252y0313 9/29/03
ECO252 QBA2
FIRST HOUR EXAM
October 2, 2003
Name ________KEY_______
Hour of class registered _____
Show your work! Make Diagrams!
I. (8 points) Do all the following Normal distribution problems. Before you take another exam please read
‘Things that you should never do on an exam or anywhere else’ in your syllabus supplement.
1)
The owner of a fish market determined that the average weight for a catfish is 3.2 pounds with a standard deviation of 0.8
pounds. Assuming the weights of catfish are normally distributed, the probability that a randomly selected catfish will
weigh more than 4.4 pounds is _______.
x ~ N 3.2,0.8 . Make a diagram! Either show a Normal curve with a mean at zero and shade the
area above 1.5, or show a Normal curve with a mean at 3.2 and shade the area above
4.4.
4.4  3.2 

Px  4.4   P  z 
 Pz  1.5  Pz  0  P0  z  1.5
0.8 

 .5  .4332  .0668
2)
The owner of a fish market determined that the average weight for a catfish is 3.2 pounds with a standard deviation of 0.8
pounds. Assuming the weights of catfish are normally distributed, the probability that a randomly selected catfish will
weigh between 3 and 5 pounds is _______.
x ~ N 3.2,0.8 . Make a diagram! Either show a Normal curve with a mean at zero and shade the
area between -0.25 and 1.00, or show a Normal curve with a mean at 3.2 and shade the
area between 3 and 5.
5  3.2 
 3  3.2
P3  x  5  P 
z
 P 0.25  z  1.00 
0.8 
 0.8
 P 0.25  z  0  P0  z  2.25   .0987  .4878  .5865
3)
The owner of a fish market determined that the average weight for a catfish is 3.2 pounds with a standard deviation of 0.8
pounds. A citation catfish should be one of the top 2% in weight. Assuming the weights of catfish are normally
distributed, at what weight (in pounds) should the citation designation be established? (Find
x.02 )
Make a diagram! Draw a Normal curve with a mean at zero. By definition z .02 is the
point with 2% above it and 98% below it. Since 50% is below zero, it must be true that
P0  z  z.02   .4800 . Show this on your diagram. If you look at the Normal table, the
closest you can come to .4800 is P0  z  2.05   .4798 , though P0  z  2.06   .4803
is almost as good. So z .02 is 2.05 or 2.06. Use the formula x.02    z.02
 3.2  2.05 0.8  3.2  1.64  4.84 or x.02    z.02  3.2  2.06 0.8  3.2  1.648
 4.848 . Both of these are fine. An exact answer should be between these two, maybe
4.843.
252y0313 9/29/03
4)
The owner of a fish market determined that the average weight for a catfish is 3.2 pounds with a standard deviation of 0.8
pounds. Assuming the weights of catfish are normally distributed, the probability that a randomly selected catfish will
weigh less than 2.2 pounds is _______.
x ~ N 3.2,0.8 . Make a diagram! Either show a Normal curve with a mean at zero and shade the
area below -1.25, or show a Normal curve with a mean at 3.2 and shade the area below
2.2.
2.2  3.2 

Px  2.2   F 2.2  P  z 
 Pz  1.25   Pz  0  P 1.25  z  0 
0.8 

 .5  .3944  .1056
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II. (5 points-2 point penalty for not trying part a.)
A random sample is taken of the number of automobiles involved in fog-related accidents on the
New Jersey Turnpike. The following data is found.
Accident
1
2
3
4
5
No. of cars
30
3
2
5
4
a. Compute the sample standard deviation, s , of the number of vehicles involved. Show your
work! (3)
b. A turnpike spokesperson says that the mean number of automobiles involved in this type of
pile-up is at most 6   6 . Use the mean and standard deviation that you computed above to evaluate this
statement at the 95% confidence level.(3)
c. (Extra credit) A turnpike spokesperson says that the median number of automobiles in one of
these accidents lies between 3 and 5. Treating this as a confidence interval for the median, what is the
confidence level? What about these data makes me suspect that locating the median is a better idea than the
hypothesis test in b)? (4)
Solution: a)
index x in order
x
x2
x 44
x

 8.80
30 900
1
2
n
5
3
9
2
3
2
4
3
4
x 2  nx 2
s
5
25
4
5
n 1
4
16
5
900
44 954
954  58.80 2

 141 .70
x 2  954 , n  5.
x  44 ,
So
4




s  141 .70  11 .904
b) H 0 :   6, H 1 :   6.
From the formula table:
Interval for
Confidence
Interval
Mean (
  x  t 2 s x
unknown)
DF  n 1
H0 :   0
H1 :    0
Test Ratio
t
x  0
sx
Critical Value
xcv    t 2 s x
141 .70
 28 .34  5.3235 .
5
n
You have not done a hypothesis test unless you have stated your hypotheses, run the numbers and
stated your conclusion. You must do only one of the following.
x   0 8.80  6

 0.5260 . Since this is a one(i) Test ratio for a test of the mean. The test ratio is t calc 
sx
5.3235
Note that s x 
s
Hypotheses

4
 2.132 . The ‘reject’ zone is the area above 2.132. Since 0.526 is not
sided, right-tail test, pick tn 1  t .05
in the ‘reject’ zone, do not reject the null hypothesis.
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(ii) A critical value for the sample mean. Because the alternate hypothesis is   6, we need a critical value
4
for x above 6. Since this is a one-sided test, we use tn 1  t .05
 2.132 . The two-sided formula ,

n 1

n 1
x   t
s , becomes x    t
s  6  2.132 5.3235   6  11.350  17.350 . Make a
cv
0

2
x
cv
0

x
diagram showing an almost Normal curve with a mean at 6 and a shaded 'reject' zone above 17.350. Since
x  8.80 is not above 17.350, we do not reject H 0 .
(iii) A confidence interval for the population mean. Because the alternate hypothesis is   6, we need a
' ' confidence interval.   xcv  t  2 s x becomes   xcv  t s x  8.80  2.132 5.3235   8.8  11.350 or
  2.550 . Make a diagram showing an almost Normal curve with a mean at x  8.80 and the
confidence interval above - 2.550 shaded. Since  0  6 is above – 2.550 and thus in the confidence
interval, we do not reject H 0 . (The fact that this interval makes little sense is a clue to the answer to c).
c) The ordered sample is 2, 3, 4, 5, 900 . We have proposed a confidence interval of 2    5. Remember
that the probability of picking a number above the median is .5. This confidence interval is wrong if (i) all
five numbers in our sample are above the median, (ii) four out of the five numbers are above the median,
(iii) all five numbers in our sample are below the median, or (iv) four out of the five numbers are below the
median. To find the probability of the first two events, call a number above the median a success and find
the probability of four or more successes in 5 tries. According to the Binomial table for p  .5 and
n  5, Px  4  1  Px  3  1  .81250  .1875 . To find the probability of the other two events, find the
probability of four or more failures in 5 tries, which is the probability of zero or one successes.
Px  2  .1875 , as always, equal to the first probability. So the significance level is
  2Px  4  2.1875   .3750 , and the confidence level is 1 - .3750 = .6250, terribly low.
The use of the method in b) is based on the idea that either the sample mean is Normally distributed or the
underlying distribution is Normal. For a sample this small, the sample mean is Normally distributed only if
the underlying distribution is Normal, and the presence of one number that is considerably larger than the
rest of the numbers in the sample makes it look like the underlying distribution is highly skewed. The
median is resistant to the presence of outliers and thus a more appropriate statistic.
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III. Do all of the following Problems (17 points) Show your work except in multiple choice questions.
1. The marketing manager for an automobile manufacturer is interested in determining the proportion
of new compact car owners who would have purchased a passenger-side inflatable air bag if it had
been available for an additional cost of $300. The manager believes from previous information that
the proportion is 0.30. Suppose that a survey of 200 new compact car owners is selected and 79
indicate that they would have purchased the inflatable air bag. If you were to conduct a test to
determine whether there is evidence that the proportion is different from 0.30 and decided not to
reject the null hypothesis, what conclusion could you draw? (2)
a) There is sufficient evidence that the proportion is 0.30.
b) There is not sufficient evidence that the proportion is 0.30.
c) There is sufficient evidence that the proportion is not 0.30.
d) *There is not sufficient evidence that the proportion is not 0.30.
2.
An entrepreneur is considering the purchase of a coin-operated laundry. The present owner claims
that over the past 5 years, the average daily revenue was $675 with a population standard deviation
of $75. A sample of 30 days reveals a daily average revenue of $625. If you were to test the null
hypothesis that the daily average revenue was $675, which test would you use? (1)
a) * z -test of a population mean
b) z -test of a population proportion
c) t -test of a population mean
d)
3.
 2 -test of population variance
How many Kleenex should the Kimberly Clark Corporation package of tissues contain?
Researchers determined that 60 tissues is the average number of tissues used during a cold.
Suppose a random sample of 100 Kleenex users yielded the following data on the number of
tissues used during a cold: X = 52, s = 22. Suppose the alternative we wanted to test was
H1 :   60 . State the correct rejection region for  = 0.05. (2)
a) Reject H0 if t > 1.660.
99
b) *Reject H0 if t < – 1.660. This is a left-tailed test, so we use tn 1  t .05
.
c) Reject H0 if t > 1.984 or t < – 1.984.
d) Reject H0 if t < – 1.984.
4.
(ASW) The average monthly income of recent WCU business graduates four years ago was $3200.
I believe that the recent recession has lowered the income and think I have the data to prove it. The
data consists of 72 responses from a random sample of more recent graduates. Find the correct set
of hypotheses below. (2)
H 0:   3200
a) 
H 1 :   3200
b)
H 0:   3200
*
H 1 :   3200
c)
H 0:   3200

H 1 :   3200
d)
H 0:   3200

H 1 :   3200
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252y0313 9/29/03
5.
(Bassett et. al. p. 126) The manager of a bottling plant is anxious to reduce the variability of the
weights of bottled fruit. Over the last two years the standard deviation has held at 15.2 grams and
the distribution of the weights is believed to be normal. A new machine is introduced and the
weights of the bottles in a randomly selected sample are
 987 , 966 , 955 , 977 , 981, 967 , 975 , 980 , 953, 972  , so that n  10, x  9713 and

x
2
 9435347 . Does the new machine have better performance?
a) State your null and alternative hypotheses. (1)
b) Do a test of your null hypothesis using a test ratio and a 5% confidence level. (3)
c) Do a 95% 2-sided confidence interval for the parameter of interest. (2)
Solution: a) We want to improve performance by reducing variability, so we will switch if
H 1 :   15 .2 is true. The opposite is H 1 :   15.2. This is a small-sample test of a standard
deviation or variance.
b) According to the formula table, the test statistic is  2 
 x  9713  971 .3 and
x
n
2 
10
n  1s
2
 02

9123 .344 
15 .2 2
x
s
2
 nx 2
n 1

n  1s 2
 02
and
9435347  10 971 .32
 123 .344 . So
9
 4.805 . Since this is a left tail test and   .05 , our rejection
9
 3.3251 . Since our test ratio is not in the rejection region, we do not
region is below  .95
reject the null hypothesis or buy the new machine.
c) According to the outline, the interval is
n 1
 2
2
9 
n 1
n  1s 2
9
 22
2 
n  1s 2
12 2
. Since
  2 .025  19.0228 ,  21   2 .975  2.7004 and n  1s 2  9123.344  1110 ,
2
1110
1110
 2
or 58 .35  2 411 .05 or, taking square roots,
19 .0228
2.7004
7.639    20.274
the interval is
6.
(Bassett et. al. p121) A box of bolts comes from a manufacturer whose bolts supposedly have a
mean length of 5cm. and a variance of (not a standard deviation) of 0.05. A random sample of 10
bolts is taken with the following results:  5.68, 5.13, 5.82, 5.71, 5.36, 5.52, 5.29, 5.77, 5.45, 5.39  .
From these we calculate a sample mean of 5.512.We assume that the reported variance is a valid
population variance and do not calculate a sample variance and we use a 5% significance level.
a) Calculate a confidence interval for the mean on the assumption that the underlying
distribution is normal. Is the population mean significantly different from 5cm? (2)
b) If, instead, we found x  5.0707 find a p-value for the test . (2)
c) If in b, we found that these 10 bolts had come from a population of 50, how would this
affect the p-value? You do not need to do any new computations, but I expect a short
understandable explanation. (1)
d) Caramba! We look at the distribution of the lengths and find that it is not normal. So we
do a test that the median is 5. What p-value do you get and what are the implications if
our significance level is .01? (2)
e) (Extra credit) Find a confidence interval for the median with the lowest confidence level
that is above 95%.(2)
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252y0313 9/29/03
Solution: We are testing whether the mean is 5. We are told that x  5.512 ,
n  10,  2  0.05 and   .05 .
From the formula table we have:
Interval for
Confidence
Hypotheses
Interval
Mean (
H0 :   0
  x  z 2  x
known)
H1 :    0
Test Ratio
z
x  0
x
Critical Value
xcv    z  2  x
a) Using the formula above for a confidence interval, z  z.025  1.960 and
2

2
0.05

 0.0707 . So   5.512  1.960 0.0707   5.512  0.139 or 5.373 to
n
10
n
5.651. Since this interval does not include 5, we can say that the population mean is significantly
different from 5.
x   0 5.0707  5

 1. The p-value is
b) We are testing H 0 :   5 against H 1 :   5 . z 
x
0.0707
x 

(from the Normal table) 2Px  1  2.5  .3413   .3174 .

N n
. The finite population correction makes the standard error smaller, and
n N 1
thus makes z larger. If z is larger than 1, the probability of being above that value will be
smaller, so that the p-value will be smaller.
d) We are testing H 0 :  5 against H 1 :  5 All ten numbers are above 5, if the median is 5,
the probability of being above 5 is .5, the probability of getting 10 numbers above 5 is, according
to the Binomial table, Px  10   1  Px  9  1  .99902   Px  0  .00049  . Since this
is a 2-sided test, we say the p-value is twice this, or .00098. Since this is below .01, we reject the
null hypothesis.
d) The numbers are  5.68, 5.13, 5.82, 5.71, 5.36, 5.52, 5.29, 5.77, 5.45, 5.39  . If we put them in
c)  x 
 x , x 2 , x 3 , x 4 , x 5 , x 6 , x 7 , x8 , x 9 , x10 
order we have  1
 . Since the probability of
5.13, 5.29, 5.36, 5.39, 5.45, 5.52, 5.68, 5.71, 5.77 , 5.82 
being above the median is .5, if we use the 1st and the 10th number and call being above the
median a success the probability that we are wrong is
2Px  10   21  Px  9  21  .99902   2Px  0  2.00049   .00098 . This is the
significance level, the confidence level is 1-.00098 = .99902. If we use the second and 9th
number the significance level is 2Px  9  2Px  1  2.01074   .02148 and the confidence
level is .97852 and if we use the third and 8th number, the significance level is
2Px  8  2Px  2  2.05469   .10939 , which is obviously too high. So our 97.9%
confidence interval is 5.29    5.77.
7.
We do a test of H 0 : p  .60 against H 1 : p  .60 and find a test ratio of z  .98.
a) What is the p-value of our result? (2)
b) Assuming that your p-value in a) is correct, would we reject the null hypothesis if the
significance level was 8%? Why? (1)
Solution: a) Since this is a 2-sided test, 2Pz  0.98   2.5  .3365   .3270 .
b) We reject the null hypothesis if the p-value is below the significance level. Since
.3270 is not below .08, we do not reject the null hypothesis.
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252y0313 9/29/03
ECO252 QBA2
FIRST EXAM
October 2 2003
TAKE HOME SECTION
Name: ______Back_______________
Social Security Number: _________________________
IV. Do the first two problems (at least 10 each) (or do sections adding to at least 20 points - Anything
extra you do helps, and grades wrap around) . Show your work! State H 0 and H 1 where appropriate.
You have not done a hypothesis test unless you have stated your hypotheses, run the numbers and
stated your conclusion. Use a 95% confidence level unless another level is specified.
1. An airline president is tracking late arrivals and believes that the proportion is at most p 0 . Suppose
that a sample of 200 flights is selected and 82 were late. Do the following:
a) To find p 0 , take the third digit of your Social Security Number divide by 100 and add it to .30.
For example, my Social Security Number is 265398248 and the third digit is 5, so the value of p 0
that you would use is 30  5% or .35. (no point credit for this section.)
b) Using the value of p 0 that you found in a) prepare to conduct a test to determine whether there
is evidence that the proportion is at most p 0 by stating you null and alternative hypotheses. (1)
c) Find a critical value for the sample proportion for the hypotheses in b), using a significance
level of 1% specify what your ‘reject’ region is and use it to test the null hypothesis. (2)
d) (Extra credit) Assume that the actual population proportion is p 0  .03 . (Since my p 0 was .35,
I would assume that p1 was .38) , find the power of the test in c). If I had used a lower significance
level, explain whether the power would be higher, lower, or the same. (3.5)
e) Compute a test ratio for the hypotheses in c) and test the hypotheses using a significance level of
1% (2)
f) Use your test ratio in e) to get a p-value for the hypothesis in c) and explain whether and why
you would reject the null hypothesis if the significance level was 3%. (2)
g) Test the hypotheses in c) using an appropriate confidence level and a significance level of 1%
(2)
h) If you were doing a 2-sided 99% confidence interval for the proportion of flights that were late
and wanted the proportion to be known within .01 , how large a sample would you use if you expected
the proportion to be about 20%? What if you thought the proportion was about 4%? (2)
i) (Extra credit) do a power curve for the test in c), using a few carefully chosen values of p1 that
are above your p 0 . (4.5)
Solution: See 252y0313a. Take only the pages you need!
2. Robert N Carver presents us with a data set representing departure delays for flights between JFK
and LAX airports. The sample of 52 flights gives us a sample mean delay of 2.21 minutes. Assume a
population standard deviation with a value of 10 + the third digit of your Social Security number.
(Example: My Social Security Number is 265398248 and the third digit is 5, so the population
standard deviation will be   10  5  15 ). Test the assertion that the mean delay is less than 2.5
minutes. (Use a significance level of 10% in this problem.)
a) State your null and alternative hypotheses (1)
b) Find a critical value of the sample mean and specify your ‘reject’ region. Make a diagram. (1)
c) Do you reject the null hypothesis? Use your diagram to show why. (1)
d) Create a power curve for this test. (Note that negative values of the mean are not impossible and
would indicate an early departure.) (6)
e) Assume that you want a 90% 2-sided confidence interval for the mean delay, with an error of
1.0 . How large a sample would you need? (2)
f) Find a p-value for the null hypothesis.(2)
Solution: See 252y0313b. Take only the pages you need!
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