2/22/99 252y9912
ECO252 QBA2
FIRST HOUR EXAM
February 23, 1999
Name
Hour of Class Registered (Circle)
MWF TR 10 12 12:30 2:00
Show your work! Make Diagrams!
I. (14 points) Do all the following.
x ~ N 6,7
36 
 1 6
z
1. P1  x  3  P
  P 1.00  z  0.43 
7 
 7
 P1.00  z  0  P0.43  z  0  .3413  .1664  .1749
11  6 
  11  6
z
2. P11  x  11  P
  P 2.43  z  0.71
7 
 7
 P2.43  z  0  P0  z  0.71  .4925  .2611  .7536
96
  16  6
z
3. P16  x  9  P
  P 3.14  z  0.43 
7 
 7
 P3.14  z  0  P0  z  0.43  .4992  .1664  .6656
For 3.14 see SS Table 18.
06

4. Px  0  P z 
  Pz  0.86   P 0.86  z  0  Pz  0
7 

 .3051  .5  .8051
12 .5  6 
96
z
5. P9  x  12.5  P
  P0.43  z  0.93 
7 
 7
 P0  z  0.93  P0  z  0.43  .3238  .1664  .1574
6.
A symmetrical interval about the mean with 96% probability.
We want two points x .98 and x.02 , so that Px.98  x  x .02   .9600 . From the
diagram, if we replace x by z, P0  z  z.02   .4800 . The closest we can come is
P0  z  2.05   .4798 or P0  z  2.06   .4803 . So z.02  2.054 , though 2.05
or 2.06 are fine, and x    z.02  6  2.054 7  6  14.38 ,
or –8.38 to 20.38.
x.13
We want a point x .13 , so that Px x .13   .13 . From the diagram, if we
replace x by z, P0  z  z.13   .3700 . The closest we can come is P0  z  1.13   .3708 .
So z .13  1.13 , and x    z.11  6  1.137  6  7.91 , or 13.91 .
7.
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II. (6 points-2 point penalty for not trying part a.)
A sample of the number of days it took a broker to sell houses appears below. Assume that we
were sampling from a normal distribution.
Home
1
2
3
4
5
Days
49
69
88
99
33
a. Compute the sample variance, s , of the number of days it takes to sell a house. Show your
work.(3)
b. Compute an 80% confidence interval for the mean time,  , that it takes to sell a house.(3)
Solution: a.
Home
x (Days)
x2
x 338
1
49
2401
x 

 67 .6
n
5
2
69
4761
2
2
3
88
7744
2
x  nx
25796  567 .6
4
99
9801
s2 

n 1
4
5
33
1089
 736 .8 or s  27.1441.
Total
338
25796


.
b.
From the problem statement   .20 . From Table 3 of the syllabus supplement, if the
  x  t  sx and t n21  t .410  1.533 .
population variance is unknown
sx 
s

n
2
736 .8 27 .1441

 12 .1392 . So   67.6  1.533 12.1372   67.6  18.61 or 48.99 to 86.21.
5
5
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III. Do at least 3 of the following 5 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.
1.
The broker claims that the average time to sell a house is at most 48 days.
a. State the Hypotheses that you are testing. (2)
b. Using the data from the previous problem (page 2.), test the hypotheses (99% confidence
level) using:
(i)
A test ratio (2)
(ii)
Critical values (2)
(iii)
A confidence interval (2)
c. Find an approximate p-value for the null hypothesis. (1)
d. Now find a 99% confidence interval for the standard deviation.(3)
Solution: From Table 3 of the Syllabus Supplement:
Interval for
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
xcv   0  z 2  x
Mean (
  x  z 2  x
x  0
H0 :   0
z
Known)

H :  
1
  x  t 2 s x
Mean (
Unknown)
DF  n  1
H 0 :   48
a.
H 1 :   48
x
0
H0 :   0
t
H1 :    0
x  0
sx
xcv   0  t 2 s x
4
 0  48, DF  n  1  4,   .01, tn1  t .01
 3.747
From the previous page: x  67.6 and s x 
s
736 .8 27 .1441

 12 .1392 .
5
5

n
x   0 67 .6  48

 1.6146 . This is in the ‘accept’ region
b. (i) Test Ratio: t 
sx
12 .1392
below 3.747, so do not reject H 0 .
(ii) Critical Value: Since this is a one-sided test, xcv  0  t sx  48  3.747 12.1392   50  45.49
or 95.49. This means that we reject H 0 if the sample mean is above 95.49. since x  67.6 is below
this critical value, do not reject H 0 .
(iii) Confidence Interval: Since this is a one-sided test,
  x  t s x  67.6  3.747 12.1392   67.6  45.49
or   22.11. This does not contradict H 0 :   48 , because any mean in the range 22.11 to 48 satisfies
both statements, so do not reject H 0 .
4
4
c. From the t table t = 1.6146 is smaller than t .05
 2.132 but is larger than t .10
 1.533 , so that, for this 1-
sided test, we can say .10  pvalue  .05 .
d. From page 1 of the Syllabus Supplement:
 n  1 s 2
Since   .005 and 1    .995 , look up
2
2
n  1s 2
 .2005
 2 
n  1s 2
 .2995
or
 2
2
2 4 
 .005
2 
2
1
. DF  n  1  4 and   .01 .

2
4 
 14 .8603 and  .2995
 0.2070 . So
4736 .8   2  4736 .8
14 .8603
 n  1 s 2
0.2070
or 198 .33   2  14237 .68 . Finally, taking
square roots, 14.083    119.32 .
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2. A new broker opens in town and a sample of nine houses gives the following days to sale:
Home
Days
1
62
2
28
3
114
4
113
5
71
6
29
7
53
8
54
9
34
Assuming that the distribution is not normal,
a. Find a confidence level for the following interval for the median: 34    71 . (3)
b. Test the hypotheses that the median is at most 33 at the    10% level. (5)
c. If we have a sample of 350 numbers in order and take the 15 th from each end, what would the
confidence level be? (3)
d. (Extra credit) What numbers would we use if we wanted a 99% confidence interval for the
median and had a sample of 350 numbers? (3)
Solution:
a. If we put the numbers in order, we get 28 29 34 53 54 62 71 113 114. Thus we want the 3rd
number from the end. If n  9 and k  3 , we find from the binomial table (with p  .5 ),
1    1  2Px  k  1  1  2Px  2  1  2.08984   .82032 . Px  2  Px  7
b. From the outline:
Hypotheses about a median
Hypotheses about a proportion
If p is the proportion above  0
If p is the proportion below  0
 H 0 :   0

H 1 :   0
 H 0 : p .5

 H 1 : p  .5
H 0 : p .5

H1 : p  .5
 H :  33
So, our hypotheses are  0
and If p is the proportion above 33, our hypotheses become
 H 1 :   33
 H 0 : p .5
. There are 7 numbers above 33, and according to the binomial table

 H 1 : p  .5
pvalue  Px  7  1  Px  6  1  .91016  .08984   . Reject H 0 .
c.
From the outline:

2
 P x  k  1 
1

2

n  350 and k  15 , 1    1  2Px  k  1
d.

 np 
k  .5  .5n 
 . If
 P z 

npq
.5 n 


  P z  k  1 
1
2
 

14  .5  .5350   
 161 .5  
  1  2 P z 
 1  2 Px  14   1  2 P z 
   1  2Pz  25 .42   1



9.354  
.5 350
 

 
.
n  1  z . 2 n
From the outline: k 
. z  z.005  2.576 , so
2
2
350  1  2.576  350
k
 126 .4 . So use the 126th number from each end.
2
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3.
A consumer service desk claims that the average waiting time is less than 11 minutes. A sample of 300
customers gives us the following: x  10.25, s  6.01. Assume that the parent distribution is Normal
and that our confidence level is 95%.
a. State the Hypotheses to be tested. (1)
b. Find a critical value or values for x and test the hypotheses. (3)
c. Graph a power function for the test showing most of your calculations. (8)
d. Find a p-value for the Null Hypothesis if the value of x is 12. (2)
Solution:
H 0 :   11
299
a. 
 0  11, DF  n  1  299 ,   .05, tn 1  t .05
 z .05  1.645
H 1 :   11
6.012
 0.34699
xcv  0  z sx  11  1.645  (0.34699 )  10.4292 .
300
n
300
Since 10.25 is below the critical value, reject H 0 .
c. Since we do not reject H 0 when x is above 11.4292, the probability of a type II error is
b. s x 
s


6.01
Px  10.4292   1  , when 1 is any value below 11. Since the critical value is about 10.4, the values I
will use are 11, 10.7, 10.4292, 10.1 and 9.8.
Operating Characteristic  
Power 1   
.05
Px  10.4292   11  .95
10 .4292  10 .7 

Px  10.4292   10.7  P z 
  Pz  0.78   .2823 + .5 = .7823
0.34699


.22
Px  10.4292   10.4292  .50
.50
10 .4292  10 .1 

Px  10.4292   10.1  P z 
  Pz  0.95   .5 - .3289 = .1711
0.34699


.83
11 .3085  10 .60 

Px  10.4292   9.8  P z 
  Pz  2.01  .5 - .4778=.0222
0.35276


.98
DIAGRAM
12  11

d. Since this is a left-sided test and  0  11 , Px  12   P  z 
0.34699


  Pz  2.88   .5  .4980  .9980 .

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4.
A Senatorial candidate believes that she must get at least 78% of the vote in Philadelphia and Duquesne
County to win the election. A sample survey of 2000 voters indicates that 1499 planned to vote for her.
Using a Hypothesis-testing procedure, assess her chances of winning.
a. State the Hypotheses to be tested. (1)
b. Do a test of this hypothesis at the 99% level. (3)
c. Do the same test at the 87% level. (2)
d. Do a 2-sided 98% confidence interval for the proportion that will vote for her. (2)
e. Do a 2 sided 96% confidence interval for the proportion that will vote for her. (2)
f. (Extra Credit) If the true proportion that will vote for her is 76%, what is the power of your
test? (3)
Solution: From Table 3:
Interval for
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
Proportion
p  p  z 2 s p
p  p0
p cv  p 0  z  2  p
H 0 : p  p0
z
p
H 1 : p  p0
pq
p0 q 0
sp 
p 
n
n
q  1 p
H : p  .78
x 1499
 .7495 , so that
a.  0
Note that p 0  .78, n  2000 , p  
n
2000
H
:
p

.
78
 1
p 
.78 .22   .009263
2000
b. (i) Test Ratio: z 
, or s p 
p  p0
p

.7495 .2505   .009689 . q  1  p .
0
0
2000
.7495  .78
 3.2927 . Since   .01 and this is a one-sided test, use
.009263
z .01  2.326 . Since –3.2927 is less than  z .01 , reject H 0 . Or pvalue  Pz  3.2927   .5  .4995
 .0005   , so reject H 0 . Or:
(ii) Critical Value: pcv  p0  z  p  .78  2.326.009263  .7585 . Since p  .7495 is below this
value, reject H 0 . Or:
(iii) Confidence Interval: p  p  z s p  .7495  2.326.009689  .7720 . Since p  .7720 contradicts
H 0 : p  .78 , reject H 0 .
c. If   .13 , we know that z .13  1.13 from the first page of this exam.
(i) Test Ratio: z  3.2927 . Since –3.2927 is less than  z .13 , reject H 0 . Or pvalue  .0005   , so
reject H 0 . Or:
(ii) Critical Value: pcv  p0  z  p  .78  1.13.009263  .7695 . Since p  .7495 is below this
value, reject H 0 . Or:
(iii) Confidence Interval: p  p  z s p  .7495  1.13.009689  .7604 . Since p  .7604 contradicts
H 0 : p  .78 , reject H 0 .
d. Since   .02 and z   z.01  2.326 , p  p  z s p  .7495  2.326.009689  .7495  .025 , or .727
2
2
to .772.
e. Since   .04 , z   z.02 and z.02  2.054 from the first page of this exam, p  p  z s p
2
2
 .7495  2.054 .009689   .7495  .0199 or .730 to .769.
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f. If we use the critical value from b., we do not reject H 0 if p is above .7585. So the power is
.7585  .76 

 1  Pz  0.16   1  .5  .0636   1  .5636   .4364 .
1  P p  .7585 p  .76  1  P  z 
.009263 

Actually  p 
.76 .24   .009550
2000
, but it does not affect the power. .
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2/22/99 252y9912
5.
a.
b.
c.
d.
e.
f.
g.
Solution:
A Student wishes to estimate the mean amount that members of city governments earn. The
student guesses that the standard deviation is about $1000. What sample size is required to
keep the error in estimating the mean at $100 if a confidence level of 98% is used. (3)
The student also wants to estimate the proportion of cities that have private refuse collectors.
What is the required sample size if no estimate of the proportion is available before the study,
the confidence level is to be 98% and the permitted error is .05. (3)
A finance expert wants to test whether new legislation has reduced the volatility of stock
prices. He assumes that the appropriate parent distribution is Normal and that the variance of
stock prices before the legislation was 9.0, and on the basis of 23 daily measurements after
the legislation finds that s 2  5.3 . State the Hypotheses to be tested. (1)
Test the Hypotheses in c. (   .01 ) (3)
What if the sample variance in c. were taken from a sample of 85 days? (3)
If we claim that a population has a Poisson distribution with a mean of 12 and an actual value
is found to be 5, do a 2-sided 95% hypothesis test on the claim. (3)
If in f. the mean was 120 and the number found was 50, repeat the test. (3)
a. Since   .02 , z  z.01  2.326 . From the outline n 
2
z 2 2
e2

2.326 2 1000 2
100 2
 541 .02 . So use
n  542 .
b. Since the significance level is the same, z  z.01  2.326 . We do not know p , so we use p  .5 .
2
From the outline n 
pqz 2
e2

.5.52.326 2
.05 2
 541 .02 . So use n  542 .
H 0 :  2  9.0
c. 
H 1 :  2  9.0
d.  02  9.0 , s 2  5.3 , n  23 and   .01 . Since DF  n  1  22 and the test is one-sided, use
 .29922  9.5425 . From the outline,  2 
n  1s 2
 02

22 5.3  12.956 . Since this is above 9.5425, do not
9.0
reject H 0 .
e.  02  9.0 , s 2  5.3 , n  85 and   .01 . DF  n  1  84 and the test is one-sided. From the outline,
2 
n  1s 2
 02

84 5.3  49 .467 . Since this sample is too large for the
9.0
 2 Table, we must use
z  2  2  2DF   1  249 .467   284   1  9.9465  12 .8228  2.88 . We compare this to
z .01  2.326 . Since it is below –2.326, reject H 0 .
f. Since this is a two-sided test, with   .05 , use the Poisson (12.0) table to compute a p-value.
pvalue  2Px  5  2.02034   .04068   . Since pvalue  , reject H 0 .
g. This is still a Poisson problem, but because of the large mean and the lack of a suitable Poisson table,


use the Normal approximation to the Poisson distribution, i.e. x ~ N m, m , where m is the mean. The
easiest way to do this is z 
z.025
xm

50  120
 6.39 . Since this is a two-sided test, z should be between
m
120
 1.960 . It is not, so reject H 0 .
8