252y0112
2/28/01
ECO252 QBA2
FIRST HOUR EXAM
February 20, 2001
Name ____key___________
Hour of class registered _____
Class attended if different ____
Show your work! Make Diagrams!
Note that since relatively few people took this exam, commentary on 252y0111 is more complete.
I. (14 points) Do all the following.
x ~ N 3,7
93
  9`3
z
1. P9  x  9  P
  P 1.71  z  0.86 
7
7 

 P1.71  z  0  P0  z  0.86   .4564  .3051  .7615
93
 0`3
z
2. P0  x  9  P
  P 0.43  z  0.86 
7 
 7
 P0.43  z  0  P0  z  0.86   .1664  .3051  .4715
28  3 
 9`3
z
3. P9  x  28   P
  P0.86  z  3.57 
7
7 

 P0  z  3.57   P0  z  0.86   .4998  .3051  .1947
53

4. Px  5  P z 
  Pz  0.29 
7 

 Pz  0  P0  z  0.29   .5  .1141  .6141
03
 53
z
5. P5  x  0  P
  P 1.14  z  0.43 
7
7 

 P1.41  z  0  P0.43  z  0  .3729  .1664  .2065
6.
A symmetrical interval about the mean with 85% probability. We want
two points x .075 and x.925 , so that Px.075  x  x.925   .8500 . Make a diagram.
If you do a diagram for z , it will show two points, z .075 and z .925 . z .925 (which
has 92.5% above it!) is below zero and z .075 is above zero. Since zero is halfway
between these two points, the diagram will show 85% split between the two sides
of zero, so that 42.5% is between z .925 and zero, and 42.5% is between zero
and z .075 . The probability below z .925 and the probability above z .075 are both 7.5%.
From the diagram, if we replace x by z, P0  z  z.075   .4250 . The closest we can come is
P0  z  1.44   .4251 . So z.075  1.44 , and x    z.075  3  1.447  3  10 .08 or -7.08 to 13.08.
13 .08  3 
  7.08  3
z
To check this note that P7.08  x  13.08   P
  P 1.44  z  1.44 
7
7


 2P0  z  1.44   2.4251   .8502  85%.
x.135 We want a point x .135 , so that Px x .135   .135 . Make a diagram.
The diagram for z will show one point , z .135 which has 13.5% above it (and 86.5%
below it!) and is above zero because zero has only 50% below it. Since zero has 50%
above it, the diagram will show 36.5% between zero and z .135 .
From the diagram, if we replace x by z, P0  z  z.135   .3650 . The closest we can come is
P0  z  1.10   .3643 . So z.135  1.10 , and x    z.135  3  1.107  3  7.70 , or 10.70.
7.
10 .70  3 

To check this note that Px  10 .70   P z 
  Pz  1.10   Pz  0  P0  z  1.10  
7


 .5  .3643  .1357  13.5%.
252y0112
2/21/01
II. (6 points-2 point penalty for not trying part a.)
The Watched Potts Bank of Pottstown wishes to monitor the debt-to-equity ratio of the firms to
which it has made commercial loans. A random sample of 8 of the firms is taken. Assume that the bank is
sampling from a population with a normal distribution.
firm
1
2
3
4
5
6
7
8
Debt/equity ratio
1.34
1.04
1.46
1.22
1.20
1.79
1.38
1.42
a. Compute the sample standard deviation, s , of the debt-to-equity ratios. Show your work! (3)
b. Compute a 95% confidence interval for the mean debt-to-equity ratio,  .(3)
Solution: a)
x
 x  10.85  1.35625

n
x
Firm
1
2
3
4
5
6
7
8
8
15 .0621  81.35625 2
n 1
7
 0.0495411 or s  0.2226 .
s2 
2
Total
 nx
2

10.85
b)
x (Ratio)
1.34
1.04
1.46
1.22
1.20
1.79
1.38
1.42
x2
1.7956
1.0816
2.1316
1.4884
1.4400
3.2041
1.9044
2.0164
15.0621
From the problem statement   .05 . From Table 3 of the syllabus supplement, if the
  x  t  sx and t n21  t .7025  2.365 .
population variance is unknown
2
s
0.0495411 0.2226
sx 


 0.07869 . So   1.35625  2.365 0.07869   1.356  0.186 or 1.170
8
n
8
to 1.542.
2
252y0112 2/16/01
III. Do at least 3 of the following 4 Problems (at least 10 each) (or do sections adding to at least 30 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.
Error: On this page, 1.43 should be changed to 1.33. Remove section f!!!
1. The Watched Potts Bank of Pottstown is a subsidiary of the Umongous Bancorp. The Bancorp is
threatening to remove the current officers of the bank if the firms to which they are making loans have an
average debt-to-equity ratio that exceeds 1.33. Are the officers in trouble? The data are on the previous
page. For your convenience the data are repeated below.
firm
1
2
3
4
5
6
7
8
Debt/equity ratio
1.34
1.04
1.46
1.22
1.20
1.79
1.38
1.42
Test to see if the mean is above 1.33 using the sample mean and standard deviation you found in part II.
a. Choose the correct null hypothesis from the list below and write the alternative hypothesis. (2)
H 0 : x  1.33
H 0 :   1.33
H 0 :   1.33
H 0 : x  1.33
H 0 :   1.33
H 0 : x  1.33
H 0 : x  1.33
H 0 : p  1.33
H 0 :   1.33
H 0 : p  1.33
H 0 :   1.33
H 0 : x  1.33
b. Find a critical value appropriate for this problem, using a confidence level of 95%.(3)
c. Use your critical value to test the hypothesis. State clearly whether you reject the null
hypothesis. (2)
d. Repeat the test using (i) a test ratio (2) and (ii) a confidence interval. (2)
e. Test the hypothesis that the average debt-to-equity ratio is exactly 1.33 at the 95% confidence
level. (2)
Solution: From Table 3 of the Syllabus Supplement:
Interval for
Confidence
Hypotheses
Interval
Mean (
  x  z 2  x
H0 :   0
Known)
H :  
1
Mean (
Unknown)
  x  t 2 s x
Test Ratio
z
0
H0 :   0
t
Critical Value
x  0
xcv   0  z 2  x
x  0
sx
xcv   0  t 2 s x
x
H1 :    0
DF  n  1
a) We are afraid that   1.33 . Because this does not contain an equality, it cannot be a null hypothesis. So
we must test its opposite H 0 :   1.33 and our alternate hypothesis is H1 :   1.33
7
8
 1.895 (If you chose a 2-sided test, t .025
 2.306 )
b) n  8,  0  1.32, DF  n  1  7,   .05, tn1  t .05
From the previous page: x  1.35625 and s x 
s

n
0.0495411
0.2226

 0.07869 .
8
8
3
252y0112
2/21/01
Critical Value: Since this is a one-sided test, xcv  0  t sx  1.33  1.895 0.07869   1.33  0.1491
or 1.4791. It might help to remember that x   0 is always in the 'accept' region.
c) The critical value in b) means that we reject H 0 if the sample mean is above 1.4791.
Since x  1.35625 is not above this critical value, do not reject H 0 .
d) (i) Test Ratio: t 
x  0 1.35625  1.33

 0.3336 . This is in the ‘accept’ region
sx
0.07869
7
below tn 1  t .05
 1.895 , so do not reject H 0 . It might help to remember
that zero is always in the 'accept' region for t.
(ii) Confidence Interval: Since this is a one-sided test, and the alternate hypothesis
is H1 :   1.33 ,   x  t sx  1.35625  1.950 0.07869   1.35625  0.14911
or   1.207 . This does not contradict H 0 :   1.33 , because any mean in the
range 1.207 to 1.33 satisfies both statements, so do not reject H 0 .
7
e) H 0 :   1.33 , H 1 :   1.33 . n  8,  0  1.33, DF  n 1  7,   .05, tn1  t.025
 2.365 .
2
x  1.34625 and sx  0.07741 . Use one of the three methods below.
(i) Critical Value: Since this is a two-sided test, xcv  0  t sx  1.33  2.365 0.07741   1.33  0.1831
2
or 1.1469 and 1.5131. Since x  1.34625 is between these two values, do not reject H 0 .
x  0 1.34625  1.32

 0.3391 . This is in the ‘accept’ region
sx
0.07741
7
7
between  tn1  t.025
 2.365 and tn1  t.025
 2.365 , so do not reject H 0 .
(ii) Test Ratio: t 
2
2
(iii) Confidence Interval: Since this is a two-sided test, use   x  t s x
2
 1.346  2.365 0.29477   1.346  0.1831 . We can thus say that 1.163    1.529.
Since this interval includes  0  1.33, we cannot reject H 0 .
4
252y0112
2/21/01
2. Ken Black tells us that, nationwide, 17% of Americans drink milk as their primary breakfast beverage. A
milk producer in Wisconsin believes that the proportion of Wisconsin residents that drink milk for breakfast
is above 17%. She takes a survey of 500 Wisconsin residents and finds that 105 drink milk as their primary
beverage. Use a 90% confidence level in all parts of this problem.
a. Do a 90% confidence interval for the proportion of residents that use milk as their primary
beverage. (3)
b. Assume that the milk producer wants to estimate the proportion of Wisconsin residents that use
milk as their primary beverage within .005 , how large a sample does she need? Assume that the
correct proportion is 17% in this section. (3)
c. Her original purpose in collecting the data presented above was to test the hypothesis that more
than 17% of the people in Wisconsin drink milk as their primary breakfast beverage. Test this
hypothesis now.
(i) State your null and alternative hypotheses. (2)
(ii) Do a test of your null hypothesis using a test ratio and the 99% confidence level (3)
(iii) Using your test ratio find a p-value for your null hypothesis. (2)
(iv) Do a test of your null hypothesis using a critical value for the observed proportion (2)
(v) Do a test of your null hypothesis using an appropriate confidence interval (2)
d. (extra credit - this was covered in your text, but not in class) How would you modify your work
and conclusion in either (ii) or (iii) in c) if you found that your sample of 500 residents had been
taken in a state with only 3000 residents? (3)
Solution: From Table 3:
Interval for
Confidence
Interval
Proportion
p  p  z 2 s p
pq
n
q  1 p
sp 
Hypotheses
Test Ratio
H 0 : p  p0
H 1 : p  p0
z
p  p0
p
Critical Value
p cv  p 0  z  2 
p 
p
p0 q 0
n
.210 .790   .01831 , z  z  1.645 and
x 105

 .210 , s p 

.05
2
500
n 500
p  p  z s p  .21 1.645.01831  .210  .030 or .180 to .240.
a)   .10 , n  500 , p 
2
b) From the outline n 
n
pqz 2
e2

pqz 2
e2
.17 .83 1.645 2
.005 2
. p  .17, e  .005 and since   .10 , use z  z.05  1.645 . Thus
2
 15272 .805 and we must use a sample of at least 15273.
H 0 : p  .17
c) (i) 
H 1 : p  .17
p 
.17 .83  
500
(ii) Test Ratio: z 
Note that p 0  .17 , n  500 , p 
x 105

 .210 , so that
n 500
.0002822  .016799 . q 0  1  p 0 .
p  p0
p

.210  .17
 2.381 . Since   .10 , and this is a one-sided test,
.016799
use z .10  1.282 . Since 2.381 is more than z .10 , reject H 0 . Make a diagram. Show a normal curve with
a mean at 0 and a 'reject' region starting at z .10  1.282 and show where 2.381 falls on this diagram.
(iii) pvalue  P p  .210   Pz  2.38   .5  .4613  .0087   , so reject H 0 .
5
252y0112
2/21/01
(iv) Critical Value: pcv  p0  z  p  .17  1.282.016799  .1915 . Since p  .210 is
above this value, reject H 0 . Make a diagram. Show a normal curve with a mean at 0.17 and a 'reject'
region starting at .1915 and show where .210 falls on this diagram.
(v) Confidence Interval: Since the alternate hypothesis is H1 : p  .17 , the one-sided confidence interval
will be p  p  z s p  .210  1.282.01831  .1865 . Since p  .1865 contradicts H0 : p  .17 , reject
H 0 . This is an extremely rare case. Make a diagram. Show a normal curve with a mean at .210 and
represent the confidence interval with a region above .1865. Represent the null hypothesis with a region
below .17. show that they do not overlap.
d) Because the population size is now N  3000 , which is less than 20 times the sample size, n  550 , we
need a finite population correction factor. With this factor, the formula for the variance of the sample
proportion becomes
p0 q0 N  n
.17 .83 3000  500
.1411 2500


 .000235245  .01534 .
n
N 1
550
3000  1
550 2999
Note that the text on page 324 omits the -1.
p  p 0 .210  .17

 2.607 . Since   .10 and this is a one-sided test,
(ii) Test Ratio: z 
p
.01534
p 
use z .10  1.282 . Since 2.607 is more than z .10 , reject H 0 . Or
(iii) pvalue  P p  .210   Pz  2.61  .5  .4955  .0045   , so reject H 0 . The conclusions only
became stronger.
6
252y0112
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3. Ken Black says that a soft drink manufacturer takes a sample of 50 12-ounce cans of soda to check that
the mean is at least 12 ounces and finds that the sample mean is 11.985 ounces. From long experience, the
producer assumes that the population standard deviation is 0.10 ounces. Use a 99% confidence level in a) d) in this problem.
a. State the null hypothesis and find the p-value for it. (3)
b. Find a critical value for the sample mean. (2)
c. What would the power of the test in b) be if the mean was, in fact, 11.98 ounces? (3)
d. Doing appropriate calculations, find and sketch the power curve for the test in b). (5)
e. Do a 73% confidence interval for the mean based on the sample mean of 11.985 (3)
Solution: a) First, state the problem and find a critical value or values.
H 0 :   12

0.10
  0.10, n  60,   .01 so  x 

 0.014142 . z  z.01  2.327 .

n
50
H 1 :   12
x  0
11 .985  12
 1.06 and the p-value is Px  11.985   Pz  1.06   .5  .3554  .1446 .
.014142
Note: Since last time you had to use the t table to find the p-value, most of you assumed that you had to do
that this time too. Think! Since you were given a population variance, you should have used z . Moreover,
just because you have to double a p-value in a 2-sided test, doesn't mean you should double it in a 1-sided
test.
z
x

b) Since this is a one-sided test, the formula for a two-sided critical value x cv   0  z   x becomes
2
xcv   0  z  x , so that xcv  12  2.327 0.014142   11.9671 . So we will not reject H 0 if the sample
mean x is greater than or equal to 11.9671. Make a diagram. Put 12 in the middle and show a 'reject'
region below 11.9671. Shade this area to represent the significance level.
c) Since a type II error is wrongly ‘accepting’ the null hypothesis, we compute the probability that the
sample mean will be above or equal to the critical value for each value of 1 . Our computations are below.

x  1 
Note that, in general, for this one-sided hypothesis   P  z  cv
 . Here 1  11 .98 , so
x 


x  1 
 11 .9671  11 .98 
  P  z  cv
  Pz 
  Pz  0.91  .5  .3186  .8186 . The power is 1 - .8186
x 
0.014142



= .1814.
d) Decide on what values of 1 to use to compute  , the probability of a type II error. The usual set
of values includes the mean from the null hypothesis, (12), the critical value (11.9671), a point about
midway between these values (11.98) and two points, one further out beyond the critical value by a distance
about equal to the distance between the null hypothesis mean and the critical value (11.93), and another
roughly halfway between this point and the critical value (11.95).
Compute  for each value of 1 . Since a type II error is wrongly ‘accepting’ the null hypothesis,
we compute the probability that the sample mean will be above or equal to the critical value for each value
of 1 . Our computations are below. Note that, in general, for this one-sided hypothesis

xcv  1 
.
x 

We know that if 1  12 , we will find   1    .99 and the power will be .01.
At the critical value,   .5 and the power is also .5.
  Pz 
7
252y0112
2/28/01
1  11 .95   Px  11 .9671   11 .95   P  z 


11 .9671  11 .95 
  Pz  1.21  .5  .3869  .1131
.014142

power  1    .8869
1  11 .93   Px  11 .9671   11 .93   P  z 


11 .9671  11 .93 
  Pz  2.62   .5  .4956  .0044
.014142

power  1    .9956
You now have the power for 5 points below or equal to 12. Make a diagram.
e) From page 3 of this document,   x  z  x , and we know that
2
x 


x  11.985
and
0.10
 0.014142 .   1  .73  .27 . z  z.135 . To find z.135 (You found this on page 1!),
2
n
50
make a diagram. Show the mean at zero, that the probability above z.135 is .135, so that the probability
between zero and z.135 must be 50% - 13.5% = 36.5%. So P0  z  z.135   .3650 . The best we can do on
the
Normal
table
is
So
and
P0  z  1.10   .3643 .
z.135  1.10 ,

  x  z 2 x  11.985  1.100.014142  11.985  0.016 or 11.969 to 11.1156.
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4. (Use a 99% confidence level in all parts of this problem.) My old friend Ken Black talks about a worker
who has an average of 20 tubes at his work station. To satisfy the requirements of just-in-time inventory, the
superintendent wants the standard deviation of the number of tubes at the station to be less than or equal to
3. The station is visited 8 times and, from the eight measurements taken, she calculates a sample mean
number of tubes at the station of 20.875 and a sample standard deviation of 4.5806.
a. For starters, do a 2-sided test of the statement that the population standard deviation is 3. (3)
b. Do a confidence interval for the standard deviation. (3)
c. Actually the test in a) should have been a 1-sided test, so do it again. (2)
d. The superintendent visits the work station 61 times during the month. She now calculates the
mean at 20.475 and the sample standard deviation at 3.4520. Repeat the test in a) with these new
numbers. (2)
e. Now repeat the test in c) with these new numbers. (2)
f. A paper company will cut farmed timber only if the median height of the trees is over 40 feet. A sample
of 40 trees is taken and 27 of them are over 40 feet. State your null and alternate hypotheses and tell
whether we would reject the null hypothesis at the 99% confidence level. Finally, tell whether your
conclusion means that they do or do not cut. (4)
g. I am sure that the Friendlyville, Alabama police department is writing an average of more than 25
speeding tickets a day on its one-block-long main street. Sho' 'nuff , in a one-day visit I see 38 tickets
written. Assuming that the Poisson distribution applies, state the null and alternative hypotheses and test the
null hypothesis. (3)
h. (Extra Credit) We agree that one day is not enough time to test the hypotheses in g) and so I stay for a
full week. In that time I see 185 tickets written. Repeat the test. (3)
H 0 :  2  9
H :   3
n  8, DF  n  1  7, s  4.5806 ,   .01, and  2  .005 .
Solution: a)  0
or 
H1 :  2  9
H1 :   3
From the outline, the test ratio formula is  2 
n  1s 2
 02

74.5806 2
 16 .31925 . From the chi-square
9
7  20.2778 and  27  0.9893 . Make a diagram. Show a chi-squared distribution with a mean
table, .2005
..995
at about DF  7 , and a 99% 'accept region between the two values of chi-squared just given. Show one
2.5% 'reject' region below the 99.5% value and a second 'reject' region above the 0.5% value. Since
16.31925 falls in the 'accept' region, do not reject H 0 .
b) According to the outline, a two-sided confidence interval would be
74.5806 2   2  74.5806 2
n  1s 2
 22
2 
n  1s 2
12 2
. If we fill
or 7.2430   2  148 .4618 . If we take square
20 .2778
0.9893
roots, we get 2.6913    12.1895 .
H 0 :  2  9
H 0 :   3
n  8, DF  n  1  7, s  4.5806 and   .01 . Make a diagram. Show a
c) 
or 
H1 :  2  9
H1 :   3
chi-squared distribution with a mean at about DF  7 , and a 99% 'accept region below  27  18.4753 .
in the blanks, we get
Show the 1% 'reject' region above the 1% value of chi-squared. Since  2 
n  1s
 02
.01
2
= 16.31925 falls in
the 'accept' region, do not reject H 0 .
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2/21/01
H 0 :  2  9
H :   3
d)  0
or 
H1 :  2  9
H1 :   3
2 
n  1s 2
 02

n  61, DF  n  1  60 , s  3.4520   .01, and

2
 .005 .
60 3.4520 2
 79 .4420 . For large samples z  2 2  2DF   1
9
 279 .4420   260   1  158 .88  119  12 .6049  10 .9081  1.6920 . Make a diagram. Show a
normal curve with a mean at 0 and 'reject' regions below z.005  2.576 and above z.005  2.576 and
show where 1.6920 falls on this diagram. Do not reject H 0 .
H 0 :  2  9
H :   3
e)  0
or 
. Make a diagram. Show a normal curve with a mean at 0 and a 'reject'
H1 :  2  9
H1 :   3
region above z.01  2.327 and show where 1.6920 falls on this diagram. Do not reject H 0 .
f) From outline point B6a for the sign test for a median:
Hypothesis about
Hypotheses about a proportion
a median
If p is the proportion
If p is the proportion
above  0
 H 0 :   0

H 1 :   0
 H 0 : p .5

 H 1 : p  .5
below  0
 H 0 : p .5

 H 1 : p  .5
 H :   40
 H : p .5
We are told that x  27 are above 40 feet and that n  40 . So  0
becomes  0
and
 H1 :   40
 H 1 : p  .5
there are several methods to do this in B6 and the problem solutions. If we use the formula used by the text
2x  1  n
(quoted as z 
in the outline) without a continuity correction we find pval  Px  27 
n

227   40 
  Pz  2.21  .5  .4864  .0136 . Since this is below   .01, we reject H 0 and cut
 P z 
40 

the timber.
 H :   25
g)  0
From the Poisson(25) table, pval  Px  38   1  Px  37   1  .99076  .00924 . Since
 H1 :   25
this is below   .01, we reject H 0 .
 H :   175
h)  0
This time we follow Problem B4d) and say (using the Normal approximation to the
 H1 :   175

185  175 
  Pz  0.74   .5  .2704  .2296 . Since this
Poisson distribution) pval  Px  185   P z 
175 

is not below   .01, we do not reject H 0 .
10