2/27/98 252y9812
ECO252 QBA2
FIRST HOUR EXAM
February 19, 1998
Name
Hour of Class Registered (Circle)
MWF 10 11 TR 12:30 2:00
Hour of Class Attended (If Different)
______________
I. (14 points) Do all the following.
x ~ N  2,6.5
1.
2.
30  2 
 1 2
z
.  z  4.31
P1  x  30  P
  P 015
 6.5
6.5 
 P 015
.  z  0  P 0  z  4.31 .0596.5000 .5596
3  2
 3  2
z
. 
P 3  x  3  P
  P 0.77  z  015
 6.5
6.5 
 P 0.77  z  0  P 0  z  015
.  .2794.0596 .3390
3.
6  2
22
z
P 2  x  6  P
  P 0  z  0.62 .2324
 6.5
6.5 
4.
3  2 

P x  30
.   P z 
  P z  0.77

6.5 
 P 0.77  z  0  P z  0 .2794.5000 .7794
0  2
 3  2
z
P 30
.  x  0  P
  P 0.77  z  0.31
 6.5
6.5 
 P 0.77  z  0  P 0.31  z  0 .2794.1217 .1577
5.
6.
A symmetrical interval about the mean with 93% probability.
We want two points x .0350 and x.9650 , so that P x.9650  x  x .0350  .9300 . From the
diagram, if we replace x by z, P 0  z  z.0350  .4650 . The closest we can come is
. , and x    z.0350  2  181
P 0  z  181
.  .4649 . So z.0350  181
.  65
.   2  1177
. ,
or -9.77 to 13.77.
7.
x .07
We want a point x .07 , so that P x  x .07  .07 . From the diagram, if we replace x by z,
P 0  z  z.07  .43 . The closest we can come is P0  z  148
.  .4306 or
, and
.
P0  z  147
.  .4292 . Use something between the two. So z.07  1475
x    z.07  2  1475
.  65
.   2  9.59 , or 11.59.
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II. (6 points-2 point penalty for not trying part a.) Show your work!
A sample of seven printheads were tested for durability. The number of characters (in millions)
printed before failure are listed below. Assume that we were sampling from a normal distribution.
Printer Characters
1
0.8
2
1.4
3
1.3
4
1.1
5
0.9
6
1.7
7
1.3
a. Compute the sample variance of the above numbers. Show your work.(3)
Solution:
Printer
x
x2
x 8.5
(Characters)
x

 12143
.
1
0.8
0.64
n
7
2
1.4
1.96
2
2
2
x  nx
10.89  712143
.

2
3
1.3
1.69
s 

n 1
6
4
1.1
1.21
Unfortunately, with
 009476
.
5
0.9
0.81
a bit too much rounding, you can get
6
1.7
2.89
7
1.3
1.69
.
.
s 2 .1967 , which gives s x  01235
Total
8.5
10.89


b. Compute a 98% confidence interval for the mean time.(3)
From the problem statement  .02 . From page 10 of the syllabus supplement, if the
.
population variance is unknown   x  t  s x and t n21  t .601  3143
.
Solution:
2
sx 
s
n

0.09436 0.3078
. So   12143

 0116350
.
.
 3143
.  0116350
.
.  0.37 or 0.85 to 1.58.
  121
7
7
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III. Do at least 2 of the following 4 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!
1. Using the data from the previous problem (page 2.), test the hypothesis that the population mean is 1.0 at
the 90% confidence level using:
a. A test ratio (2)
b. Critical values (2)
c. A confidence interval (2)
d. Find an approximate p-value for the null hypothesis. (1)
e. Now find a 90% confidence interval for the standard deviation.(3)
Solution: From page 10 of the Syllabus Supplement:
Interval for
Confidence
Interval
Hypotheses
Mean (
known)
  x  z  x
Mean (
unknown)
  x  t  sx
2
2
DF  n  1
H 0:   1
H 1:   1
Test Ratio
H0 :   0
H1 :    0
H 0 :   0
H1:   0
t
x  0
x
x  0
x cv   0  z 2  x
x cv   0  t  2 s x
sx
n 1
6
 0  1 DF  n  1  6  .10 t    t .05  1943
.
2
From the previous page: x  12143
and s x 
.
a. Test Ratio: t 
z
Critical Value
s

n
0.09436 0.3078
.

 0116350
.
7
7
x   0 12143
.
1
= 1.8419. This is in the 'accept region'

sx
.116350
between -1.943 and +1.943, so accept H 0 .
b. Critical Value: x cv 
 0  t  s x  1  1943
.  0116350
.
  1  0.226 or 0.774
2
to 1.226. This includes x  12143
, so accept H 0 .
.
c. Confidence Interval:
  x  t  sx  12143
.
 1943
.  011635
.
.
 0.226
  1214
2
or 0.988 to 1.440. This includes  0  1 , so accept H 0 .
 6
 6
.
.
d. From the t table t = 1.842 is between t .05  1943
and t .10  1440
. Since this is 2-sided, we double pvalue, and .10   p  value  .20 .
e. From page 1 of the Syllabus Supplement:
 n  1 s 2
 2
2 
2
2  6
2  6
Since  .05 and 1   .95 , look up  .05 and  .95 . So
2
2
 6 0.09476
2 
12.5916
0.2124    05496
.
 6 0.09476
16354
.
 n  1 s 2
2
1
. DF  n  1  6 and  .10 .

2
 n  1 s 2
 .205
2 
 n  1 s 2
 .295
or
or 0.04515   2  0.3427 . Finally, taking square roots,
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2. Data from section II is repeated below.
Printer Characters
1
0.8
2
1.4
3
1.3
4
1.1
5
0.9
6
1.7
7
1.3
a. Test that the population mean is 1 assuming that the parent population is normally distributed and that
the sample of seven is taken from a population of 50. (3)
b. Drop the assumption that the parent distribution is normal ( and that the populate size is 50) and figure
out a confidence level for the confidence interval ( 08
.    17
. ) (3)
c. Put a p-value on the statement that the median is 1, again assuming that the distribution is not normal. (2)
d. Looking at your result in c, would you reject the hypothesis that the median is 1 at the 90% confidence
level? Explain. (1)
e. Repeat c-d assuming that the parent distribution is normal.(2)
Solution: n  7, N  50 , and from Part II s 2  0.09476 and x  12143
.
.
a. s x 
s
n
N n

N 1
s2  N  n

 
n  N  1
0.09476  50  7 
.

  0.011880  010899
 50  1 
7
x   0 12143
.
1
= 1.966. This is not in the 'accept region'

sx
.10899
(i) Test Ratio: t 
between -1.943 and +1.943, so reject H 0 .
or (ii) Critical Value: x cv 
 0  t  s x  1  1943
.  010899
.
  1  0.212 or 0.788
2
to 1.212. This does not include x  12571
, so reject H 0 .
.
or (iii) Confidence Interval:
  x  t  sx  12571
.
 1943
.  010899
.
.
 0.212
  1257
2
or 1.045 to 1.469. This does not include  0  1 , so reject H 0 .
b. Since 0.8 and 1.7 are the highest and lowest numbers, the interval is wrong if all seven numbers are
below the median or all seven are above the median. The first is equivalent to getting seven heads in seven
tosses of a fair coin, and the second is equivalent to getting seven tails but both these probabilities are the
same . So, using the binomial table,


  2 P x  7 n  7, p .5  21  P x  6  21.99219  2.0078 .01562 and the confidence level is
1   .9843 .
H0 :  1
c. 
There are two numbers below 1 and five numbers above 1. A p-value for the statement that
 H1 :   1
the median is 1 would be the probability of 5 or more numbers above the median, equivalent to the
probability of getting 5 or more heads.. This is doubled because the statement is 2-sided.
 
 

p  value  2 P x  5 n  7, p .5  2 1  P x  4  21.63672  2.3633 .7266
d. Since the p-value (.7266) is greater than the significance level (.10) accept H 0 .
e. Since , if the distribution is normal, the mean and the median are equal, this is the same as a on this or the
previous page.
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3. A manufacturer of condensers says that fewer than 4% of the condensers produced are defective. A
sample of 300 is selected from a very large shipment and 10 defective items are found.
a. Do a 2-sided 95% confidence interval for the proportion that are defective.(2)
b. Do a 2-sided 93% confidence interval for the proportion that are defective.(2)
c. State the Hypotheses to be tested in the problem above.(1)
d. Do a test of the hypothesis in c) at a 95% confidence level. (3)
e. Repeat d) at the 93% confidence level.(2)
f. If the true proportion defective is 3%, what is the power of the test? (3)
g. The New York Times/CBS News poll says that 10% of the population believe the President Clinton is
telling the whole truth about Paula Jones. On how large a sample would this have to be based to be accurate
within 0.5%? (2)
Solution: From page 10 of the Syllabus Supplement:
Interval for
Confidence
Interval
Hypotheses
Test Ratio
Proportion
p  p  z 2 s p
H 0 : p  p0
H 1 : p  p0
z
pq
n
q  1 p
sp 
p  p0
p
Critical Value
p cv  p 0  z  2 
p 
p
p0 q 0
n
.0333.9667
pq
pq
x 10
q  1  p  1.0333 .9667 s p 

.0333 s p 

.010364 .
n
n 300
n
300
. .010364 .0333  .0203 or .0130 to .0536.
a.   .05 .025 p  p  z.025 s p .0333  1960
2
2
You do not need a null hypothesis for a confidence interval unless you are specifically using
it to test the hypothesis!

.07 .035 p  p  z.035 s p .0333  181
. .010364 .0333  .0187 or .015 to .052. This value
b.
2
2
of z was found on page 1.
H : p  .04
c.  0
H 1 : p  .04
p
d.  .05
p0 .04
q 0  1  p0 .96  p 
p0 q 0

n
.04.96
300
.0113137 . This is a one-sided test
with  .05, z  z.05  1645
.
.
p  p 0 .0333.04

 0.5893 . Since this is above
(i) Test Ratio: z 
p
.0113137
-1.645, accept H 0 .
. .0113137 .02138 .
or (ii) Critical Value: pcv  p0  z  p .04  1645
Since p .0333 is above this, accept H 0 .
. .013137 or
or (iii) Confidence interval: p  p  z s p .0333  1645
p .05194 . This does not contradict H 0  p .04 , so accept H 0 .
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2/24/98 252y9812
e. This is a one-sided test with  .09, z  z.09  1475
. This value of z was found on page 1.
.
p  p 0 .0333.04

 0.5893 . Since this is above
(i) Test Ratio: z 
p
.0113137
-1.475, accept H 0 .
. .0113137 .0233 .
or (ii) Critical Value: pcv  p0  z  p .04  1475
Since p .0333 is above this, accept H 0 .
. .013137 or
or (iii) Confidence interval: p  p  z s p .0333  1475
p .0500 . This does not contradict H 0  p .04 , so accept H 0 .
f. Use a critical value from d or e. In d, we accept H 0 if p is greater than
. .0113137 .02138 . We will compute  , the probability of a type II error
. pcv  p0  z  p .04  1645
 p  p1 
.03.97
p1q1
by finding P p  p cv p  p1  P  z  cv

 where p1 .03 and  p1 
 p1 
n
300

 .02138.03 
.0.00985.   P p  .02138 p .03  P z 
 P z  088
.   P 088
.  z  0  P z  0
.00985 

.3106.5000 .8106 .




g. If  .05 , z   z.025  1960
and n 
.
2
pqz 2
e2

2
. 
.10.90196
.005 2
 13829.8 , so use a sample of size
13830.
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4. A sample variance is found to be 250. The parent distribution is approximately normal.
a. Do a 98% confidence interval for the population variance if the sample size is 28. (2)
b. Do a 98% confidence interval for the population variance if the sample size is 125. (2)
c. If the sample size is 125, do a test of the statement that the population variance is 200 at the 98%
confidence level using the method taught in class. (3)
d. Repeat the test in c) using critical values for the sample variance as explained on page 10 of the syllabus
supplement. (2)
e. Looking at what you did in d), do a test of the statement that the population variance is at least 200. (3)
f. If we claim that a population has a Poisson distribution with a mean of 7 , and the actual value found is
4, do a hypothesis test at the 95% level. (3)
Solution: s 2  250 and from page 10 of the Syllabus Supplement:
Interval for
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
VarianceSmall Sample
2 
VarianceLarge Sample
 
n  1s 2
H 0 :  2   02
 .25.5 2 
H1: :  2   02
s 2DF 
H 0 :    02
H1 :  2   02
2 
2
scv
 02

 .25.5   20
2
n 1
z
2
 z 2  2DF 
n  1s 2
2  2  2DF   1
a. The variance formula is explained on pages 1 and 2 of the Syllabus Supplement.  .02 and there are 27
degrees of freedom.
 27 250
n  1s 2
 2
2 
n  1s 2
2 
1
2
 27 250
becomes
 n  1 s 2
 .201
2 
 n  1 s 2
 .299
and then
2
or 143.73   2  524.31 .
2 
46.9631
12.8785
b.  .02 and there are 124 degrees of freedom. Since this is too many degrees of freedom for the  2
table, use the large sample formula.
s 2 DF
z   2 DF
2
250 2124
2.326  2124
 
250 2124
2.326  2124
 
s 2 DF
 z   2 DF
where z   z.01  2.326 .
2
2
.
   18551
. . If we square this, it becomes
or 13776
.
189.794   2  344157
.
2
H 0 :   200 2  n  1 s 2 124 250
 

 155 . This is a large sample. z  2  2  2 DF   1
c. 
2
2
200

0
H 1 :   200
 2155  2124  1  17.6068  157162
.
 189
. .  .01 , z   z.01  2.326
2
and z is between 2.326 , so accept H 0 .
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2/24/98 252y9812
 .25.5   20
H 0 :  2  200
2
2
2
d. 
so  0  200 . As the table (above) says, scv 
. The two values of
2
n 1
H 1 :   200
 .25.5  are  12   .299  12.8785 and  2   .201  46.9631 .One of these values is
2
2
2
2
scvL

12.8785 200
2
 95.40. , and the other is scvU

27
between them, accept H 0 .
 46.9631 200
27
 347.81 . Since s 2  250 is
H 0 :  2  200
e. 
H 1 :  2  200
(i) Critical Value: Use
2
scv
2
with  .01 , so scv


 .25.5   20
2
n 1
 12  20

, but this is a one-sided test
 .298 20

14.01 250
 130 .
n 1
n 1
27
There was trouble here, but no one called it to my attention. The
2
2
table does not give  .98
, but does give  .975
, which is about 14.5, and
2
would be acceptable. I looked at  .99
and settled on 14.01, which is
about two-thirds of the way between them. Since s 2  250 is above
130, accept H 0 .
or(ii) Test Ratio:  2 
 n  1 s 2

2
0

 27 250
200
 33.7500 . Since this is above
 .298  14.01 , accept H 0 .
H 0 : Poisson 7

f. This is the easiest problem on the exam. This is a 2-sided test of 
. Actually, we are

H 1 : not Poisson 7
assuming Poisson and checking for a mean of 7. The p-value of the null hypothesis when x  4 (4 is below
the mean) is (from the Poisson table with a parameter of 7) 2 P x  4  2 .17299 = .3460. Since this is
above the significance level (.05), accept H 0 .
8
Note: Last year we used a take-home exam, which we probably will never do again. This is one of the
10 versions of the exam. A shorter version of these problems is fair game.
252y981b 2/20/98
Name __________________
Hour of Class Registered (Circle)
MWF 10 11 TR 12:30 2
Hour of Class Attended (If Different)
__________
First Take-home Exam - QBA2
PART IV. Show your work! Neatness counts (Especially in Graphs)
1. The Trash Bag Case: (Bowerman and O'Connell) A television network refuses to air a commercial that
claims that a trash bag has a breaking strength of 50 lbs. without proof. A sample of 250 shows that
 x  12644 and  x
2
 640156 . Do a 2-sided test of the claim using a confidence level of 99%.
a. State the H 0 and H1 (1)
b. Using critical values for the sample mean, test the Hypotheses. (2)
c. Construct a power curve for the test. (7)
Solution:
x 12644
x 2  nx 2 640156  25050.5760 2
2
.
x

 50.5760 . s 

 2.7030 . s  16441
.
n
250
n 1
249
H 0 :   50
a. 
 n  250,  .01 
H 1 :   50
b. From page 10 of Syllabus supplement x cv   0  t  s x .  0  50 . Since this is a large sample


2
t  2  z 2  z.025  2.576 . s x 
s

2.7030
. So xcv   0  t  s x
 .010812  0103981
.
2
250
n
is not between
 50  2.576 0103981
.
.
  50  0.2679 or xcvl  49.732 to x cvu  50.268. Since x  505760
these values, reject H 0 .
c. We accept H 0 if x is between 49.732 and 50.268. We will compute  , the probability of a type II error
 x  1
x  1 
by finding P x cvl  x  x cvu    1  P  cvl
 z  cvu
 for the values  1   0 ,

x 
x

 0   0.5 t  2 s x ,  0  10
.  t  2 s x ,  0  15
.  t  2 s x and  0   2.0 t  2 s x , where


t  2 s x   2.576 0103981
.
  0.2679 .
The values for this appear below. All computations are for  1 upper .
 1 lower
 1 upper
zL 
x cvl   1
x
zU 
50.00
50.00
-2.58
49.87
50.13
-3.86
49.73
50.27
-5.15
49.60
50.40
-6.44
49.46
50.54
-7.72
*   Pz L  z  zU  with values from the z or t table.
x cvu   1
x
2.58
1.29
0.00
-1.29
-2.58
 *
power  1  
.9900
.9014
.5000
.0985
.0050
.0100
.0980
.5000
.9015
.9950
9
252y981b 2/20/98
2. The Cheese Spread Case: (Bowerman and O'Connell) (Extra credit) A cheese spread producer will not
adopt a new package unless there is strong evidence that fewer than 10% of users will stop buying the
product if the new package is used. A sample of 1000 is taken and 63 consumers say that they will not buy
the product. State the hypotheses, do the test at the 95% confidence level and construct the power curve for
the test. (5)
You can't answer a problem you haven't read. If you can't answer the questions below, you
didn't read the problem!
1. Where in this problem is a mean mentioned?
2. Which of your hypotheses is 'fewer than 10%' (not 'at most 10%')?
Solution:
H : p  .10
a)  0
H 1 : p  .10

 p0 .10 n  1000

x
63


.063  .05

n 1000
Common error: the statement says 'fewer than.' This must be in your hypotheses and since it does not
contain an equality, it cannot be a null hypothesis.
p0 q 0
b) From page 10 of Syllabus supplement pcv  p0  z 2  p where  p 
. This must become
n
and
.
pcv  p0  z  p because the hypothesis is one sided. z.05  1645
p
.10.90
p0 q 0
. .00949 .0844 . We will accept

 0.00949 . So pcv  p0  z  p .10  1645
n
1000
H 0 if p is greater than .0844. Since p .063 , reject H 0 .
c) We accept H 0 if p is greater than .0844. We will compute  , the probability of a type II error by
p 
 p  p1 
finding P p  p cv p  p1  P  z  cv
 where 
 p1 



p1

p1q1
for the values p1  p0 ,
n
.  z 2  p and p0   2.0 z 2  p , where
p0   0.5 z 2  p , p0  10
.  z 2  p , p0  15
z 2  p  1645
. .00949 .0156 . The values for this appear below.
p1
 p1
.1000
.0922
.0844
.0766
.0688
.00949
.00915
.00879
.00841
.00800
*   P z  z1 
z1
 *
power  1  
-1.645
.9500
.0500
-0.853
.8023
.1977
0.000
.5000
.5000
0.928
.1762
.8238
1.950
.0256
.9744
p  p1
where z1  cv
computed from rounded z 1 using the z or t table.
 p1
10