2/25/00 252y0012
ECO252 QBA2
FIRST HOUR EXAM
February 22, 2000
Name ______KEY______
Hour of class registered __
Class attended if different _
Show your work! Make Diagrams! Note that all diagrams are available on a separate sheet.
I. (14 points) Do all the following.
x ~ N 2,7
3 2 
 1 2
z
1. P1  x  3  P
  P 0.43  z  0.14 
7 
 7
 P0.43  z  0  P0  z  0.14   .1664  .0557  .2221
11  2 
  11`2
z
2. P11  x  11  P
  P 1.86  z  1.29 
7 
 7
 P1.86  z  0  P0  z  1.29   .4686  .4015  .8701
22
  24  2
z
3. P24  x  2  P
  P 3.71  z  0  .4999
7
7 

02

4. Px  0  P z 
  Pz  0.29   Pz  0  P0.29  z  0
7 

 .5  .1141  .3859
12 .5  2 
 9`2
z
5. P9  x  12.5  P
  P1.00  z  1.50 
7
7 

 P0  z  1.50   P0  z  1.00   .4332  .3413  .0919
6.
A symmetrical interval about the mean with 83% probability.
We want
two points x .915 and x.085 , so that Px.915  x  x .085   .8300 . From the diagram,
if we replace x by z, P0  z  z.085   .4150 . The closest we can come is
P0  z  1.37   .4147 . So z.085  1.37 , and x    z.085  2  1.37 7  2  9.59 ,
11 .59  2 
  7.59  2
z
or -7.59 to 11.59. To check this note that P7.59  x  9.59   P

7
7


 P1.37  z  1.37   2P0  z  1.37   2.4147   .8294  .83
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  2  1.137  2  7.91 ,
or 9.91 . To check this note that
9.91  2 

Px  9.91  P z 
  Pz  1.13   Pz  0  P0  z  1.13  .5  .3708
7 

 .5  .3708  .1292  .13
7.
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II. (6 points-2 point penalty for not trying part a.)
A truck dealer asserts that a new model of truck has lower gas consumption than your present fleet
model. A random sample of 7 runs is made with the new model with gas consumption as below. Assume
that we were sampling from a normal distribution.
run
1
2
3
4
5
6
7
Consumption
11.56
10.97
12.74
12.08
12.66
12.84
11.15
a. Compute the sample standard deviation, s , of gasoline consumption. Show your work! (3)
b. Compute a 90% confidence interval for the mean gas consumption,  .(3)
Solution: a.
x 
s
 x  84.00  12.00
n
Home
1
2
3
4
5
6
7
Total
7
x

1011 .6722  712 .00 2
n 1
6
 0.612033 or s  0.782326 .
2
2
 nx
2

x (Days)
11.56
10.97
12.74
12.08
12.66
12.84
11.15
84.00
x2
133.6336
120.3409
162.3076
145.9264
160.2756
164.8656
124.3225
1011.6722
.
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 .
b.
2
sx 
s

n
to 12.575.
0.612033
0.782326

 0.29569 . So   12.00  1.943 0.29569   12.00  0.5745 or 11.425
7
7
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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. Use a 95% confidence level unless another level is specified.
1. A truck dealer asserts that a new model of truck has lower gas consumption than your present fleet
model. The data is on the previous page. On the basis of a great deal of experience, you know that the mean
for your present model is 12.3 miles per gallon. For your convenience the data are repeated below.
run
1
2
3
4
5
6
7
Consumption
11.56
10.97
12.74
12.08
12.66
12.84
11.15
Evaluate the dealer’s statement 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 :   12.3
H 0 :   12.3
H 0 : x  12 .3
H 0 :   12.3
H 0 : x  12 .3
H 0 : x  12 .3
H 0 :   12.3
H 0 : x  12 .3
H 0 :   12.3
H 0 : x  12 .3
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)
e. Test the hypothesis that the mean consumption of gas is exactly 12.1 for the dealer's fleet at the
90% confidence level. (2)
f. Assume that the data does not come from a normal distribution and evaluate the statement that
the median is 12.8. (4)
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
Mean (
Unknown)
  x  t 2 s x
x
0
H0 :   0
H1 :    0
t
x  0
sx
xcv   0  t 2 s x
DF  n  1
a. The dealer's statement is   12.3 . Because this does not contain an equality, it cannot be a null
hypothesis. So we must test its opposite H 0 :   12 .3 and our alternate hypothesis is H 1 :   12 .3
6
 1.440
b. n  7,  0  12 .2, DF  n  1  6,   .10, tn1  t .10
s
0.612033
0.782326

 0.29569 .
7
n
7
Critical Value: Since this is a one-sided test, xcv  0  t sx  12.3  1.440 0.29569   12.3  0.4258
or 11.87. It might help to remember that x   0 is always in the 'accept' region.
From the previous page: x  12 .00 and s x 

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2/25/00 252y0012
c. The critical value in b) means that we reject H 0 if the sample mean is below
11.87. Since x  12 .00 is not below this critical value, do not reject H 0 .
d. (i) Test Ratio: t 
x   0 12 .00  12 .3

 1.015 . This is in the ‘accept’ region
sx
0.29569
6
above tn1  t.10
 1.440 , so do not reject H 0 . It might help to remember
that zero is always on the 'accept' region for t.
(ii) Confidence Interval: Since this is a one-sided test, and the alternate hypothesis
is H 1 :   12 .2 ,   x  t s x  12.00  1.440 0.29569   12.0  0.4258
or   12.4258 . This does not contradict H 0 :   12 .3 , because any mean in the
range 12.3 to 12.4258 satisfies both statements, so do not reject H 0 .
6
e. H 0 :   12 .1 , H1 :   12 .1 . n  7, 0  12.1, DF  n  1  6,   .01, tn1  t.05
 1.943 ,
2
x  12 .00 and sx  0.29569 . Use one of the three methods below.
(i) Critical Value: Since this is a two-sided test, xcv  0  t sx  12.1  1.943 0.29569 
2
 12.1  0.5745 or 11.525 and 12.675. Since x  12 .00 is between these two values,
do not reject H 0 .
x  0 12 .00  12 .1

 0.338 . This is in the ‘accept’ region
sx
0.29477
6
6
between tn1  t.05
 1.943 and tn1  t.05
 1.943 , 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
 12.0  1.943 0.29569   12.0  0.5745 . We can thus say that 11.425    12.575
Since this interval includes 0  12 .1, we cannot reject H 0 .
f. H 0 :   12 .8 , H 1 :  12.8 . If we subtract 12.8 from the values of x, we get the following:
x
x 
run
1
11.56
-1.24
2
10.97
-1.83
3
12.74
-0.06
4
12.08
-0.72
5
12.66
-0.14
6
12.84
+0.04
7
11.15
-1.65
For n  7 , there is 1 value above the alleged median and 6 values below. The p-value is thus
either 2Px  6 or 2Px  1 . From the binomial table for p  .5, p  value  2Px  6
 2Px  1  2.06250   0.12500 Since this is above the significance level   .10 , we cannot reject H 0 .
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2. Using data from part II
a. Do a 90% confidence interval for the standard deviation for your fleet (3)
b. Test the hypothesis that the population standard deviation is less than 1 mile per gallon at the
90% confidence level. (4)
c. Using the sample standard deviation that you got in part II, test the hypothesis that the
population standard deviation is 1.2 at the 90% confidence level if the sample size is 62. (3)
d. Assuming that the population standard deviation is 0.9 miles per gallon and using the sample
mean that you found in part II, test again the null hypothesis about the mean that you found in
question III-1a on the previous page. (2)
e. Repeat 2d above assuming that the data come from a sample of 7 out of a population of 70. (2)
Solution: From part II x  12 .00 , s 2  0.612033 or s  0.782326 . Also n  7, DF  n 1  6,
and   .10 .
a. From page 1 of the Syllabus Supplement:
n  1s 2
 2
2
look up
 .2056 
 12 .5916 and
 .2956 
 1.6354 . So
60.612033    2  60.612033 
12 .5916
1.6354
0.54004    1.49848 .
 2 
n  1s 2
2 
1
n  1s
 .205
2
2
 
2
. Since 
n  1s 2
 .295
2
 .05 and 1  
2
 .95 ,
or
or 0.29164   2  2.24544 . Finally, taking square roots,
b. H 0 :   1.00 H 1 :   1.00 . From the outline  2 
n  1s 2
 02
is the test ratio. If  02  1.00
and H 1 :   1.00 is true, s 2 will tend to be below 1, and the ratio will be on the small side.
 
Since this is a 1-sided test and   .10 ,compare the test ratio with  .2906  2.2041 .
Since the mean of this chi-squared distribution is DF  6 and the mean must be in the 'accept'
region, our test is to reject the null hypothesis if  2  2.2041 . We compute
2 
n  1s 2
 02

60.612033 
 3.6722 , so we cannot reject H 0 .
1.00
c. H 0 :   1.2 H 1 :   1.2 . s 2  0.612033 or s  0.782326 , but n  62 From the outline
2 
61 0.612033  25.9264 is the test ratio. Since this is a large sample,
1.22
2 DF   1  225 .9264   261  1  7.2009  12 .0000  3.799 .   .10 ,
n  1s 2
 02
z  2 2 

z   z .05  1.645 - we reject the null hypothesis if z is not between 1.645 . Since -3.799
2
is below that range, reject H 0 . You could also use a confidence interval here.
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2/25/00 252y0012
d. H 0 :   12 .3 , H 1 :   12 .3 . n  7,  0  12 .3,   .10 , z  z .10  1.282 ,
x  12 .00 and  x 

0.9
 0.34017 . Use one of the three methods below.
n
7
(i) Critical Value: Since this is a one-sided test, x cv   0  z  x  12 .3  1.282 0.34017 
= 11.88. Since x  12 .00 is above the critical value, do not reject H 0 .
(ii) Test Ratio: z 

x  0
x

12 .00  12 .3
 0.882 . This is not in the ‘reject’ region
0.34017
below  z   z .10  1.282 , so do not reject H 0 .
(iii) Confidence Interval: Since this is a one-sided test, use   x  z  x
 12.0  1.282 0.34017   12 .436 . But   12.436 does not contradict
  12.2 , so we cannot reject H 0 .
e. H 0 :   12 .2 , H 1 :   12 .2 . n  7, N  70,  0  12 .2,   .01, z  z .01  2.327 ,

N  n 0.9 63

 0.32504 . Use one of the three methods below.
n N 1
7 69
(i) Critical Value: Since this is a one-sided test, x cv   0  z  x  12 .3  1.282 0.32504 
= 11.883. Since x  12 .00 is above the critical value, do not reject H 0 .
x   0 12 .00  12 .3

 0.923 . This is not in the ‘reject’ region
(ii) Test Ratio: z 
x
0.32504
x  12 .00 and  x 
below  z   z .10  1.282 , so do not reject H 0 .
(iii) Confidence Interval: Since this is a one-sided test, use   x  z  x
 12.0  1.282 0.32504   12 .417 . But   12.417 does not contradict
  12.2 , so we cannot reject H 0 .
Remember! My most common comment in grading was "Did you read the question?" If you answer
a question about the standard deviation with an answer appropriate for a proportion or a mean, no
amount of knowledge of the formulas will help you!
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3. We have the following data (Assume in this question that the original data comes from the normal
distribution.):
x  970 ,   60 and n  225 .
a. What is the smallest sample size that we could use so that a 90% confidence interval for the mean could
be given with an error of  3 ? (3)
b. If we test the hypothesis that the population mean is 975, what is the p-value for this hypothesis? (3)
c. If our confidence level is 94%, would we reject the hypothesis in 3b? Why? (1)
d. If instead your data reads x  968 , s  15 .3 and n  36 , find a range for the p-value and tell
whether you would reject the null hypothesis if the confidence level is 94%. (3)
e. Due to the advent of EasyPass the mean number of cars passing a Garden State Parkway toll booth with
a human collector is falling. Assume that the average number of motorists passing a given location was
known to be 10.5 per minute. Do a one sided hypothesis test using a confidence level of 95% if an
observer reports that the number that actually passed in a minute was 5. (3)
f. To get a more powerful test note that 10.5 cars a minute means 630 per hour. Would you reject the null
hypothesis if 300 cars passed in an hour? (3)
z 2 2 1.645 2 60 2

Solution: a.   .10, z  z.05  1.645, e  3 and   60 . From the outline n 
2
e2
32
 1082 .41 . So use a sample size that is at least 1083.

60

 4.00 .
b. H 0 :   975 , H 1 :   975 . n  225 ,  0  975 , x  970 and  x 
n
225
x   0 970  975
z

 1.25 . p  value  2Pz  1.25  2.5  .3944   .2012 .
x
4.00
c. If 1    .94,   .06. We reject a hypothesis if the p-value is below the significance level.
Since this is not true here, do not reject H 0 .(To do it the hard way use z.06  1.55 .)
d. . H 0 :   975 , H 1 :   975 . x  968 , s  15 .3 and n  36 , s x 
t
s
n

15 .3
 2.550 ,
36
x   0 968  975
35

 2.745 From the t table t = 2.745 is smaller than t .001
 3.340 ,
sx
2.55
35
 2.724 , so that, for a 1-sided test, we can say .005  pvalue  .001 .
but is larger than t .005
But for this 2-sided test we must say .01  pvalue  .002 . We reject H 0 if pvalue   .
Since   .06 , we reject H 0 .
e. Since this is a problem involving a comparison of the number per unit time to the average per unit time
we use the Poisson distribution. H 0 : Poisson(mean  10.5) , H 1 : Poisson(mean  10 .5) . This is a 1-sided
test, so look up Px  5 on the Poisson (10.5) table. You should get a p-value of .05038. Since this is
above the significance level of .05, do not reject the null hypothesis.
f. Since this is a problem involving a comparison of the number per unit time to the average per unit time
we use the Poisson distribution. H 0 : Poisson(mean  630 ) , H 1 : Poisson(mean  630 ) . This is a 1-sided
test, so we would like to look up Px  230  on the Poisson(630 ) table. Because we do not have this table
we approximate the Poisson distribution with a Normal distribution with   630 and   630 . Thus

300  630 
  Pz  13.15   .5  .5000  0 . Since this is below the
pvalue  Px  300   P z 

630 

significance level of .05, reject the null hypothesis
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4. An electronics supplier claims that at least 96% of parts supplied to a government agency meet
specifications. A sample of 400 parts is tested and 36 parts are rejected.
a. State the null and alternative hypotheses (2).
b. Test the hypothesis at the 95% level. (3)
c. If we wish to create a 2-sided confidence interval for the proportion that meet specifications with an
error of not more than .01 , what is the minimum sample size required? (3)
d. To check the correctness of your solution in 4c, create the 2-sided confidence interval. (If you did not
do 4c, assume a sample size of 237.) (2)
e. Remembering what you did on page 1, do a 74% confidence interval for the proportion. (2)
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
Critical Value
p  p0
p
p cv  p 0  z  2 
p 
p
p0 q 0
n
H : p  .96
x 364
 .91 , so that
a.  0
Note that p 0  .96 , n  400 , p  
n
400
H
:
p

.
96
 1
.96 .04  
p 
400
b. (i) Test Ratio: z 
.000096  .009798 , or s p 
p  p0
p

.91.09   .0143 . q  1  p .
0
0
400
.91  .96
 5.103 . Since   .05 and this is a one-sided test,
.009798
use z .05  1.645 . Since –5.103 is less than  z .05 , reject H 0 . Or
pvalue  Pz  5.103   .5  .5000  0   , so reject H 0 . Or:
(ii) Critical Value: pcv  p0  z  p  .96  1.645.009798  .9439 . Since p  .91 is
below this value, reject H 0 . Or:
(iii) Confidence Interval: p  p  z s p  .91 1.645.0143  .9335 . Since p  .9335
contradicts H 0 : p  .96 , reject H 0 .
c. From the outline n 
n
pqz 2
e2

pqz 2
e2
.91.09 1.960 2
.012
. p  .91, e  .01 and since   .05 , use z .025  1.960 . Thus
 3146 .01 and we must use a sample of at least 3147.
x 364

 .91,   .05 and z .025  1.960 . s p 
n 400
p  p  z s p  .91 1.960.005101  .91 .010 .
d. n  3147 , p 
.91.09   .005101
3147
2
e. Note that 1    .74 , so that   .26 and, from the front page, z  z.13  1.13
2
p  p  z 2 s p  .91 1.13.005101  .91 .006 or .904 to .916.
8