252y0211 2/27/02
ECO252 QBA2
FIRST HOUR EXAM
February 19, 2002
Name _____Key______________
Hour of class registered _____
Class attended if different ____
Show your work! Make Diagrams! How many of you looked at "Things You Should Never Do"
before this exam?
I. (14 points) Do all the following.
x ~ N 9,6
28 .2  9 
14  9
z
 P0.83  z  3.20 
1. P14  x  28 .2  P 
6 
 6
 P0  z  3.20   P0  z  0.83  .4993  .2967  .2026
9 9
7 9
z
 P 2.67  z  0  .4962
2. P7  x  9  P 
6
6 

13  9 
  13  9
z
 P 3.67  z  0.67 
3. P13  x  13  P 
6 
 6
 P 3.67  z  0  P0  z  0.67   .4999  .2486  .7485
0  9
  13  9
z
 P 3.67  z  1.50 
4. P13  x  0  P 
6
6 

 P 3.67  z  0  P 1.50  z  0  .4999  .4332  .0667
17  9 

5. F 17  (The cumulative probability up to 17) Px  17   P  z 
6 

 Pz  1.33  Pz  0  P0  z  1.33  .5  .4082  .9082
6. A symmetrical interval about the mean with 58% probability.
We want two points x.79 and x.21 , so that Px.79  x  x 21   .5800 . From the diagram,
if we replace x by z, P0  z  z.21   .2900 . The closest we can come is
P0  z  0.81  .2910 . So z.21  0.81 , and x    z.21  9  0.816  9  4.86 ,
13 .86  9 
 4.14  9
z
or 4.14 to 13.86. To check this note that P4.14  x  13 .86   P

6
6


 P0.81  z  0.81  2P0  z  0.81  2.2910   .5820  58%
7. x.095
We want a point x.095 , so that Px x.095   .095 . (This is the 90.5 percentile)
From the diagram, if we replace x by z, P0  z  z.095   .4050 . The closest we can come is
P0  z  1.31  .4049 . So z.095  1.31 , and x    z.095  9  1.316  9  7.86 , or 16.86 .
16 .86  9 

To check this note that Px  16.86   P z 
  Pz  1.31
6


 Pz  0  P0  z  1.31  .5  .4049  .0951  .095
252y0211 2/21/02
II. (6 points-2 point penalty for not trying part a.)
A new product is tried on seven patients. Their breathing capacity after using the product is shown
below (Note: You may want to move the decimal point to the left and work in thousands.).
Patient capacity
1
2
3
4
5
6
7
2850
2380
2800
2860
2300
2650
2640
a. Compute the sample standard deviation, s , of the breathing capacity. Show your work! (3)
b. Compute a 90% confidence interval for the mean breathing capacity,  .(3)
Solution: a)
Original data
Data divided by 1000
2
Row
Row
x
x
x
x2
1
2850
8122500
1
2.85
8.1225
2
2380
5664400
2
2.38
5.6644
3
2800
7840000
3
2.80
7.8400
4
2860
8179600
4
2.86
8.1796
5
2300
5290000
5
2.30
5.2900
6
2650
7022500
6
2.65
7.0225
7
2640
6969600
7
2.64
6.9696
18480 49088600
18.48 49.0886
n  7,
n  7,
x 2  49088600
x 2  49 .0886
x  18480 ,
x  18 .48,


 x  18480  2640

x
n


x
7
49088600  72640 
s

n 1
6
 50233 .33 or s  224.13 .
2
2
2
n


7
49 .0886  72.640 2
n 1
6
 0.05023333 or s  0.22413 .
s
2
x 2  nx
2

s
50233 .33 224 .13
0.05023333
0.22413

 84 .712
sx 


 0.084712
7
7
n
7
n
7
b. From the problem statement   .10 . From Table 3 of the syllabus supplement, if the population
variance is unknown   x  t s x and t n21  t .605  1.943 .
sx 
s
x  nx
2


 x  18.48  2.640 (thousands)


2
So   2640  1.943 84.712   2640  164 .6 or
2475.4 to 2804.6.
So   2.640  1.943 0.084712   2.640  0.1646
or 2.4754 to 2.8046 (thousands).
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252y0211 2/21/02
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.
1. The population mean for similar patients to those mentioned on the previous page who had not used the
new medicine was 2628. For your convenience the data are repeated below.
Patient
1
2
3
4
5
6
7
capacity
2850
2380
2800
2860
2300
2650
2640
Test to see if the mean breathing capacity is now above 2628 using the sample mean and standard deviation
you found in part II.
a. State the null and alternate hypothesis (2)
b. Find a critical value appropriate for this problem, using a confidence level of 90%.(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). Each time state clearly
whether you reject the null hypothesis and why.
e. Do a 90% two - sided confidence interval for the variance. (3)
f. (Extra credit) Assume that the data does not come from a normal distribution. (i) State a
confidence interval for the median using the highest and lowest values and give the confidence
level.(4) (ii) Do the same using the second highest and second lowest values and give the
confidence level. (3)
Solution: From Table 3 of the Syllabus Supplement:
Interval for
Confidence
Hypotheses
Interval
Mean (
  x  t 2 s x
H0 :   0
Unknown)
H :  
DF  n 1
1
0
Test Ratio
t
x  0
sx
Critical Value
xcv   0  t 2 s x
a) You were asked to see if mean breathing capacity is now above 2628 (or 2.628 thousand). This gives us
  2628. Since this does not contain an equality, it must be an alternative hypothesis, so we have
H 0 :   2628 and H 1 :   2628 .
b) Our facts, from the previous page are: n  7, x  2640, s  224.13, or better, s x  84.712 (or
0.0847 thousand). In addition,   .10 and  0  2628 . Since our alternate hypothesis is   2628 and is
6
 1.440 .
one-sided, our one -sided critical value must be above 2628. Our value of t is tn 1  t .10
xcv  0  t sx  2628  1.440 84.712   2750 .0 or 2.750 thousand .
c) Make a diagram . Show an almost Normal curve with a center at 2628 and a 10% 'reject' zone above
2750.0. Since x  2640 is not in the 'reject' zone, do not reject H 0 .
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252y0211 2/21/02
d) (i) The test ratio is t 
x   0 2640  2628

 0.1417 . Make a diagram . Show an almost Normal
sx
84 .712
6
curve with a center at 0 and a 10% 'reject' zone above t .10
 1.440 . Since t  0.1417 is not in the 'reject'
zone, do not reject H 0 . (ii) The confidence interval will have the same direction as the alternate hypothesis,
so it will be   x  ts x  2640  1.440 84.712 . or   2518.0 . Make a diagram . Show an almost
Normal curve with a center at 2640 and a shaded confidence interval above 2518. If you now represent
H 0 :   2628 by shading the area below 2628, you will have double-shaded the area between 2518 and
2628, so the confidence interval and the null hypothesis test do not contradict one another, and we do not
reject H 0 .
e ) From the outline, the small-sample confidence interval is
formula table,  2 
n  1s 2
.25 .5 2 
n  1s 2
 22
2 
n  1s 2
12 2
or from the
. The table says  .2056   12.5916 and  .2956   1.6354 We know that
s 2  50233 .33 or 0.050233 million. So
650233 .33   2  650233 .33
or 23937   2  184297 . In millions, these limits would be 0.023 and
12 .5916
1.6354
0.184.
f) (i) We can use the binomial tables to answer this. The numbers in order are 2300, 2380, 2640, 2650,
2800, 2850 and 2860. Since n  7 and p  .5, the probability that all 7 numbers are above the median or
that all seven numbers are below the median is 2Px  7  21  Px  6  21  .99219   .01562 So
P2300    2860   1  .01562  .98438 . (ii) The probability that 6 or more number are above the median
or six or more are below the median is 2Px  6  21  Px  5  21  .93750   .125 . So
P2380    2850   1  .125  .875 .
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252y0211 2/21/02
2. You have heard that, though the mean wealth held by US households is at least 270 (thousand dollars),
the median wealth held by households is no more than 61 ( thousand dollars). You take a sample of 300
households in your state and find that the mean wealth is 210(thousand) with a sample standard deviation of
707(thousand), Out of the 300 households, 161 have wealth above 61 (thousand dollars). Use a 95%
confidence level.
a. Test the statement about median wealth using a critical value. (4)
b. Test the same statement using a test ratio and a p-value (3)
c. Test the statement about the mean wealth using a critical value (3)
d. Test the same statement using a test ratio and a p-value. (3)
e. Do a 95% 2-sided confidence interval for the proportion of houses with wealth above 61 (thousand
dollars.) (3)
f. How large a sample would I need, if I wanted the proportion in the confidence interval to be correct to
.01 ? (3)
Solution: a) The outline explains that hypotheses about a median are hypotheses about a proportion. The
correspondence is below:
Hypotheses about
Hypotheses about a proportion
a median
If p is the proportion
If p is the proportion
 H 0 :   0

H 1 :   0
above  0
below  0
 H 0 : p .5

 H 1 : p  .5
 H 0 : p .5

 H 1 : p  .5
Since we are told that the median wealth is no more than 61 and the proportion above 61 is 161 out of 300,
Our hypotheses are
 H : p .5
 H 0 :  61
x 161
 .53667 .
and  0
Note that n  300 , x  161 . so that p  

n 300
 H 1 : p  .5
 H1 :   61
From the formula table we have:
Interval for
Confidence
Interval
Proportion
p  p  z 2 s p
Hypotheses
Test Ratio
H 0 : p  p0
z
H1 : p  p0
pq
n
q  1 p
sp 
And we can also use the sign test formula, z 
2x  n
n
.
p  p0
p
Critical Value
pcv  p0  z 2  p
p0 q0
n
q0  1  p0
p 
  .05 , so we use z  z.05  1.645 .
p0 q0
.5.5

 .02886 , so the critical value
n
300
is pcv  p0  z p  .5  1.645.02886  .5475. Make a diagram of a normal curve with a mean at .5 and a
The standard deviation of the sample proportion is  p 
reject zone above .5475. Since p  .53667 is not in the 'reject' zone, do not reject H 0 .
b) If we use a test ratio we get z 
p  p0
p

.53667  .5
2 x  n 2161   300
 1.271 or z 

 1.270 . To
.02886
n
300
get a p-value use pval  Pz  1.27  .5  .1064  .3936 . Since pval    .05, do not reject H 0 .
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252y0211 2/21/02
c) Our hypotheses are now H0 :   270 and H1 :   270 . The problem says n  300 , x  210 , s  707
and   .05 , DF  n  1  299 , so we can use z  z.05  1.645 in place of t .
From the formula table we have:
Interval for
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
Mean (
x  0
  x  t 2 s x
xcv    t 2 s x
H0 :   0
t
unknown)
s
H
:



DF  n  1
x
1
0
sx 
s

707
 40 .82 . Because of the alternate hypothesis, we want a critical value below 270, so we
n
300
use xcv    t s x  270  1.645 40.82   202 .85. Make a diagram of a normal curve with a mean at 270
and a reject zone below 202.85. Since x  210 is not in the 'reject' zone, do not reject H 0 .
d) If we use a test ratio we get t 
x   0 210  270

 1.470 . To get a p-value use
sx
40 .82
pval  Pz  1.47  .5  .4292  .0708 . Since pval    .05, do not reject H 0 . It would also be
acceptable to note that 1.470 is between t.10  1.282 and t.05  1.645 so that the p-value is between
.05 and .10.
e) Using the confidence interval formula from a) above with z  z.025  1.960 and s p 
2

pq
n
.53667 .46233 
 .0273 . p  p  z s p  .53667  1.960.0273  .537  .053 or .484 to .590.
2
300
f) From the outline n 
pqz2

.53667 .46333 1.960 2
 9552 .3 . So use 9553 or more. It would also be
e2
.012
acceptable to use p  q  .5 in this formula to get a slightly larger sample size.
Most of you did not state hypotheses in this problem, making grading it a nightmare!
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252y0211 2/21/02
3. In the previous problem you tested the mean wealth held by US households is at least 270 (thousand
dollars), and, based on the fact that the sample standard deviation is 707 (thousand dollars), created a
critical value for the sample mean. Since the sample was so large, you can assume that 707 is the population
standard deviation. ( The sample mean is still 210.)
a. Assume that the mean wealth is actually 260 (thousand dollars), and using the critical value you found on
the last page, what is the power of the test? (3)
b. Find the power curve for your test showing the necessary calculations.(6)
c. Do a 2-sided 95% confidence interval for the mean in the problem on the last page and figure out how
large a sample you would need to get the error part of the confidence interval down to 10 (thousand
dollars). (4)
d. Do a 2-sided 58% confidence interval for the mean, assuming that your sample of 300 came from a
population of 1000. (3)
e. (Extra credit) If you place the 300 values in the sample in order, which numbers would you use to find a
95% confidence interval for the median? (4).
Solution: a) Our hypotheses were H0 :   270 and H1 :   270 . The problem said n  300 , x  210 ,
s  707 and   .05 , DF  n  1  299 , so we can use z  z.05  1.645 in place of t .
sx 
s

707
 40 .82 . Because of the alternate hypothesis, we wanted a critical value below 270, so
n
300
we used xcv    t sx  270  1.645 40.82   202 .85. We will not reject H 0 if x  202 .85 . This problem


says that 1  260 . From the outline Then   P ' Accepting' H 0 H 0 is false

x  1 
202 .85  260 

 Px  202 .85   260   P  z  cv
  Pz 
  Pz  1.40  .5  .4236  .9236 .

40 .82


x


Power  1    1  .9236  .0764. Pathetic!
b) Half of the distance between 270 and 202.85 is, roughly, 34, so I found the power at 1  270 ,
1  236 , 1  202 .85, 1  169 , and 1  135 .

x  1 
Note that, in general, for this one-sided hypothesis   P  z  cv
.
x 

From previous examples we know that   1    .95
1  270
power  1    .05
1  236
  Px  202 .85   236   P  z 


202 .85  236 
  Pz  0.81  .5  .2910  .7910 .
40 .82

power  1    .2090
1  202 .85 From previous examples, we know that at the critical value   .5
power  1    .5
1  169
  Px  202 .85   169   P  z 


202 .85  169 
  Pz  0.83  .5  .2967  .2033 .
40 .82

power  1    .7967
1  135
  Px  202 .85   135   P  z 

202 .85  135 
  Pz  1.66  .5  .4545  .0485 .
40 .82


power  1    .9515
You now have the power for 5 points above or equal to 270. Make a diagram.
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252y0211 2/21/02
c) From the last question x  210 , s  707 , sx  40.82 and   x  t s x .   .05 , and
2
z 2  z.025  1.960. Again, we are using z in place of t because of the large number of degrees of
freedom.   x  t s x  210  1.960 40.82   210  80.00 or 130.0 to 290.0. From the outline
2
n
z 2 2
e2

1.960 2 707 2
10 2
 19202 .2 . So use a sample of size 19203 or larger.
d) The confidence level is 58% and   1  .58  .42 . From the first page z  z.21  0.81. From the
2
outline if n  .05 N , use  x 
x 
707
300

n
N n
. Since N  1000 and n  300 , we compute
N 1
1000  300
 34.168.   x  t 2 s x  210  0.8134.168   210  27.7 or 182.3 to 237.7.
1000  1
n  1  z . 2 n
300  1  1.960 300
 134 We will put the
2
2
numbers in order and use the 134th from the bottom and the 134th from the top (the 167th number).
e) The formula from the outline was k 

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252y0211 2/21/02
4. According to Ronald Weirs, the Veterans Administration closed the cardiac units of VA hospitals that
had mortality rates above 5%. In one hospital, their were 7 deaths in 102 operations (Consider this a sample
proportion). Using a 99% confidence level, can you say that the hospital's mortality rate (proportion that
died) was significantly above 5%? (Be specific about your significance level, especially in parts e and f.)
a. State your null and alternate hypotheses (2)
b. Do the problem using a critical value for the proportion that died. State clearly whether you reject the
null hypothesis and why. Does this mean that the mortality rate was significantly above 5%? (3)
c. Do the problem using a test ratio. (2)
d. Do the problem using an appropriate confidence interval. (3)
e. According to Weirs, a high school running back can expect an average of 8 injuries per 100 games. A
Poisson distribution applies. You believe that your league is particularly rough. In the last 100 games there
were 15 injuries. Formulate a test of your belief and tell me if you are right. Explain why. (4)
f. If, instead, you had data for 1000 games, what would be a critical value for the number of injuries? (3)
Solution: a) In parts a) through d) we are doing tests on a proportion. The material from the formula table is
quoted in question 2. The important words in the question for hypothesis formulation are " Can you say that
the hospital's mortality rate (proportion that died) was significantly above 5%?" This translates as
H1 : p  .05 . It is an alternate hypothesis because it does not contain an equality. The null hypothesis is
thus H 0 : p  .05. The problem says that   .01, n  102 , x  7. so that p 
x
7

 .0687 .
n 102
p0 q0
.05.95 

 .02158 . This is a one-sided test and z  z.01  2.327 .
n
102
b) Since the alternative hypothesis says p  .05, we need a critical value that is above .05. We use
p 
pcv  p0  z p  .05  2.327.02158  .1002. Make a diagram of a normal curve with a mean at .05 and
a reject zone above .1002. Since p  .0687 is not in the 'reject' zone, do not reject H 0 . We cannot say that
the mortality rate is significantly above 5%.
p  p0
.0687  .05

 0.867 . Make a diagram of a normal curve with a mean at
c) The test ratio is z 
p
.02158
zero and a reject zone above z  z.01  2.327 . Since z  0.867 is not in the 'reject' zone, do not reject
H 0 . We cannot say that the mortality rate is significantly above 5%.
pq
.0687 .9313 

 0.025045 . To make the 2-sided
n
102
confidence interval, p  p  z s p , into a 1-sided interval, go in the same direction as H1 : p  .05 . We get
2
d) To do a confidence interval we need s p 
p  p  z s p  .0687  2.327.025045  .0687  .0583  .0104. To see that p  .0104 does not contradict
H 0 : p  .05, make a diagram. Represent the confidence interval under a normal curve with a mean at
.0687 by shading the area above .0104. Represent the null hypothesis by shading the area below .05. these
areas overlap, indicating that both can be true at the same time.
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252y0211 2/21/02
e) Our null hypothesis is H 0 : Poisson(mean  8) . The alternative is most safely stated as
H1 : not Poisson(mean  8) , though we will test for a mean greater than 8. According to the problem
x  15. The easiest way to do this is to assume that the null hypothesis is true and to compute a p-value
which would be defined as Px  15 . Using the Poisson table with a parameter of 8, we find that
pval  Px  15  1  Px  14  1  .98214  .01786 . If we use a significance level of   .05 , we reject
H 0 , since the p-value is below the significance level, but if we use a significance level of   .01, we do
not reject H 0 , since the p-value is above the significance level. If you rejected the null hypothesis, you can
say that this is a rough league.
f) In the unlikely case that you have data on 1000 games instead of 100, you would expect a Poisson
distribution with a mean of 10 times 8 or 80. Our hypotheses are now H 0 : Poisson(mean  80) . The
alternative is most safely stated as H1 : not Poisson(mean  80) . The question is 'how far above 80 can x
be before we reject the null hypothesis.' We do not have a table for the Poisson distribution with a mean of
80, but we do know that this distribution is very similar to a Normal distribution with a mean of   80 and
a standard deviation of   80. We know that the probability of x being below   z  80  z 80
is 1   .
These numbers imply that:
If   .05 , xcv  80  1.645 80  94 .91 . We reject H 0 if x is 95 or larger.
If   .01, xcv  80  2.327 80  100 .81 . We reject H 0 if x is 101 or larger.
10