252y0711 2/22/07
ECO252 QBA2
Name ________________
FIRST HOUR EXAM
February 23, 2007
Version 1
Show your work! Make Diagrams! Exam is normed on 50 points. Answers without reasons are not
usually acceptable.
I. (8 points) Do all the following.
x ~ N 2, 7
2  2
  23 .4  2
z
 P 3.63  z  0  = .4999
1. P23 .4  x  2  P 
7
7 

For z make a diagram. Draw a Normal curve with a mean at 0. Indicate the mean by a vertical line!
Shade the area between -3.63 and zero. Because this is completely on the left of zero, but ends at zero, this
is exactly the kind of probability given by the standard Normal table. Look up the probability between zero
and 3.63 on the table and you are done. If you wish, make a completely separate diagram for x . Draw a
Normal curve with a mean at 2. Indicate the mean by a vertical line! Shade the area between -23.4 and
the mean (2). Since this area ends at the mean we do not need to add or subtract.
7  2
 7  2
z
 P 1.29  z  0.71
2. P 7  x  7   P 
7 
 7
 P1.29  z  0  P0  z  0.71  .4015 + .2611 = .6626
For z make a diagram. Draw a Normal curve with a mean at 0. Indicate the mean by a vertical line!
Shade the area between -1.29 and 0.71. Because this is on both sides of zero, we must add the area between
-1.29 and zero to the area between zero and 0.71. If you wish, make a completely separate diagram for x .
Draw a Normal curve with a mean at 2. Indicate the mean by a vertical line! Shade the area between -7
and +7. These numbers are on either side of the mean (2), so we add the area between -7 and the mean to
the area between the mean and 7.
0  2

 Pz  0.29   Pz  0  0.29  z  0 = .5 - .1141 = .3859
3. Px  0  P  z 
7 

For z make a diagram. Draw a Normal curve with a mean at 0. Indicate the mean by a vertical line!
Shade the area below -0.29. Because this is entirely on the left side of zero, we must subtract the area
between -0.29 and zero from the area below zero. This is identical to way you get the p-value for a leftsided test. If you wish, make a completely separate diagram for x . Draw a Normal curve with a mean at
2. Indicate the mean by a vertical line! Shade the area below zero. This area is entirely on the left side of
the mean (2), so we subtract the area between 0 and the mean from the area below the mean.
4. x.085 (Do not try to use the t table to get this.) For z make a diagram. Draw a Normal curve with a
mean at 0. z .085 is the value of z with 8.5% of the distribution above it. Since 100 – 8.5 = 91.5, it is also
the .915 fractile. Since 50% of the standardized Normal distribution is below zero, your diagram should
show that the probability between z .085 and zero is 91.5% - 50% = 41.5% or P0  z  z.085   .4150 . If
we check this against the Normal table, the closest we can come to this is P0  z  1.37   .4147 . (1.38 is
also acceptable here.) So z .085  1.37 . This is the value of z that you need for an 83% confidence
interval. To get from z .085 to x.085 , use the formula x    z , which is the opposite of z 
x
.

x  2  1.37 7  11.59 . If you wish, make a completely separate diagram for x . Draw a Normal curve
with a mean at 2. Show that 50% of the distribution is below the mean (2). If 8.5% of the distribution is
above z .085 , it must be above the mean and have 41.5% of the distribution between it and the mean.
11 .59  2 

Check: Px  11 .59   P  z 
  Pz  1.37   Pz  0  P0  z  1.37 
7


 .5  .4147  .0853  .085 . This is identical to the way you get a p-value for a right-sided test.
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II. (9 points-2 point penalty for not trying part a.) Edwin Mansfield tells us that in order to compute
information about the average expenditure by honors students at Semiconscious State University, a sample
of four students’ expenditures is taken. The results are as below. I should not have to tell you that parts b c,
and f require hypothesis tests.
x
192
150
168
90
a. Compute the sample standard deviation, s , of expenditures. Show your work! (2)
b. Is the population mean significantly above 130? (Use a 90% confidence level.) Show your
work! (3)
c. Redo b) when you find out that there are only 10 honors students at Semiconscious State
University.
d. (Extra Credit) Find an approximate p value for your null hypothesis. (2)
e. Assume that the population standard deviation is 40 and create an 83% two-sided confidence
interval for the mean. (2)
f. (Extra Credit) Given the data, test the hypothesis that the population standard deviation is above
40?
Solution: a) Compute the sample standard deviation, s , of expenditures
Row
1
2
3
4
x
x2
xx
 x  x 2
192
150
168
90
600
36864
22500
28224
8100
95688
42
0
18
-60
0
1764
0
324
3600
5688
The first two columns are needed for the computational (shortcut) method. The first, third and fourth are
needed for the definitional method.
 x  600 ,  x  95688 ,  x  x   0 (a check),  x  x   5688 and
 x  600  150 s x2   x 2  nx 2  95688  4150 2  5688  1896 s
x
2
2
n  4.
x  1896  43 .5431
n 1
3
3
n
4
b) Is the population mean significantly above 130? (Use a 90% confidence level.)
Note that ‘Above 130’ does not contain an equality, so that it must be an alternative hypothesis. Our
hypotheses are H 0 :   130 and H 1 :   130 .
Given:  0  130 , s  43 .5431 , n  4. df  n  1  3, x  150 and   .10 . So
sx 
s

43 .5431
n

4
1896
3
3
 1.638 and t n1  t.05
 2.353 .
 474  21 .7715 . Note that tn 1  t .10
2
4
There are three ways to do this. Do only one of them.
x   0 150  130

 0.9186 . This is a right-sided test - the larger the sample mean is,
(i) Test Ratio: t 
sx
21 .7715
the more positive will be this ratio. We will reject the null hypothesis if the ratio is larger than
3
tn 1  t .10
 1.638 . Make a diagram showing a Normal curve with a mean at 0 and a shaded 'reject' zone
above 1.638. Since the test ratio is below 1.638, we cannot reject H 0 .
(ii) Critical value: We need a critical value for x above 130. Common sense says that if the sample mean is
too far above 130, we will not believe H 0 :   130 . The formula for a critical value for the sample mean
is x    t n1 s , but we want a single value above 130, so use x    t n1 s
cv
0

2
x
cv
0

x
2
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 130  1.638 21 .7715   165 .66 . Make a diagram showing an almost Normal curve with a mean at 130
and a shaded 'reject' zone above 165.66. Since x  150 is below 165.66, we do not reject H 0 .
(iii) Confidence interval:   x  t sx is the formula for a two sided interval. The rule for a one-sided
2
confidence interval is that it should always go in the same direction as the alternate hypothesis. Since the
alternative hypothesis is H 1 :   130 , the confidence interval is   x  tn1 s x or
  150  1.638 21.0402   115 .5333 . Make a diagram showing an almost Normal curve with a mean at
x  150 and, to represent the confidence interval, shade the area above 115.5333 in one direction. Then, on
the same diagram, to represent the null hypothesis, H 0 :   130 , shade the area below 130 in the opposite
direction. Notice that these overlap. What the diagram is telling you is that it is possible for   115.5333
and H 0 :   130 to both be true. (If you follow my more recent suggestions, it is actually enough to show
that 130 is on the confidence interval.) So we do not reject H 0 .
c) Redo b) when you find out that there are only 10 honors students at Semiconscious State University.
s
We now have s x 
N  n 43 .5431

N 1
4
10  4
1896  6 

   316  17 .7764 .
10  1
4 9
n
x   0 150  130

 1.1251 . This is a right-sided test - the larger the sample mean is,
(i) Test Ratio: t 
sx
17 .7764
the more positive will be this ratio. We will reject the null hypothesis if the ratio is larger than
3
tn 1  t .10
 1.638 . Make a diagram showing a Normal curve with a mean at 0 and a shaded 'reject' zone
above 1.638. Since the test ratio is below 1.638, we cannot reject H 0 .
(ii) Critical value: We need a critical value for x above 130. Common sense says that if the sample mean is
too far above 130, we will not believe H 0 :   130 . The formula for a critical value for the sample mean
is x    t n1 s , but we want a single value above 130, so use x    t n1 s
cv
0

2
x
cv
0

x
 130  1.638 17.7764   159 .12 . Make a diagram showing an almost Normal curve with a mean at 130
and a shaded 'reject' zone above 165.66. Since x  150 is below 159.66, we do not reject H 0 .
(iii) Confidence interval:   x  t sx is the formula for a two sided interval. The rule for a one-sided
2
confidence interval is that it should always go in the same direction as the alternate hypothesis. Since the
alternative hypothesis is H 1 :   130 , the confidence interval is   x  tn1 s x or
  150  1.638 17.7764   120 .8823 . Make a diagram showing an almost Normal curve with a mean at
x  150 and, to represent the confidence interval, shade the area above 120.8823 in one direction. Then, on
the same diagram, to represent the null hypothesis, H 0 :   130 , shade the area below in the opposite
direction. Notice that these overlap. What the diagram is telling you is that it is possible for   120.8823
and H 0 :   130 to both be true. (If you follow my more recent suggestions, it is actually enough to show
that 130 is on the confidence interval.) So we do not reject H 0 .
d) (Extra Credit) Find an approximate p value for your null hypothesis. (2)
In b) t  0.9186 and in c) t  1.1251 . Remember that df  n  1  3 . The 3 line of the t-table is below.
df .45
.40
.35
.30
.25
.20
.15
.10
.05 .025 .01
.005 .001
3 0.137 0.277 0.424 0.584 0.765 0.978 1.250 1.638 2.353 3.182 4.541 5.841 10.21
3
3
3
t  0.9186 falls between t .20
 0.978 and t .15
 1.250 . t  1.1251 also falls between t .20
 0.978 and
3
t .15
 1.250 . So whichever one you used .15  pvalue  .20.
e) Assume that the population standard deviation is 40 and create a 83% two-sided confidence interval for
the mean. (2) . On the first page we found z .085  1.37 . According to Table 3, if the population variance is
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known,   x  z  x , where x  150 and  x 
2


n
40
 20 . . 1    1  .17  .83 , so
4
  .17  .085 .   150  1.37 20   150  27.4.
2
2
f) (Extra Credit) Given the data, test the hypothesis that the population standard deviation is above 40?
This is an alternate hypothesis, H 1 :   40 . The null hypothesis is H 0 :   40 Remember n  4 ,
s x2 
x
2
 nx 2
n 1
 1896 . Table 3 says  2 
n  1s 2
 02

31896 
40 2
 3.550 .Recall df  n  1  3, and
  .10 . To do a one-sided test, compare this with  .103 6.2514. Our computed  2 is less than the table
value, so do not reject the null hypothesis.
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III. Do as many of the following problems as you can.(2 points each unless marked otherwise adding to
13+ points). Show your work except in multiple choice questions. (Actually – it doesn’t hurt there
either.) If the answer is ‘None of the above,’ put in the correct answer.
1.Which of the following is a Type 1 error?
a) Rejecting the null hypothesis when the null hypothesis is false.
b) *Rejecting the null hypothesis when the null hypothesis is true.
c) Not rejecting the null hypothesis when the null hypothesis is true.
d) Not rejecting the null hypothesis when the null hypothesis is false.
e) All of the above
f) None of the above.
2. In a survey, 540 out of 1000 respondents indicated that they would rather have $150 than a day off.
You believe that a majority (over 50%) will prefer the money to the day off. Do the data support your
opinion? Your hypotheses are:
a) H 0 : p  .50 , H 0 : p  .50
b) H 0 : p  .50 , H 0 : p  .50
c) H 0 : p  .50 , H 0 : p  .50
d) H 0 : p  .50 , H 0 : p  .50
e) H 0 : p  .50 , H 0 : p  .50
f) *None of the above.
Explanation: p  .50 is an alternative hypothesis since it does not contain an equality, the opposite of
H 1 : p  .50 is H 0 : p  .50 . These are not mentioned in the question.
3. The chi-squared distribution is used in which of the following circumstances.
a) We are testing a single mean, the underlying distribution is Normal and the population
variance is unknown.
b)* We are testing a single variance, the underlying distribution is Normal and the population
variance is unknown.
c) We are testing a single proportion, the underlying distribution Binomial and we have a
large sample.
d) We are testing a single median, the underlying distribution is Normal and the population
variance is unknown.
e) None of the above.
[6]
4. According to Dummeldinger, The Florida Association of Retired Teachers (FART) is thinking of
changing its name. It takes a survey of 60 members and finds that many wish to change the name.
Assuming that the alternative hypothesis is H1 : p  .6 and that   .01 , how many people (what
proportion of 60) would be required to reject the null hypothesis? (Show your work.) (3) [9]
Solution: The Outline formula for proportions is below.
Interval for
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
Proportion
p  p  z 2 s p
pq
n
q  1 p
sp 
H 0 : p  p0
H1 : p  p0
z
p  p0
p
pcv  p0  z 2  p
p0 q0
n
q0  1  p0
p 
5
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So pcv  p0  z  p becomes pcv  p0  z  p . Since   .01 , we need a z.01  2.327 .
2
p0 q0
.6.4

 .0040  .06325 pcv  p0  z  p  .6  2.327.06325  .74717. Since
n
60
.74717 60   44.83 , we need 45 out of 60.
p 
5. According to Dummeldinger, The Florida Association of Retired Teachers (FART) is thinking of
changing its name. It takes a survey of 60 members and finds that 45 wish to change the name. Assuming
that the alternative hypothesis is p  .6 , what is the p-value for the null hypothesis? (3) [12]
Solution: Since p 
x 45

 .75 , the p-value is
n 60
.75  .6 

P p  .75   P z 
  Pz  2.37   .5  .4911  .0089
.06325 

6. Given the numbers in problem 5) create a 95% two sided confidence interval for the proportion that
favors a change. (3)
[15]
pq
.75 .25 
x 45

 .00313  .0559 . Since
 .75 p  p  z  s p , where s p 
Solution: p  
2
n
60
n 60
z .025  1.960 , we have p  .75  1.960 .0559   .75  .1096 or .6404 to .8596.
7. Assuming that about 60% of the old FART members favor a name change, how many members would
we have to poll to know what the proportion is within .01 ? (2)
[17]
Solution: The usually suggested formula is n 
covered. n 
.60 .40 1.960 2
.012
pqz 2
e2
. This is the formula everyone forgets that we
 9219 .84 . So at least 9220.
8. While we are playing with old folks in Florida, assume that a Florida town has had a median age of 55.
Because they are afraid that the median age is now higher the planning board takes a sample of 200 people.
How many of the 200 people would have to be over 55 for the planning board to conclude that the median
has risen? (3)
[20]
Solution: From the outline we have the table below.
Hypotheses about
a median
 H 0 :   0

 H 1 :   0
 H 0 :   0

H 1 :   0
 H 0 :   0

H 1 :   0
Hypotheses about a proportion
If p is the proportion
If p is the proportion
above  0
below  0
 H 0 : p .5

 H 1 : p .5
 H 0 : p .5

 H 1 : p .5
 H 0 : p .5

 H 1 : p .5
 H 0 : p .5

 H 1 : p  .5
 H 0 : p .5

 H 1 : p  .5
 H 0 : p .5

 H 1 : p  .5
So if our hypotheses are H 0:  55 and H 1:  55 and p is the proportion that are over 55, they become
H 0: p  .5 and H 1: p  .5 . pcv  p0  z  p . Since   .05 , we need a z .05  1.645 .
p0 q0
.5.5

 .00125  .03536 . pcv  p0  z  p  .5  1.645.03536  .55817. The number
n
200
is .55827 200   111 .63 , which rounds up to 112.
p 
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252y0711 2/22/07
ECO252 QBA2
FIRST EXAM
February 23, 2007
TAKE HOME SECTION
Name: _________________________
Student Number and class: _________________________
IV. Do at least 3 problems (at least 7 each) (or do sections adding to at least 20 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.) Answers without
reasons usually are not acceptable. Neatness and clarity of explanation are expected. This must be
turned in when you take the in-class exam. Note that answers without reasons and citation of
appropriate statistical tests receive no credit. Many answers require a statistical test, that is, stating or
implying a hypothesis and showing why it is true or false by citing a table value or a p-value. If you haven’t
done it lately, take a fast look at ECO 252 - Things That You Should Never Do on a Statistics Exam (or
Anywhere Else). All problems are based on Anderson, Sweeny and Williams except 4f and 4g.
1.
The average sales of a grocery store have been $80000 per day. After new advertising policies are
adopted, sales in thousands of dollars are tabulated for 64 days. Test to see if the average sales are
now above 80000.
x
69.07
109.29
84.73
75.51
78.99
72.72
76.23
52.80
81.67
88.30
86.43
81.79
68.74
91.19
83.13
87.95
72.69
75.33
103.18
64.20
92.14
87.70
78.09
74.41
72.73
86.89
90.65
86.96
68.47
85.43
85.03
92.67
82.26
83.34
63.91
83.54
72.60
78.85
66.50
69.59
83.59
80.67
65.65
91.22
81.44
82.75
91.43
91.48
87.44
83.41
93.22
91.05
82.13
91.84
97.30
75.19
94.47
96.50
71.65
91.50
76.37
95.71
91.49
86.80
To personalize the data below take the last digit of your student number, divide it by 10 and add it to the
numbers above. If the last digit of your student number is zero, add 1.00. Label the problem ‘Version 1,’
‘Version 2,’ … ‘Version 10’ according to the number that you used. (For example, Seymour Butz’s
student number is 976502, so he will add 0.20 and change the data to 69.27, 81.87, 72.89 etc. – but see the
hint below, you do not need to write down all the numbers that you are using, just your computations.)
Hint - if you use the computational formula: For the original numbers n  64 ,
and
x
2
 x  5280 .00
 442387 .8608 . If you add a quantity a to a column of numbers,
 x  a   x na,  x  a    x  2a x na
2
2
2
Assume that the Normal distribution applies to the data and use a 99% confidence level.
a. Find the sample mean and sample standard deviation of the incomes in your data, showing
your work. (1) (Your mean should be above 82 (thousand) and your sample standard deviation
should be above 10 (thousand))
b. State your null and alternative hypotheses (1)
c. Test the hypothesis using a test ratio (1)
d. Test the hypothesis using a critical value for a sample mean. (1)
e. Test the hypothesis using a confidence interval (1)
f. Find an approximate p-value for the null hypothesis. (1)
g. On the basis of your tests, have sales increased? Why? (1)
h. How do your conclusions change if the random sample of 64 days is taken from a population of
100 days? (2)
7
252y0711 2/22/07
i. Assume that the Normal distribution does not apply and, using your data, test that the median is
above 80000. (3)
[12]
j. (Extra credit) Use your data to get an approximate 99% 2-sided confidence interval for the
median.
Solution: a) Find the sample mean and sample standard deviation of the incomes in your data, showing
your work. (1) (Your mean should be above 82 (thousand) and your sample standard deviation should be
above 10 (thousand)).
Here are the computations for my original x (Version 0) and for x  0.9 (Version 9).
x
Row
1
69.07
2 109.29
3
84.73
4
75.51
5
78.99
6
72.72
7
76.23
8
52.80
9
81.67
10
88.30
11
86.43
12
81.79
13
68.74
14
91.19
15
83.13
16
87.95
17
72.69
18
75.33
19 103.18
20
64.20
21
92.14
22
87.70
23
78.09
24
74.41
25
72.73
26
86.89
27
90.65
28
86.96
29
68.47
30
85.43
31
85.03
32
92.67
33
82.26
34
83.34
35
63.91
36
83.54
37
72.60
38
78.85
39
66.50
40
69.59
41
83.59
42
80.67
43
65.65
44
91.22
45
81.44
46
82.75
47
91.43
48
91.48
49
87.44
50
83.41
51
93.22
x2
x  0.9
x  0.92
4770.6649
11944.3041
7179.1729
5701.7601
6239.4201
5288.1984
5811.0129
2787.8400
6669.9889
7796.8900
7470.1449
6689.6041
4725.1876
8315.6161
6910.5969
7735.2025
5283.8361
5674.6089
10646.1124
4121.6400
8489.7796
7691.2900
6098.0481
5536.8481
5289.6529
7549.8721
8217.4225
7562.0416
4688.1409
7298.2849
7230.1009
8587.7289
6766.7076
6945.5556
4084.4881
6978.9316
5270.7600
6217.3225
4422.2500
4842.7681
6987.2881
6507.6489
4309.9225
8321.0884
6632.4736
6847.5625
8359.4449
8368.5904
7645.7536
6957.2281
8689.9684
69.97
110.19
85.63
76.41
79.89
73.62
77.13
53.70
82.57
89.20
87.33
82.69
69.64
92.09
84.03
88.85
73.59
76.23
104.08
65.10
93.04
88.60
78.99
75.31
73.63
87.79
91.55
87.86
69.37
86.33
85.93
93.57
83.16
84.24
64.81
84.44
73.50
79.75
67.40
70.49
84.49
81.57
66.55
92.12
82.34
83.65
92.33
92.38
88.34
84.31
94.12
4895.8009
12141.8361
7332.4969
5838.4881
6382.4121
5419.9044
5949.0369
2883.6900
6817.8049
7956.6400
7626.5289
6837.6361
4849.7296
8480.5681
7061.0409
7894.3225
5415.4881
5811.0129
10832.6464
4238.0100
8656.4416
7849.9600
6239.4201
5671.5961
5421.3769
7707.0841
8381.4025
7719.3796
4812.1969
7452.8689
7383.9649
8755.3449
6915.5856
7096.3776
4200.3361
7130.1136
5402.2500
6360.0625
4542.7600
4968.8401
7138.5601
6653.6649
4428.9025
8486.0944
6779.8756
6997.3225
8524.8289
8534.0644
7803.9556
7108.1761
8858.5744
8
252y0711 2/22/07
52
53
54
55
56
57
58
59
60
61
62
63
64
91.05
8290.1025
91.95
8454.8025
82.13
6745.3369
83.03
6893.9809
91.84
8434.5856
92.74
8600.7076
97.30
9467.2900
98.20
9643.2400
75.19
5653.5361
76.09
5789.6881
94.47
8924.5809
95.37
9095.4369
96.50
9312.2500
97.40
9486.7600
71.65
5133.7225
72.55
5263.5025
91.50
8372.2500
92.40
8537.7600
76.37
5832.3769
77.27
5970.6529
95.71
9160.4041
96.61
9333.4921
91.49
8370.4201
92.39
8535.9121
86.80
7534.2400
87.70
7691.2900
5280.00 442387.8608 5337.60 451943.7008
For the original numbers
x
s x2
 x  5280.00,  x
2
 442387.8608, n  64 . This means
 x  5280 .00  82.50 . Using the computational or definitional formula, we get the following.
n
x

64
2
 nx 2
n 1
442387 .8608  64 82 .50 2


63
So s x  107 .7438  10 .3800 .
sx 
 x  x 
2

63
6787 .8608
 107 .7438
63
2
s
107 .7439

 1.6835  1.2972
n
64
 x  0.9  5337.60,  x  0.92  451943.7008. These numbers could have been
gotten by using  x  a    x  na  5280 .00  64 (.9)  5337 .60 and  x  a 2
  x 2  2a x na 2  442387.8608  2.95280.00  64.92  442387.8608  9504.0000  51.8400
 x  .9  5337 .60  83.40 . Using the computational or
 451943 .7008 n  64 . This means x  .9 
For Version 9
n
definitional formula, we get the following.
s x2 
x
2
 nx 2
n 1

451943 .7008  64 83 .40 2

63
64
 x  x 
63
2

6787 .8608
 107 .7438
63
s2
107 .7439

 1.6835  1.2972
n
64
b) . State your null and alternative hypotheses (1)
The problem statement is “Test to see if the average sales are now above 80000.” This translates
as   80 (in thousands), and is an alternative hypothesis because it contains no equality. Thus our
hypotheses are H 0 :   80 and H 1 :   80 . Because we suspect that the mean is above 80, this
is a right-side test.  0  80 and   .01 .
For Version 0 x  82.50 , s x  1.2972 and n  64 .
For Version 9 x  83.40 , s x  1.2972 and n  64 .
Note that Table 3 has the following.
Interval for
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
So s x  107 .7438  10 .3800 .
Mean (
unknown)
  x  t 2 s x
DF  n 1
sx 
H0 :   0
H1 :    0
t
x  0
sx
xcv   0  t  2 s x
sx 
s
n
9
252y0711 2/22/07
63
63
 2.387 and note that using t n1  t.005
So tn 1  t .01
 2.656 is inappropriate in a 1-sided test.
2
c) Test the hypothesis using a test ratio (1)
x   0 82 .50  80
83 .40  80
 2.6210

 1.9350 . Version 9: t 
Version 0: t 
1.2972
sx
1.2972
The larger the sample mean is, the more positive will be this ratio. We will reject the null
63
hypothesis if the ratio is larger than tn 1  t .01
 2.387 . Make a diagram showing a Normal
curve with a mean at 0 and a shaded 'reject' zone above 2.387. Since the test ratio is below 2.387
for Version 0, we do not reject H 0 . Since the test ratio is above 2.387 for Version 9, we reject
H0 .
d) Test the hypothesis using a critical value for a sample mean. (1)
We need a critical value for x above 80. Common sense says that if the sample mean is too far
above 128, we will not believe H 0 :   80 . The formula for a critical value for the sample mean
is x    t n1 s , but we want a single value above 80, so use x    t n1 s
cv
0

2
x
cv
0

x
 80  2.387 1.2972   83 .096 . Make a diagram showing an almost Normal curve with a mean
at 80 and a shaded 'reject' zone above 83.096. For Version 0, since x  82.50 is below 83.096, we
do not reject H 0 . For Version 10, since x  83.40 is above 83.096, we reject H 0 .
e) Test the hypothesis using a confidence interval (1)
  x  t sx is the formula for a two sided interval. The rule for a one-sided confidence interval
2
is that it should always go in the same direction as the alternate hypothesis. Since the alternative
hypothesis is H 1 :   80 , the confidence interval for Version 0 is   x  tn1 s x
 82.50  2.387 1.2972   79.404 . Make a diagram showing an almost Normal curve with a
mean at x  82.50 and, to represent the confidence interval, shade the area below above 79.404
in one direction. Then, on the same diagram represent H 0 :   80 by shading the area below 80
or simply showing 80. Since there are values below or equal to 80 that are also above or equal to
79.404, we do not reject H 0 . The confidence interval for Version 9 is   83 .40  2.387 1.2972 
 80 .3036 . Make a diagram showing an almost Normal curve with a mean at x  82.50 and, to
represent the confidence interval, shade the area below above 80.3036 in one direction. Then, on
the same diagram represent H 0 :   80 by shading the area below 80 or simply showing 80. Since
there are no values below or equal to 80 that are also above or equal to 80.3036, we reject H 0 .
f. Find an approximate p-value for the null hypothesis. (1)
x   0 82 .50  80
83 .40  80
 2.6210

 1.9350 . Version 9: t 
Version 0: t 
1.2972
sx
1.2972
There are 63 degrees of freedom. Here is the line on the t-table for 63 degrees of freedom.
df .45 .40 .35 .30 .25 .20 .15 .10 .05 .025 .01 .005 .001
63 0.126 0.254 0.387 0.527 0.678 0.847 1.045 1.295 1.669 1.998 2.387 2.656 3.225
64
64
 1.669 and t .025
 1.998 , so .025  pvalue  .05 .
For Version 0 t  1.9350 falls between t .05


64
64
 2.387 and t
 2.656 , so .005  pvalue  .01 .
For Version 9 t  2.6210 falls between t
.01
.005
g. On the basis of your tests, have sales increased? Why? (1)
If H 0 :   80 was rejected, sales have increased. If H 0 :   80 was not rejected, sales have not
increased.
10
252y0711 2/22/07
h) How do your conclusions change if the random sample of 64 days is taken from a population of 100
days? (2)
63
Recall that s x2  107.7438 , tn 1  t .01
 2.387 and   .01
For Version 0 x  82.50 , s x  1.2972 and n  64 .
For Version 9 x  83.40 , s x  1.2972 and n  64 .
sx 
s
n
N n
100  64
107 .7438  36 
 1.2972

   0.61218  0.7824 . Note that
N 1
100  1
64
 99 
36
 0.6030 . If we use the test ratio, we find the following.
99
x   0 82 .50  80
83 .40  80
 4.3456

 3.195 . Version 9: t 
Version 0: t 
0.7824
sx
0.7824
The larger the sample mean is, the more positive will be this ratio. We will reject the null
63
 2.387 . Make a diagram showing a Normal
hypothesis if the ratio is larger than tn 1  t .01
curve with a mean at 0 and a shaded 'reject' zone above 2.387. Since both test ratios are above
2.387, we reject H 0 for both versions.
i) Assume that the Normal distribution does not apply and, using your data, test that the median is above
80000. (3)
[12]
Given the question in j), it would pay me to sort the data right now.
x ordered x  0.9 ordered
Row x ordered x  0.9 ordered
Row
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
52.80
63.91
64.20
65.65
66.50
68.47
68.74
69.07
69.59
71.65
72.60
72.69
72.72
72.73
74.41
75.19
75.33
75.51
76.23
76.37
78.09
78.85
78.99
80.67
81.44
81.67
81.79
82.13
82.26
82.75
83.13
83.34
53.70
64.81
65.10
66.55
67.40
69.37
69.64
69.97
70.49
72.55
73.50
73.59
73.62
73.63
75.31
76.09
76.23
76.41
77.13
77.27
78.99
79.75
79.89
81.57
82.34
82.57
82.69
83.03
83.16
83.65
84.03
84.24
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
83.41
83.54
83.59
84.73
85.03
85.43
86.43
86.80
86.89
86.96
87.44
87.70
87.95
88.30
90.65
91.05
91.19
91.22
91.43
91.48
91.49
91.50
91.84
92.14
92.67
93.22
94.47
95.71
96.50
97.30
103.18
109.29
84.31
84.44
84.49
85.63
85.93
86.33
87.33
87.70
87.79
87.86
88.34
88.60
88.85
89.20
91.55
91.95
92.09
92.12
92.33
92.38
92.39
92.40
92.74
93.04
93.57
94.12
95.37
96.61
97.40
98.20
104.08
110.19
In both Version 0 and Version 9 we have 23 numbers below 80 and 41 above 80. Since we have
not yet studied the Wilcoxon signed rank test, we will use the sign test.
11
252y0711 2/22/07
The problem statement is now “Test to see if the median sales are now above 80000.” This
translates as   80 (in thousands), and is an alternative hypothesis because it contains no equality.
Thus our hypotheses are H 0 :    and H 1 :  80 . Because we suspect that the median is
above 80, this is a right-side test.
  .01 and n  64 . From the outline we have the table below.
Hypotheses about
a median
Hypotheses about a proportion
If p is the proportion
If p is the proportion
above  0
below  0
 H 0 :   0
 H 0 : p .5
 H 0 : p .5



 H 1 : p .5
 H 1 : p .5
 H 1 :   0
 H 0 :   0
 H 0 : p .5
 H 0 : p .5



 H 1 : p  .5
H 1 :   0
 H 1 : p .5
 H 0 :   0
 H 0 : p .5
 H 0 : p .5



H
:



H
:
p

.
5
0
 1
 H 1 : p  .5
 1
So if our hypotheses are H 0:  80 and H 1:  80 and p is the proportion that are over 80, so
that p 
41
 .6406 , they become H 0: p  .5 and H 1: p  .5 . pcv  p0  z  p . Since   .01 ,
64
p0 q0
.5.5

 .00390625  .0625 .
n
64
 p0  z  p  .5  2.327.0625  .6454. The ‘reject’ zone is above .6454. Since p is not in
we need a z.01  2.327 .  p 
pcv
the ‘reject’ zone, we cannot reject the null hypothesis. It may be more accurate to use
n
2x  1  n
n
, where the + applies if x  , and the  applies if x  . We would get
z
2
2
n
241  1  64
 2.125 . The p-value would be Pz  2.125   .5  .4832  .0168 . Since
64
  .01 is above the p-value, do not reject the null hypothesis.
But if our hypotheses are H 0:  80 and H 1:  80 and p is the proportion that are under 80,
z
so that p 
23
 .3594 , they become H 0: p  .5 and H 1: p  .5 . pcv  p0  z  p . Since
64
p0 q0
.5.5

 .00390625  .0625 .
n
64
 p0  z  p  .5  2.327.0625  .3546. The ‘reject’ zone is below .3546. Since p is not in
  .01 , we need a z.01  2.327 .  p 
pcv
the ‘reject’ zone, we cannot reject the null hypothesis. It may be more accurate to use
n
2x  1  n
n
, where the + applies if x  , and the  applies if x  . We would get
z
2
2
n
223   1  64
 2.125 . The p-value would be Pz  2.125   .5  .4832  .0168 . Since
64
  .01 is above the p-value, do not reject the null hypothesis.
z
j) (Extra credit) Use your data to get an approximate 99% 2-sided confidence interval for the median.
12
252y0711 2/22/07
In the outline, we were given a formula for k , k 
n  1  z .2 n
and the interval will be x k to
2
64  1  2.576 64
x n1k . Since   .01 , we need a z .005  2.576 . k 
 22 .196 . x 22 to x 43 is
2
78.75 to 87.44 for Version 0 and 79.75 to 88.34 for Version 9.
2.
Once again, assume that the Normal distribution applies, but assume a population standard
deviation of 10 (thousand) and that we are testing whether the mean is above 80 (thousand). (99%
confidence level)
a. State your null and alternative hypotheses (1)
b. Find a p-value for the null hypothesis using the mean that you found in 1. On the basis of your
p-value, would you reject the null hypothesis? Why? (1)
c. Create a power curve for the test. (6)
[20]
Solution: a) State your null and alternative hypotheses (1)
The problem statement is “Test to see if the average sales are now above 80000.” This translates
as   80 (in thousands), and is an alternative hypothesis because it contains no equality. Thus our
hypotheses are H 0 :   80 and H 1 :   80 . Because we suspect that the mean is above 80, this
is a right-side test.  0  80 and   .01 . We now know   10 , so  x 
For Version 0 x  82.50 and n  64 .
For Version 9 x  83.40 and n  64 .
The formula table gives the following.
Interval for
Confidence
Hypotheses
Interval
Mean (
known)
  x  z 2  x
x 

H0 :   0
H1 :    0
Test Ratio
z
x  0
x


n
10
 1.250
64
Critical Value
xcv   0  z 2  x
xcv   0  z 2  x
n
b) Find a p-value for the null hypothesis using the mean that you found in 1. On the basis of your
p-value, would you reject the null hypothesis? Why? (1)
x   0 82 .50  80

 2.00 and pvalue  Px  82.50   Pz  2.00 
For Version 0 z 
x
1.250
 Pz  0  P0  z  2.00   .5  .4772  .0228 . Since the p-value is above   .01 , do not reject
H0 .
For Version 9 z 
x  0
x

83 .40  80
 2.72 and pvalue  Px  83.40   Pz  2.72 
1.250
 Pz  0  P0  z  2.72   .5  .4967  .0033 . Since the p-value is below   .01 , reject H 0 .
13
252y0711 2/22/07
c) Create a power curve for the test. (6)
Remember that the hypotheses are H 0 :   80 and H 1 :   80 .  x  1.250 ,  0  80 ,
  .01 and this is a right-side test. Because this is a right-sided test we need a critical value
above 80, x cv   0  z  x  80  2.327 1.250   82.909 . We will not reject H 0 if the sample
mean is below 83.909.
Lay down a grid for 1 . Because this is a right-sided test, all values of 1 will be above 80, and
because the distance from 80 to the critical value is a little under 3, use an interval of half that, say
1.5. Our values of 1 will be 80, 81.5, 82.909, 86 and 89.
 1 80
82 .909  80 

P x  82 .909   80   P  z 
 Pz  2.327   .5  .4901  .9901  1  
1.250 

Power  1  .9901  .0099   . We really didn’t have to do this since we knew the
answer, but it served as a check.
 1 81.5
82 .909  81 .5 

Px  82 .909   81 .5  P  z 
  Pz  1.13   .5  .3708  .8708
1.250


Power  1  .8708  .1292
 1 82.909
82 .909  82 .909 

Px  82 .909   82 .909   P  z 
  Pz  0   .5 .
1.250


Power  1  .5  .5 Of course this was unnecessary too, since the power when the mean
is the critical value is always .5.
 1 86
82 .909  86 

Px  82 .909   86   P  z 
 Pz  2.47   .5  .4932  .0068
1.250 

Power  1  .0068  .9932
82 .909  89 

 Pz  4.87   .5  ..5000  0
 1 89 Px  82 .909   89   P  z 
1.250 

Power  1  0  1
Make a diagram: Your y-axis should go from zero to one and your x-axis from 80 to 90. It
should be reasonably accurate. See the Outline pages for a picture of the power curve.
14
252y0711 2/22/07
3.
Nationwide 16.5% of all CEOs have advanced degrees. You take a survey of a random sample of
160 CEOs in your city and find that 24  a  have advanced degrees, where a is the second to last
digit of your student number. (For example, Seymour Butz’s student number is 976512, so he will
subtract one and say that x  24  1  23 .) Label your solution ‘Version a ,’ where a is the
number that you are using. Is the fraction of executives that have advanced degrees in your city
significantly less than the national percentage?
a. Formulate your null and alternative hypotheses and do a hypothesis test with a 99% confidence
level. (2)
b. Find a p-value for the null hypothesis. (1)
c. (Extra credit) How would your answer to a) change if your sample of 160 came from a
population of 200? (1)
d. (Extra credit) Using a critical value of the proportion for testing your null hypothesis, create a
power curve for the test by using the alternate hypothesis and finding the power for values of
p1  .165 . (Up to 6 points)
e. Assume that the proportion of CEOs in your city is the observed proportion in part a, how large
a sample would you need to estimate the proportion above that have advanced degrees with an
error of .001? (2)
f. Use the proportion that you found in a) to create a 2-sided 99% confidence interval for the
proportion. Does it differ significantly from .165? Why? (2)
[28]
Solution: a) . Formulate your null and alternative hypotheses and do a hypothesis test with a 99%
confidence level. (2)
The problem asks “Is the fraction of executives that have advanced degrees in your city
significantly less than the national percentage?” This can be written as p  .165 , and because it
does not contain an equality must be an alternate hypothesis. We thus have H 0 : p  .165 and
H1 : p  .165 . Since we are concerned about the proportion being below .165, this is a left-sided
test. We are given p 0  .165 ,   .01 and n  160 . So q 0  1  p 0  .835 and
p0 q0
.165 .835 

 .00086109  0.029344 . z  z.01  2.327
n
160
The formula table says the following.
Interval for
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
p 
Proportion
p  p  z 2 s p
pq
n
q  1 p
sp 
Version 0: x  24 , p 
H 0 : p  p0
H1 : p  p0
z
p  p0
p
pcv  p0  z 2  p
p0 q0
n
q0  1  p0
p 
x 24

 .1500 , q  1  p  .8500 and
n 160
pq
.1500 .8500 

 .000796875  0.028229
n
160
x 15
 .09375 , q  1  p  .90625 and
Version 9: x  24  9  15 , p  
n 160
sp 
sp 
pq
.09375 .90625 

 .000531006  0.023044
n
160
15
252y0711 2/22/07
There are three ways to do this. Do only one of them.
p  p 0 .1500  .165

 0.511 and for Version 9
(i) Test Ratio: For Version 0 z 
p
0.029344
z
p  p0
.09375  .165

 2.428 . This is a left-sided test - the smaller the sample proportion
p
0.029344
is, the more negative will be this ratio. We will reject the null hypothesis if the ratio is smaller than
 z   z.01  2.327 . Make a diagram showing a Normal curve with a mean at 0 and a shaded
'reject' zone below -2.327. In Version 0 the test ratio is not below -2.327, so we cannot reject H 0 .
In Version 9 the test ratio is below -2.327, so we reject H 0 .
(ii) Critical value: We need a critical value for p below .165. Common sense says that if the sample
proportion is too far below .165, we will not believe H 0 : p  .165 . The formula for a critical
value for the sample mean is pcv  p0  z  x , but we want a single value below .165, so use
2
p cv  p 0  z  x  .165  2.327 .029344   .0967 . Make a diagram showing a Normal curve
with a mean at .165 and a shaded 'reject' zone below .0967. For Version 0 p  .1500 is not below
.0967, so we do not reject H 0 . For Version 9 p  .09375 is below .0967, so we reject H 0 .
(iii) Confidence interval: p  p  z  s p is the formula for a two sided interval. The rule for a one-sided
2
confidence interval is that it should always go in the same direction as the alternate hypothesis.
Since the alternative hypothesis is H1 : p  .165 , the confidence interval is p  p  z s p . For
Version 0 make a diagram showing an almost Normal curve with a mean at p  .1500 . The
confidence interval will be p  .1500  2.327 .028229   .2157 . To represent the confidence
interval, shade the area below .2157 in one direction. Then, on the same diagram, to represent the
null hypothesis, H 0 : p  .165 , shade the area above .165 in the opposite direction. Notice that
these overlap. What the diagram is telling you is that it is possible for p  .2157 and
H 0 : p  .165 to both be true. (If you follow my more recent suggestions, it is actually enough to
show that .165 is on the confidence interval.) So we do not reject H 0 .
For Version 9 make a diagram showing an almost Normal curve with a mean at p  ..9375 . The
confidence interval will be p  .09735  2.327 .023044   .1510 . To represent the confidence
interval, shade the area below .1510 in one direction. Then, on the same diagram, to represent the
null hypothesis, H 0 : p  .165 , shade the area above .165 in the opposite direction. Notice that
these do not overlap. What the diagram is telling you is that it is impossible for p  .1510 and
H 0 : p  .165 to both be true. (If you follow my more recent suggestions, it is actually enough to
show that .165 is not on the confidence interval.) So we reject H 0 .
b) Find a p-value for the null hypothesis. (1)
This is a left-sided test - the smaller the sample proportion is, the more negative will be the z-ratio.
p  p 0 .1500  .165

 0.511 pvalue  P p  .1500   Pz  0.511 
For Version 0 z 
p
0.029344
 .5  .1950  .3050 .
p  p 0 .09375  .165

 2.428 . pvalue  P p  .09375   Pz  2.428 
For Version 9 z 
p
0.029344
 .5  .4925  .0075
We will reject the null hypothesis if the p-value is smaller than   .01 .
16
252y0711 2/22/07
c) (Extra credit) How would your answer to a) change if your sample of 160 came from a population of
200? (1)
p 0  .165 ,   .01 , n  160 and N  200 . So q 0  1  p 0  .835 and
N  n p0 q0
200  160 .165 .835 

 .000173084  0.013156 . z  z.01  2.327 .
N 1
n
200  1
160
It’s probably easiest to use a critical value. p cv  p 0  z  x  .165  2.327 .013156   .1344 .
Make a diagram showing a Normal curve with a mean at .165 and a shaded 'reject' zone below
.1344. For Version 0 p  .1500 is still not below .1344, so we do not reject H 0 . For Version 9
p  .09375 is below .1344, so we reject H 0 . In general p-values will be smaller and rejection
more likely with this finite population correction.
p 
d) (Extra credit) Using a critical value of the proportion for testing your null hypothesis, create a power
curve for the test by using the alternate hypothesis and finding the power for values of p1  .165 . (Up to 6
points)
Remember H1 : p  .165 , so we will consider possible means below .165. In a) we rejected
H 0 : p  .165 if the observed proportion was below p cv  p 0  z  x
 .165  2.327 .029344   .0967 .
.165 - .0967 is roughly equal to .07, and half of .07 is .035. The operating characteristic is the
probability that we will not reject the null hypothesis for various levels of p1 . We can write




p  p1 
and with a grid of .035, we will evaluate it at .165,
  P p  .0967 p  p1   P  z  cv

p1 q1 


n 

.130, .0967, .06 and .025. Remember n  160
p1  .165
p 
.165 .835 
 .00086109  0.029344
160
.0967  .165 

P  p  .0967 p  .165   P  z 
 Pz  2.33   .5  .4901  .9901  1  
.029344 

p1  .130
p 
.130 .870 
 .000706875  .02659
160
.0967  .130 

P p  .0967 p  .130   P  z 
 Pz  0.12   .5  .0478  .5478
.02659 

p1  .0967
P p  .0967 p  .0967  .5
p1  .06
p 
.06 .94 
 .0003525  .01877
160
.0967  .06 

P  p  .0967 p  .06   P  z 
 Pz  1.95   .5  .4744  .0256
.01877 

p1  .025
p 
.025 .975 
 .000152344  .01234
160
.0967  .025 

P  p  .0967 p  .025   P  z 
 Pz  3.81  .5  .4999  .0001
.01877 

17
252y0711 2/22/07
Make a diagram: Your y-axis should go from zero to one and your x-axis from 0 to .165. It
should be reasonably accurate. See the Outline pages for a picture of the power curve for a test for
a mean. This should look somewhat different.
e). Assume that the proportion of CEOs in your city is the observed proportion in part a, how large a
sample would you need to estimate the proportion above that have advanced degrees with an error of .001?
(2)
The usually suggested formula is n 
pqz 2
e2
. This is the formula everyone forgets that we
covered. z  z.01  2.327 . e  .001 .
Version 0: p 
x 24
.1500 .8500 2.327 2

 .1500 , q  1  p  .8500 and n 
 690404
n 160
.001 2
Version 9: x  24  9  15 , p 
n
.09375 .90625 2.327 2
.001 2
x 15

 .09375 , q  1  p  .90625 and
n 160
 460058
f) Use the proportion that you found in a) to create a 2-sided 99% confidence interval for the proportion.
Does it differ significantly from .165? Why? (2)
[28]
z 2  z.005  2.576
Version 0: x  24 , p 
x 24

 .1500 , q  1  p  .8500 and
n 160
pq
.1500 .8500 

 .000796875  0.028229
n
160
p  p  z s p  .1500  2.576.028229  .150  .073 or .077 to .223
sp 
2
This interval includes .165 so the results of Version 0 are not significantly
different from .165.
x 15
 .09375 , q  1  p  .90625 and
Version 9: x  24  9  15 , p  
n 160
pq
.09375 .90625 

 .000531006  0.023044
n
160
p  p  z s p  .09375 2.576.023044  .094  .059 or .035 to .153.
sp 
2
This interval does not include .165 so the results of Version 0 are significantly
different from .165.
18
252y0711 2/22/07
4.
Standard deviation is often a measure of reliability. A manufacturer is claiming that the equipment
in use to fill bottles fills the bottles so that the standard deviation is 0.1 oz or less. You take a
sample of 20 bottles and get a sample standard deviation of 0.110  a . To get a take the third to
last digit of your student number and multiply it by 0.001. (For example, Seymour Butz’s student
number is 976502, so he will add .005 and say that s  0.110  0.005  0.115 . ) Label your
solution Version a .
a. Formulate the null and alternative hypotheses necessary to see if the standard deviation is not
more than 0.1 and test the hypothesis using a 99% confidence level and a test ratio. (2)
b. What assumptions are necessary to perform this test? (1)
c. Try to get a rough p-value. Interpret its meaning (1)
d. Do a 99% two- sided confidence interval for the standard deviation (1)
e. (Extra credit) Redo 4a) using an appropriate confidence interval. (2)
f. (Extra credit) Find a critical value for s in 4a). (1)
g. Do a. again assuming that you took a sample of 110 bottles. (2)
h. A machine supposedly requires adjustment at most once a month. Last month, however it
needed to be adjusted 3 times. Assuming that the Poisson distribution applies and using a 5%
significance level, is there something wrong? (2)
i. Over a period of 5 years the average number of accidents per year on a stretch of highway was
30. The speed limit was reduced to 45 mph and there were only 4 accidents in a 3-month period.
Use the Poisson distribution to test if there has been a significant reduction in accidents. Use a 5%
significance level. (2)
[42]
Solution: n  20 ,   .01 , Version 0 has s  .110 and Version 9 has s  .110  .009  .119
a) Formulate the null and alternative hypotheses necessary to see if the standard deviation is not more than
0.1 and test the hypothesis using a 99% confidence level and a test ratio. (2)
The formula table has the following.
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 :  2   02
 z 2  2DF 
2 
n  1s 2
 02
z 
2   2DF   1
2
H1 :  2   02
2
s cv

s cv 
 .25 .5 2  02
n 1
 2 DF
 z  2  2 DF
H0 :   0.1 and H1 :   0.1 . Since we are concerned about the proportion being above .165,
this is a right-sided test. The test ratio  2 
n  1s 2
 02
 
would be tested against  .2995n 1 and
n 1
 
2 19
 .2005
in a two sided test. In a right sided test we only test against  .201n 1 .  .01  36 .1907 .
Version 0:  2 
Version 9:
n  1s 2

19 .110 2
.12
n  1s 2  19.119 2
2 
 02
.12
 02
 22 .9900
 26 .9059
Make a diagram. Show a curve skewed to the left with a mean at df  19 . Shade the ‘reject’ zone
2 19
above  .01  36 .1907 . Since the test ratio is below 36.1907 for all versions, do not reject the
null hypothesis.
b) What assumptions are necessary to perform this test? (1)
This test assumes that the parent population is Normal.
19
252y0711 2/22/07
c) Try to get a rough p-value. Interpret its meaning (1)
The 19 df column of the chi-squared table is shown below as two lines.

0.005 0.010 0.025 0.050 0.100 0.900 0.950
0.975 0.990 0.995
2 19
38.5820 36.1907 32.8523 30.1435 27.2036 11.6509 10.1170 8.9065 7.6327 6.8440

Version 0:  2  22.9900
Version 9:  2  26.9059
Note that both of these fall between  .21019  27 .2036 and  .29019  11 .6509 . So we could say
.10  pvalue  .90 . Actually the median is about df  13  19  .3333  18 .6667 . Since the median
is  .25019 , we could say .50  pvalue  .90
d) Do a 99% two- sided confidence interval for the standard deviation (1)
19
19
 6.8440 . According to the outline
 .2005
 38 .5820 and  .2995
n  1s 2
 22
 2 
n  1s 2 .
2
 1 2
Version 0 has s  .110 and Version 9 has s  .110  .009  .119
19 .110 2
So for version 0
19 .110 2
 2 
38 .5820
square roots 0.0772    0.1833
So for version 9
19 .119 2
6.8440
 2 
19 .119 2
38 .5820
square roots 0.0835    0.1983 .
6.8440
or 0.005959   2  .033591 . So, if we take
or 0.006974   2  0.039313 . So, if we take
e) (Extra credit) Redo 4a) using an appropriate confidence interval. (2)
H0 :   0.1 and H1 :   0.1 . Since we are concerned about the proportion being above .165,
this is a right-sided test. The confidence interval should be in the same direction as the alternate
 
hypothesis and include the sample variance or standard deviation.  .20119  36 .1907 and
 .29919  7.6327 . Rule out the .99 value because it will produce a number above the sample
standard deviation. The one-sided interval is  2 
19 .110 2
So for version 0  2 
36 .1907
n  1s 2 .
 2
or  2  .006352 . So, if we take square roots   0.0797 .
19 .119 2
or  2  .007434 . So, if we take square roots   0.0862 .
36 .1907
f) (Extra credit) Find a critical value for s in 4a). (1)
So for version 9  2 

 .25 .5 2  02
 22  02
 12 2  02
or
and
.


n 1
n 1
n 1
Since our alternate hypothesis is H1 :   0.1 , we want a single critical value above 0.1. This
The formula for a 2-sided interval is
2
s cv
2
s cv
2
s cv
36 .1907 0.12
 0.01900 or
n 1
19
s cv  0.1378 . Our ‘reject’ zone is above 0.1378. Since Version 0 has s  .110 and Version 9 has
s  .119 , we cannot reject the null hypothesis for either.
2
would be s cv

 2  02
 
2
. We will use  .20119  36 .1907 and s cv

20
252y0711 2/22/07
g) Do a) again assuming that you took a sample of 110 bottles. (2)
H0 :   0.1 and H1 :   0.1
n  110 ,   .01 , Version 0 has s  .110 and Version 9 has s  .110  .009  .119 . The test ratios
are still  2 
2 
n  1s 2
 02
must use z 
n  1s 2
 02


109 .110 2
109 .119 2
0.12
 131 .89 for Version 0 and
 154 .35 . Since we have no table for 109 degrees of freedom, we
0.12
2  2  2df   1 . Since this is a 1% right sided test, the ‘reject zone is the area
under the Normal curve above z .01  2.327 .
For Version 0 z  2139 .89   2109   1  279 .78  217  16 .7266  14 .7309  1.996 . This
is below 2.327, so we do not reject the null hypothesis.
For Version 9 z  2154 .35   2109   1  308 .70  217  17 .5699  14 .7309  2.839 . This
is above 2.327, so we reject the null hypothesis.
h) A machine supposedly requires adjustment at most once a month. Last month, however it needed to be
adjusted 3 times. Assuming that the Poisson distribution applies and using a 5% significance level, is there
something wrong? (2)
For the Poisson distribution with a mean of 1, Px  3  1  Px  2  1  .91970  .0803
This would only result in a rejection of the null hypothesis H 0 : mean  1 if we use a 10%
significance level. We cannot reject the null hypothesis if we use a significance level of 5% or 1%.
i) Over a period of 5 years the average number of accidents per year on a stretch of highway was 30. The
speed limit was reduced to 45 mph and there were only 4 accidents in a 3-month period. Use the Poisson
distribution to test if there has been a significant reduction in accidents. Use a 5% significance level. (2)
If the mean was 30 for a year, it should be 7.5 for a 3-month period. The Poisson table says
Px  3  .05915 . Since this is not quite below 5%, we cannot reject the null hypothesis.
However, if we can keep accidents down to 8 in a 6-month period, the mean will be 15 and
Px  8  .03745 will result in a rejection of H 0 : yearly mean  30.
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