of companies that offer stock options.
~ 73  1
P1 
= .796
91  2
~
SE (D P ) =
75  1
~
P2 
= .685
109  2
~
~
~
~
P1 (1  P1 ) P2 (1  P2 )

=
n1  2
n2  2
=
(.796)(1  .796) .685(1  .685)

91  2
109  2
.00369 = .061
Z* = Z .95 = 1.645
~ ~
~
P1  P2  Z * SE ( DP )
(.796 - .985)  1.645 (.061) => .111  .09993
( .01 , .21)
b)
Test the null hypothesis that the two types of companies are equally likely
to offer this kind of benefit to their employees. State appropriate null and
alternative hypotheses, compute the test statistic, and report the P-value
and your conclusion use  = 5%.
73
75
Pˆ1 
 .80 Pˆ2 
 .69
H o : P1  P2
91
109
73  75
Pˆ 
 .74
H a : P1  P2
91  109
Z
Pˆ1  Pˆ2
.80  .69

 1.77
SE ( DP )
.062
SE(DP) =
=
.74(1  .74)(
1
1

)
91 109
.003879 = .062
P-Value = 2 * P(z > 1.77) = 2(1-.9616) = .0768
Do not reject H o . The two types of companies are equally likely to offer thiskind
of benefit to their employees.
3
Multiple choice questions
Select the best answer
1.
The number z* with probability 0.1 falling below it in the standard distribution is
(a) 4.032
(b) 3.365
E
(c) 2.571
(d) 1.476
(e) None of the above.
2.
If the P-value is smaller than the level of significance , the null hypothesis is
rejected.
(a) True
(b) False
(c) Not enough information
A
3.
Which of the following is not true regarding the advantages of the mean as a
measure of location?
(a)
The mean is relatively insensitive to skewness in data?
(b)
The mean uses all the values in the data set.
(c)
The mean is uniquely defined for a given set of data.
(d)
All of the above are true.
4.
If Z has a standard Normal distribution, then P (Z > 1) = P (Z < -1).
(a) True
(b) False
A
A
5.
You are told that your score on an examination was at the 35th percentile. Your
score was
a) above the third quartile
b) between the median and the third quartile
C
c) between the first quartile and the median
d) below the first quartile
e) non of the above.
6.
The number of television sets sold in a day at an appliance store is an example of
a continuous random variable.
7.
(a) True
(b) False
Smaller p-values indicate more evidence in support of:
a.
b.
c.
d.
the null hypothesis
the alternative hypothesis
the quality of the researcher
further testing
B
B
4
Use the following problem for questions 8 and 9.
Parking violations in a particular city result in a fine of either $5, $10, or $25, depending
on the type of violation. Suppose that X = amount of fine has the following probability
distribution.
X
5
10
25
P(X)
.2
.70
.10
8.
The mean value of X is.
B
a) 5.22
b) 10.50
c) 27.25
d) 6300
e) None of these
9.
The standard deviation of X is.
a) 5.22
b) 10.50
c) 27.25
d) 6300
e) None of these A
10.
An opinion poll asks a random sample of adults whether they favor banning
ownership of handguns by private citizens. A commentator believes that more
than half of all adults favor such a ban. To assess this claim, you would test
a.
H0:  = 0 against Ha:  > 0
b.
H0: P = 0.5 against Ha: P > 0.5
B
c.
H0: P = 0.5 against Ha: P 0.5
d.
H0: P > 0.5 against Ha: P = 0.5
e.
None of the above
11.
The hypothesis H0: P1 = P2 is equivalent to H0: P1-P2 = 0.
(a) True
(b) False
(c) Not enough information
A
12.
As the sample size increases, the t distribution becomes more similar to the
__ distribution.
a. Normal
A
b. Poisson
c. Exponential
d. Binomial
e. none of the above
13.
In a large population of college students, 20% of the students have experienced
feeling of math anxiety. If you take a random sample of 10 students from this
population, the probability that exactly 2 students have experienced math anxiety
is
10!
(.2) 2 (1  .2) 8  .3019
a. .3020
P(x=2) =
2!8!
b. .2634
A
c. .2013
d. .5
e. 1
14.
Refer to the previous problem. The standard deviation of the sample proportion of
students who have experienced math anxiety is
a. .0160
b. .1265
B
c. .2530

P(1  P)
(.2)(.8)

 .1265
n
10
d. 1
5
15.
e. .2070
You want to estimate the mean SAT score for a population of students with a 90%
confidence interval. Assume that the population standard deviation is  = 100. If
you want the margin of error to be approximately 10, you will need a sample size
of

a. 16
10 = z*
b. 271
10 = 1.645
n
100
n
 10 n = 164.5
B
c. 38
2
d. 1476
16.
17.
18.
19.
 164.5 
n= 
 = 270.6
 10 
e. none of these
You want to compute a 90% confidence interval for the mean of a population with
unknown population standard deviation. The sample size is 30. The value of t*
you would use for this interval is
a. 1.96
b. 1.645
c. 1.699
C
d. .90
e. 1.311
A 90% confidence interval for the mean reading achievement score for a
population of third grade students is (44.2, 54.2). The margin of error of this
interval is
a. 95%
54.2  44.2
5
b. 5
B
2
c. 2.5
d. 54.2
e. The answer cannot be determined from the information given
Refer to the previous problem. The sample mean is
a. 44.2
b. 54.2
D
c. .95
d. 49.2
e. The answer cannot be determined from the information given
A randomly selected student is asked to respond yes, no, or maybe to the
question: “Do you intend to vote in the next presidential election?” The sample
space is { Yes, no, maybe}. Which of the following represent a legitimate
assignment of probabilities for this sample space?
a. .4, .4, .2
b. .4, .6, .4
A
c. .3, .3, .3
d. .5 .3, -.2
6
20.
21.
22.
e. none of these
In a statistics course a linear regression equation was computed to predict the final
exam score from the score on the first test. The equation was ŷ = 10 + .9 x where
y is the final exam score and x is the score on the first exam. Eve scored 95 on the
first test. What is the predicted value of her score on the final exam?
(a)
95
(b)
85.5
= 10 + .9 (95) =
D
(c)
90
(d)
95.5
(e)
none of these
The heights of American men aged 18 to 24 are approximately normally
distributed with mean 68 inches and standard deviation 2.5 inches. Half of all
young men are shorter than
(a)
65.5 inches
(b)
68 inches
B
(c)
70.5 inches
(d)
can’t tell, because the median height is not given
(e)
none of these
You want to compute a 95% confidence interval for a population mean. Assume
that the population standard deviation is known to be 25 and the sample size is
100. The value of z to be used in this calculation is
a. 1.645
b. 2.009
C
c. 1.960
d. .8289
7