252y0612 10/18/06 (Open in ‘Print Layout’ format)
ECO252 QBA2
FIRST HOUR EXAM
October 17-18 2005
Version 2
Name _____KEY___________
Hour of class registered _____
Class attended if different ____
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 7, 7.5
7  7
6  7
z
 P 0.13  z  0  .0517
1. P6.00  x  7.00   P 
7.5 
 7.5
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 zero and 0.15. Because this is completely on the left of zero and touches zero, we
can simply look up our answer on the standardized Normal table. If you wish, make a completely separate
diagram for x . Draw a Normal curve with a mean at 7. Indicate the mean by a vertical line! Shade the
area between 6 and 7. This area includes the mean (7), but does not include any points to the right of the
mean, so that we neither add nor subtract.
10 .22  7 

 Pz  0.43   Pz  0  P0  z  0.43  .5  .1664  .6664
2. Px  10.22   P  z 
7.5 

For z make a diagram. Draw a Normal curve with a mean at 0. Shade the entire area below 0.43.
Because this is on both sides of zero, we must add. If you wish, make a completely separate diagram for x .
Draw a Normal curve with a mean at 7. Shade the entire area below 10.22. Since the area is on both sides
of the mean, we add.
0  7
  17 .5  7
z
 P 3.27  z  0.93 
3. P17.5  x  0  P 
7.5 
 7.5
 P3.27  z  0  P0.93  z  0  .4995  .3238  .1757
For z make a diagram. Draw a Normal curve with a mean at 0. Shade the area between -3.27 and -0.93.
Because this is completely on the left of zero, we must subtract. If you wish, make a completely separate
diagram for x . Draw a Normal curve with a mean at 7. Shade the area between -17.5 and 0. Note that zero
is below the mean. These numbers are both to the left of the mean (7), so we subtract.
4. x.04 (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.04 is the value of z with 4% of the distribution above it. Since 100 – 4 = 96, it is also the 96th
percentile. Since 50% of the standardized Normal distribution is below zero, your diagram should show
that the probability between z.04 and zero is 96% - 50% = 46% or P0  z  z.04   .4600 . The closest we
can come to this is P0  z  1.75   .4599 . (1.76 is also acceptable here.) So z.04  1.75 . To get from z.04
x
. x  7  1.757.5  20.125 . If

you wish, make a completely separate diagram for x . Draw a Normal curve with a mean at 7. Show that
50% of the distribution is below the mean (7). If 4% of the distribution is above z.04 , it must be above the
mean and have 46% of the distribution between it and the mean.
to x.04 , use the formula x    z , which is the opposite of z 
20 .125  7 

Check: Px  20.125   P  z 
  Pz  1.75   Pz  0  P0  z  1.75 
7.5


 .5  .4599  .0401  4%
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II. (5 points-2 point penalty for not trying part a.) In order not to violate the truth in labeling law a teabag
must contain at least 5.5 oz of tea. A sample of 9 items is taken from a large number of tea bags. The data
below is found. (Recomputing what I’ve done for you is a great way to waste time.) b) and d) require
statistical tests.
a. Compute the sample standard deviation, s , of the waiting times. Show your work! (2)
b. Is the population mean significantly below 5.5 (Use a 95% confidence level)? Show your
work! (3)
[13]
c. (Extra Credit) Find an approximate p value for your null hypothesis. (2)
d. Assume that the population standard deviation is 0.10 and create a 94% confidence interval for
the mean. (2)
e. (Extra Credit) Given the data, is 0.10 a reasonable value for the population standard deviation?
Original data
Solution:
x 2  266 .4394 , n  9
x  48 .96 ,
x
Row
a)
x2
1
2
3
4
5
6
7
8
9


 x  48.96  5.4400
x
5.35 28.6225
5.68 32.2624
5.35 28.6225
5.47 29.9209
5.31 28.1961
5.49 30.1401
5.41 29.2681
5.48 30.0304
5.42 29.3764
48.96 266.4394
n
sx2 

x
9
2
 nx 2
n 1

266 .4394  95.4400 2
8
0.0965
 0.0120625
8
sx  .0120625  0.109828 .
b) This material is edited from the solution to problem 9.54. Given: 0  5.5, s  0.109828 n  9,
s
df  n  1  8, x  5.4400 and   .05. So s x 
n

0.109828
9
 0.03661 . Note that
tn 1  t..805  1.860 .
Note that ‘Below 5.5’ does not contain an equality, so that it must be an alternative hypothesis. Our
hypotheses are H 0 :   5.5 and H1 :   5.5 . There are three ways to do this. You should have used one.
x  0 5.4400  5.5

 1.639 . The smaller the sample mean is, the more
sx
0.03661
negative will be this ratio. We will reject the null hypothesis if the ratio is smaller than
 tn1  t..805  1.860 . Make a diagram showing a Normal curve with a mean at 0 and a
shaded 'reject' zone below -1.860. Since the test ratio is not below -1.860, we do not reject H 0 .
(ii) Critical value: We need a critical value for x below 5.5. Common sense says that if the
sample mean is too far below 5.5, we will not believe H 0 :   5.5 . The formula for a critical
value for the sample mean is x    t n1 s , but we want a single value below 5.5, so use
(i) Test Ratio: t 
cv
0

2
x
xcv  0  tn 1sx  5.5  1.860 0.03661   5.4319 . Make a diagram showing an almost
Normal curve with a mean at 5.5 and a shaded 'reject' zone below 5.4319. Since x  5.4400 is
not below 5.4319, we do not reject H 0 .
(iii) Confidence interval:   x  t sx is the formula for a two sided interval. The rule for a one2
sided confidence interval is that it should always go in the same direction as the alternate
hypothesis. Since the alternative hypothesis is H1 :   5.5 , the confidence interval is
  x  t n1s  5.4400  1.860 0.03661   5.5081 . Make a diagram showing an almost

x
Normal curve with a mean at x  5.4400 and, to represent the confidence interval, shade the area
below 5.5081 in one direction. Then, on the same diagram, to represent the null hypothesis,
H 0 :   5.5 , shade the area above 5.5 in the opposite direction. Notice that these do overlap.
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What the diagram is telling you is that it is possible for   5.5081 and H 0 :   5.5 to both be
true. (If you follow my more recent suggestions, it is actually enough to show that 5.5 is on the
interval.) So we do not reject H 0 .
c) (Extra credit) If you wish to use a p-value approach, since this is a left-side test, the p-value is the
probability of getting a value below 5.4067 when the null hypothesis is true. If we use the test ratio above
we get. pval  Px  5.4400   Pt  1.639  . To find this approximately, go to the df  8 line of the t
table. 1.639 is between t 8  1.860 and t 8  1.397 . This means that P t 8  1.860  .05 and


..05

.. 10

P t 8  1.397  .10 . Since -1.639 is between these values, we can conclude that 05  pval  .10. If we
want to use this in b), we can say that since the p-value is not below   .05, we do not reject the null
hypothesis.
Note that this is confirmed by Minitab, with more accurate computations. The column that you are using is
t1, its mean is 5.40667, a confidence interval is   5.48239 , the t-ratio is equal to -2.29 and it gives a pvalue of .026.
One-Sample T: t2
Test of mu = 5.5 vs < 5.5
Variable
t1
N
9
Mean
5.44000
StDev
0.11011
SE Mean
0.03670
95%
Upper
Bound
5.50825
T
-1.63
P
0.070
d) In this section, we assume that the population standard deviation is 0.10 and create a 92% confidence
interval for the mean. Here   1  .92  .08. We do this by observing that the formula for a 2-sided
confidence interval when the population standard deviation is known is   x  z  x . Fortunately, we
2

0.10

 0.0333 . The confidence interval is thus
n
9
  5.4400  1.750.0333   5.4400  0.0583 or 5.38 to 5.50.
e) (Extra Credit) Given the data, we want to see if 0.10 a reasonable value for the population standard
deviation? H 0 :   0.10 and H1 :   0.10 . We will assume   .05 and use the chi – squared formula,
found out on Page 1 that z .04  1.75 .  x 
2 
n  1s 2
 02

80.0120625 
2
8
8
 9.684 . We test this against  2.975  2.1797 and  2..025  17.5346.
.10
Since it is between them we cannot 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
18+ 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.
A librarian provides a confidence interval estimate for the mean number of books checked out
daily. The estimate is 330 to 740. The point estimate that this interval is based on is.
a) 1070.
b) 740
c) *535
d) 330
e) None of the above
f) There is not enough information to tell.
Explanation: Since the sample mean sits in the middle of a confidence interval, it must be an
average of the two end points.
3.
(BLK8.30) We want a 99% confidence interval for the average income of in a town, and the
population standard deviation is known to be $1000. We have taken a sample of size 50 earlier and
found that the sample mean is $20000. What sample size should we take if the width of the
interval is to be no more than $100?
a) *2655
b) 2654
c) 665
d) 664
e) 50
f) 40
g) 20
h) None of the above.
z 2 2 2.576 2 1000 2
Explanation: The formula in the outline is n  2 
 2654 .3. This is always
e
50 2
rounded up – in this case to 2655.

4.

An entrepreneur is considering the purchase of a coin-operated laundry. The present owned claims
that over the past 6 years the average daily income was $700. A sample is taken of daily revenue
over a period of 28 days. Statistics are computed from the sample. If we want to test the statement
that the mean is $700, which of the following tests is most appropriate? (1 point changed to 2)
a) z -test of a population mean.
b) z -test of a population proportion.
c) * t -test of a population mean.
d)  2 -test of a population variance.
e) F -test.
f) All of the above could be used.
g) We do not have enough information.
[7]
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5.
We are trying to estimate the median income in a region. We wish to test if the median is over
$30000. We do not know the population variance. We can compute a statistic or statistics from a
sample of 27 incomes. To do this test, the statistic or statistics we need is (are)
a) The proportion of incomes that are above the sample mean, x .
b) The proportion of incomes that are above the sample median x.50
c) *The number of incomes in the sample that are above $30000.
d) The sample mean, x .
e) The sample mean, x and the sample variance s 2 .
f) The sample median x.50 .
g) The sample variance s 2 .
[9]
6.
A restaurant company only opens restaurants in areas with a mean family income above $45000.
The company takes a sample of 144 families in Hotzeplotz and finds a sample mean income of
$46396 and a sample standard deviation of $5400. Should they open?   .05 
a) State your null and alternative hypotheses. (1)
b) Test your hypothesis in a) by finding a critical value for the sample mean. Can we say
that the result of the sample is above $45000? Why? (3)
c) Do a 2-sided confidence interval for the mean. (2)
d) Do a 2-sided confidence interval for the variance. (2)
[17]
e) (Extra credit) Explain in as much detail as reasonable how you would find a confidence
interval for the median. (2)
Solution: This material is Part II all over again.
a) Note that ‘above $45000’ does not contain an equality, so that it must be an alternative hypothesis. Our
hypotheses are H 0 :   45000 and H1 :   45000 .
Given: 0  45000 s  5400 , n  144 , df  n  1  143 , x  46396 and   .05 . So
5400

 450 . tn 1  t..143
05  1.656 .
12
n
144
b) We need a critical value for x above 45000. Common sense says that if the sample mean is too far
above 45000, we will not believe H 0 :   45000 . The formula for a critical value for the sample mean is
x    t n1 s , but we want a single value above 45000, so use x    t n1s
s
sx 
cv

5400

0

x
2
cv
0

x
 45000  1.656 450   45745 .20 . Make a diagram showing an almost Normal curve with a mean at
45000 and a shaded 'reject' zone above 45745.20. Since x  46396 is above 45745.20, we reject H 0 .
c) Confidence interval for the mean:   x  t sx is the formula for a two sided interval. We need
2
143
t.025
 1.977 .   45000  1.977 450   45000  899 .65 . We can say that
P44110 .35    45899 .65   .95 .
d) Confidence interval for the variance: The outline says that for small samples the formula is
n  1s 2
 2
2
 2 
s 2DF 
z 2  2DF 
Since
n  1s 2
12 2
 
, but if the degrees of freedom are too large for the chi-square table use
s 2DF 
 z 2  2DF 
. Using the second formula
2143   286  16 .912 , we have
5400 2143 
1.96  2143 
 
5400 2143 
 1.96  2143 
.
5400 16.912 
5400 16 .912 
, which becomes
 
1.96  16.912 
 1.96  16.912 
5400 16 .912 
5400 16 .912 
 
and finally 4839 .18    6107 .87 .
18 .872
14 .952
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As I have said in class, no one really wants a confidence interval for the variance, but if you square the
above you get 23417558   2  37306017
e) Confidence interval for the median: If we remember n  144 ,
n  1  z . 2 n
we can use the formula given in the
144  1  1.960 144
 60 .74 . To be conservative, round
2
2
this down to 60. Put the numbers in order by size as x1 to x144 . The interval will be x60 to x85 . (Note that
the index of the second number is 144 + 1 – 60 = 85.
outline for large samples. k 
7.

Return to the problem in Question 6. Let’s say that the sample mean payment of $46396 and the
sample standard deviation of $5400 come from a sample of 81. Assume that your null hypothesis
is correct and that you get a t-ratio of 2.327. What should you say that the p-value is? (3 points).
Note that showing your calculations here could get you partial credit.
a) Exactly .01
b) Exactly .02
c) *Between .025 and .01
d) ***Between .05 and .02
e) Between .01 and .005
f) Between .02 and .01
g) Exactly .99
h) Exactly .98
i) **Between .975 and .99
j) Between .95 and .98
k) Between .98 and .99
l) None of the above – Show your answer!
[20]

80
Solution: If you have 80 degrees of freedom, the t-table says t..80
025  1.990 , t..01  2.374 ,


80
80
t..005  2.639 , and t..001  3.195 . This means that Pt  1.990   .025 and Pt  2.374   .01 . Since
2.327 lies between these values, .01  Pt  2.327   .025
*If you said that your hypotheses were H 0 :   45000 and H1 :   45000 , you have a right
sided test and pvalue  Pt  2.327  , which will be between .01 and .025.
**However, if you said that your hypotheses were H 0 :   45000 and H1 :   45000 , you
have a left-sided test and pvalue  Pt  2.327  , which will be between .975 and .99.
***If you said that your hypotheses were H 0 :   45000 and H1 :   45000 , you have a twosided test and pvalue  2Pt  2.327  , which will be between .02 and .05.
I can only promise to look at your reasoning if you have any other answer.
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8.
(Mann) According to a 1992 survey, 45% of workers say that they would change careers if they
could. You wish to show that the proportion of workers in your company that want to change
careers is below the national figure. Find the correct set of hypotheses below.
a)
 H 0: p  .45

 H1 : p  .45
b)
 H 0: p  .45

 H1 : p  .45
c)
 H 0: p  .45

 H1 : p  .45
d)
 H : p  .45
* 0
 H1 : p  .45
e)
 H 0: p  .45
*
 H1 : p  .45
f)
 H 0: p  .45
**

 H1 : p  .45
g)
 H 0: p  .45
**

 H1 : p  .45
h) None of the above. Put in your answer!
9.
[22]
(Mann) According to a 1992 survey, 45% of workers say that they would change careers if they
could. You wish to show that the proportion of workers in your union that want to change careers
is below the national figure. Assume that your null hypothesis in 8 is correct, that you take a
sample of 350 workers and that you compute the ratio z 
90%, you should do the following.
a)
b)
c)
d)
e)
f)
g)
h)
p  .45
. If your confidence level is
.45.55 
350
Reject the null hypothesis if the ratio is not between -1.96 and 1.96.
Reject the null hypothesis if the ratio is not between -1.645 and 1.645
Reject the null hypothesis if the ratio is above 1.645
Reject the null hypothesis if the ratio below -1.645
Reject the null hypothesis if the ratio is below 1.282
**Reject the null hypothesis if the ratio is above 1.282
*Reject the null hypothesis if the ratio is below -1.282
None of the above. Put in your answer!
[24]
Solution: This is clearly a one-sided test and   .10 . The table says z .10  1.282 .
 H 0: p  .45
**If you said (wrongly) 
, you have a right-sided test and you reject the null hypothesis if
 H1 : p  .45
z  1.282 .
 H 0: p  .45
*However, if you said 
, you have a left-sided test and you would reject the null hypothesis if
 H1 : p  .45
z  1.282 . Note that for a t or z ratio, zero is never in the reject region, so that e) could never be true.
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10. We wish to use a 2-sided confidence interval to test a proportion. The following things may
influence the size of a confidence interval: (.5 points for good ones, .5 off for bad ones, at worst
zero)
1. Decreasing x to make it closer to .5n
2. Increasing x from .85 n to .9n
3. Changing the null hypothesis to make p 0 closer to .5
4. Increasing the sample size. (Assume a large population)
5. Decreasing the sample size. . (Assume a large population)
6. Increasing the confidence level
7. Increasing the significance level
8. Using a continuity correction with a relatively small sample.
9. Decreasing the population size from 15 n to 14 n
10. Increasing the population size from 5n to 6 n
Put down the numbers of the things that make the confidence interval smaller. ________________
Solution: The confidence interval has the following form: p  p  z  s p , where p 
2
x
and, if we include
n
N  n pq
.
N 1 n
1. Decreasing x to make it closer to .5n . This makes p closer to .5. It was explained in class that
pq becomes larger as p approaches .5. This makes the interval larger.
2. Increasing x from .85 n to .9n . This makes p farther from .5 and makes the interval smaller.
a finite sample correction, the standard error is s p 
3. Changing the null hypothesis to make p 0 closer to .5. p 0 does not appear in the formula for a
confidence interval, so it has no effect.
4. Increasing the sample size. (Assume a large population.) Since n , the sample size, is in the
pq
, it makes the interval smaller.
n
5. Decreasing the sample size. (Assume a large population.) This makes the interval larger.
6. Increasing the confidence level. If we raise the confidence level 1    , the significance level
denominator of s p 
  gets smaller. This gives us a larger value of
z 2 , so it makes the interval larger.
7. Increasing significance level . This makes the interval smaller.
8. Using a continuity correction. This has the effect of making all intervals slightly larger. It has
little effect with a large sample.
9. Decreasing the population size from 15 n to 14 n . If we use a finite population correction, it has
the general effect of decreasing the standard error. Decreasing the population size makes its effects
stronger, so it makes the interval smaller.
10. Increasing the population size from 5n to 6 n . Let the size of the population be an , where a
is a value from 1 to 20. Then the finite population correction is
N n

N 1
an  n
a 1

. This
n 1
1 1
n
obviously grows as a grows. This makes the interval larger.
Summarizing:
Larger: 1, 5, 6, 8, 10
Smaller: 2, 4, 7, 9
No effect: 3
8
252y0612 10/18/06 (Open in ‘Print Layout’ format)
ECO252 QBA2
FIRST EXAM
October 11-12 2006
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 from now on neatness means paper neatly
trimmed on the left side if it has been torn and paper written on only one side.
1. (Moore, Notz) You are thinking that it may be desirable to start a wellness program for your
(large) company. You are told that the company will only start such a program if you can show
that the blood pressure of a group of mid-level executives is above normal. The individuals are all
between 35 and 44 years old and US statistics show that mean systolic blood pressure for men in
that age range is 128. You take a sample of 72 executives and get the following results.
x
139.60
136.21
114.18
128.45
120.68
127.51
128.07
161.43
161.09
130.61
130.15
154.74
126.29
124.48
117.51
138.22
169.05
116.14
182.53
141.96
122.14
163.18
124.61
100.03
130.77
140.35
158.74
120.02
137.23
127.22
141.54
105.67
149.55
109.52
131.40
126.54
118.77
141.15
150.30
126.93
144.71
127.32
136.69
125.06
135.21
149.44
133.89
118.37
124.80
133.00
131.74
135.69
169.61
126.71
107.30
122.73
125.35
152.64
109.62
116.59
132.00
117.84
120.01
117.47
145.25
159.94
112.34
145.10
119.39
127.67
117.97
112.40
To personalize the data below take the last digit of your student number, divide it by 10 and add it to the
numbers below. 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 139.80, 130.81, 182.73 etc. – but see
the hint below, you do not need to write down all the numbers that you are using, just your computations.)
x  9528 .41
Hint - if you use the computational formula: For the original numbers n  72 ,
and
x

2
 1280763 .7 . 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 fairly near 132 and your sample standard deviation should
be near 16 or 17.)
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, will you get a wellness program? Why? (1)
h. How do your conclusions change if the sample of 72 is taken from a population of 200? (2)
i. Assume that the Normal distribution does not apply and, using your data, test that the median is
above 128. (3)
[12]
j. (Extra credit) Use your data to an approximate 90% 2-sided confidence interval for the median.
9
252y0612 10/18/06 (Open in ‘Print Layout’ format)
2.
Once again, assume that the Normal distribution applies, but assume a population standard
deviation of 16 and that we are testing whether the mean is above 128. (90% 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 a. 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]
3.
(Moore, Notz) A recent survey said that nationwide 73% of all freshman students identified being
well-off financially as an important lifetime goal. You believe that the proportion of freshman
business majors with that goal is higher than the national figure. You take a survey of a random
sample of 200 students and find that 152  a  have being well-off as an important goal, where a
is the second to last digit of your student number. If the second to last digit of your student number
is zero, a  10 . (For example, Seymour Butz’s student number is 976502, so he will add 10 and
say that x  152  10  162 )
a. Formulate your null and alternative hypotheses and do a hypothesis test with a 95% confidence
level. (2)
b. Find a p-value for the null hypothesis. (1)
c. Find the p-value for the null hypothesis if x  162  a (1)
d(c). (Extra credit) How would your answer to a) change if your sample of 200 came from a
population of 300? (1)
e(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  .73 . (Up to 6 points)
f(d). Assume that p  .73 , how large a sample would you need to estimate the proportion above
that have being well-off with an error of .005? (2)
g(e). Use the proportion that you found in a) to create a 2-sided 95% confidence interval for the
proportion. Does it differ significantly from .73? Why? (2)
[28]
4.
Standard deviation is often a measure of reliability. A manufacturer is providing a connector with
a mean length of 2.5 mm and is getting complaints that the connector is often too large or too
small for the intended use. The previous standard deviation to the length of the part was 0.025mm,
but the manufacturer introduces a process that should make the standard deviation smaller. A
sample of 25 items is taken which yields a sample standard deviation of 0.030  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 subtract .005 and say that s  0.030  0.005  0.025 . )
a. Formulate the null and alternative hypotheses necessary to see if the goal has been achieved and
test the hypothesis using a 95% 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 (1.5)
d(c). Do a 95% two- sided confidence interval for the standard deviation (1)
e(d). (Extra credit) Redo 4a) using an appropriate confidence interval. (2)
f(e). (Extra credit) Find a critical value for s in 4a). (1)
g(f). The number of claims for missing baggage in a large metropolitan airport supposedly
follows a Poisson distribution with a mean of 72 per week. Assume that in a given week 92 are
lost. Test this hypothesis using a test ratio and a 95% confidence level. (2)
h(g). Find approximate critical values for the number of bags that could be lost in 4f. (2)
i(h). (Extra credit) Find the power of the test in 4g) if the average number of lost bags per week is
really 87. (3)
j(i). I claim that x is binomially distributed with p  .01 . Test this assertion using a 2-sided test
if there are 4 successes in 10 trials. (2)
[39.5]
10
252y0612 10/18/06 (Open in ‘Print Layout’ format)
1.
(Moore, Notz) You are thinking that it may be desirable to start a wellness program for your
(large) company. You are told that the company will only start such a program if you can show
that the blood pressure of a group of mid-level executives is above normal. The individuals are all
between 35 and 44 years old and US statistics show that mean systolic blood pressure for men in
that age range is 128. You take a sample of 72 executives and get the following results.
x
139.60
136.21
114.18
128.45
120.68
127.51
128.07
161.43
161.09
130.61
130.15
154.74
126.29
124.48
117.51
138.22
169.05
116.14
182.53
141.96
122.14
163.18
124.61
100.03
130.77
140.35
158.74
120.02
137.23
127.22
141.54
105.67
149.55
109.52
131.40
126.54
118.77
141.15
150.30
126.93
144.71
127.32
136.69
125.06
135.21
149.44
133.89
118.37
124.80
133.00
131.74
135.69
169.61
126.71
107.30
122.73
125.35
152.64
109.62
116.59
132.00
117.84
120.01
117.47
145.25
159.94
112.34
145.10
119.39
127.67
117.97
112.40
To personalize the data below take the last digit of your student number, divide it by 10 and add it to the
numbers below. 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 139.80, 130.81, 182.73 etc. – but see
the hint below, you do not need to write down all the numbers that you are using, just your computations.)
x  9528 .41
Hint - if you use the computational formula: For the original numbers n  72 ,
and
x

 1280763 .7 . If you add a quantity a to a column of numbers,
2
 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 fairly near 132 and your sample standard deviation should
be near 16 or 17.)
Solution: For version 0, the mean is 132.339. For version 10, the mean is 133.339. All
computations for the sample variance and the sample standard deviation should give identical
answers. The solution for version 0 follows.   .01 .
 x  9528 .41 ,  x
2
 1280763 .7 , n  72 . This means x 
 x  9528 .41  132 .339 .
n
72
Using the computational formula, we get the following.
x 2  nx 2 1280763 .7  72132 .339 2 19783 .71

 278 .6439 . So
sx2 

71
n 1
71

sx  278 .6439  16 .6926 . If you used the definitional formula, you should get
 x  x 
2
sx 
 19783.71 . Computations are available in an appendix.
s2

n
278 .6439
 3.870054  1.9672
72
b. State your null and alternative hypotheses (1)
Note that ‘Above normal (128)’ does not contain an equality, so that it must be an alternative
hypothesis. Our hypotheses are H 0 :   128 and H 1 :   128 .
c. Test the hypothesis using a test ratio (1)
11
252y0612 10/18/06 (Open in ‘Print Layout’ format)
x   0 132 .339  128
133 .339  128
 2.7140

 2.2057 . Version 10: t 
1.9672
sx
1.9672
The larger the sample mean is, the more positive will be this ratio. We will reject the null

hypothesis if the ratio is larger than tn 1  t ..71
01  2.380 . Make a diagram showing a Normal
curve with a mean at 0 and a shaded 'reject' zone above 2.380. Since the test ratio is below 2.2057
for Version 0, we do not reject H 0 . Since the test ratio is above 2.380 for Version 10, we reject
H . Note: t n 1  t 71  2.647 .
Version 0: t 
0

2
..005
d. Test the hypothesis using a critical value for a sample mean. (1)
Critical value: We need a critical value for x above 128. Common sense says that if the sample
mean is too far above 128, we will not believe H 0 :   128 . The formula for a critical value for
the sample mean is x    t n1 s , but we want a single value above 128, so use
cv
0

2
x
xcv   0  tn1 s x  128  2.380 1.9672   132 .682 . Make a diagram showing an almost
Normal curve with a mean at 128 and a shaded 'reject' zone above 132.682. For Version 0, since
x  132 .339 is below 132.682, we do not reject H 0 . For Version 10, since x  133 .339 is above
132.682, we reject H 0 .
e. Test the hypothesis using a confidence interval (1)
Confidence interval:   x  t sx is the formula for a two sided interval. The rule
2
for a one-sided confidence interval is that it should always go in the same direction as the
alternate hypothesis. Since the alternative hypothesis is H 1 :   128 , the confidence
interval for Version 0 is   x  t n1 s  132 .339  2.380 1.9672   127 .657 . Make a diagram

x
showing an almost Normal curve with a mean at x  132 .339 and, to represent the
confidence interval, shade the area below above 127.657 in one direction. Then, on the same
diagram represent H 0 :   128 by shading the area below 128 or simply showing 128. Since there
are values below or equal to 128 that are also above or equal to 127.657, we do not reject H 0 .
The confidence interval for Version 10 is   133 .339  2.380 1.9672   128 .657 . Make a
diagram showing an almost Normal curve with a mean at x  133 .339 and, to represent the
confidence interval, shade the area below above 128.657 in one direction. Then, on the same
diagram represent H 0 :   128 by shading the area below 128 or simply showing 128. Since there
are no values below or equal to 128 that are also above or equal to 127.657, we reject H 0 .
f. Find an approximate p-value for the null hypothesis. (1)
x   0 132 .339  128
133 .339  128
 2.7140

 2.2057 . Version 10: t 
Version 0: t 
1.9672
sx
1.9672
71
71
71
71
 1.994 , t .01
 2.647 and t .001
 2.380 , t .005
 3.209 . So
There are 71 degrees of freedom. t .025
for version 0, .01  pvalue  .025 and for version 10, .001  pvalue  .005 .
g. On the basis of your tests, will you get a wellness program? Why? (1) . If you rejected the null
hypothesis, you have shown above normal blood pressure and you get your program. Otherwise
you do not.
h. How do your conclusions change if the sample of 72 is taken from a population of 200? (2)
sx 
N n
N 1
s2

n
200  72
(1.9672 )  0.6432 1.9672   .8020 1.9672   1.5777
200  1
12
252y0612 10/18/06 (Open in ‘Print Layout’ format)
Version 0: t 
x   0 132 .339  128
133 .339  128
 3.3840

 2.7502 . Version 10: t 
1.5777
sx
1.5777

tn 1  t ..71
01  2.380 . Make a diagram showing a Normal curve with a mean at 0 and a shaded
'reject' zone above 2.380. Since both test ratios are now above 2.2057, we reject H 0 in all cases.
The Minitab output for this problem reads as follows.
One-Sample T: s0
Test of mu = 128 vs > 128
Variable
s0
N
72
Mean
132.339
99%
Lower
StDev SE Mean
16.692
1.967
Bound
127.657
T
2.21
P
0.015
Bound
128.657
T
2.71
P
0.004
One-Sample T: s10
Test of mu = 128 vs > 128
Variable
s10
N
72
Mean
133.339
99%
Lower
StDev SE Mean
16.692
1.967
i. Assume that the Normal distribution does not apply and, using your data, test that the median is
above 128. (3)
[12]
Our alternative hypothesis is H 1 :  128 , so that our null hypothesis is H 0 :  128 .
See appendix for the values in order.
Let x be the number of values above 128.
x
nx
x
p
n
Version
Version
Version
Version
Version
Version
Version
Version
Version
Version
Version
0
1
2
3
4
5
6
7
8
9
10
below 128
35
35
35
35
34
33
33
32
31
31
31
above 128
37
37
37
37
38
39
39
40
41
41
42
.5138
.5138
.5138
.5138
.5278
.5417
.5417
.5555
.5694
.5694
.5833
In the outline section on the sign test we have the following.
Hypotheses about
Hypotheses about a proportion
a median
If p is the proportion
If p is the proportion
above  0
below  0
 H 0 :   0

 H 1 :   0
 H 0 : p .5

 H 1 : p .5
 H 0 : p .5

 H 1 : p .5
 H 0 :   0

H 1 :   0
 H 0 :   0

H 1 :   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
13
252y0612 10/18/06 (Open in ‘Print Layout’ format)
 H : p .5
 H : 128
So our hypotheses  0
correspond to  0
 H 1 : p  .5
 H 1 :  128
x  .5
p  p0
To compute a test ratio, we use z 
 n

p0 q0
.25
n
n
x
n  .5
.5

2x  n
n
. Note that z.01  2.327
n
.25
.25

 .00347222  .05893 and n  72  8.48528 .
n
72
So compute the test ratio and reject the null hypothesis if the test ratio is above 2.327. With two
different formulas, it is not surprising that we get slightly different answers because of rounding
error.
x  .5
x (number
2x  n
x
n
p
above128)
n
n
.25
and that n  72 , so that
n
above 128
Version 0
37
.5138
Version 1
Version 2
Version 3
37
37
37
.5138
.5138
.5138
Version 4
38
.5278
Version 5
39
.5417
Version 6
39
.5417
Version 7
40
.5555
Version 8
41
.5694
Version 9
41
.5694
z
237   72
.5138  .5
 .2342 or z 
 .2357
.05893
8.48532
.2342
.2342
.2342
.2357
.2357
.2357
.7076
.7071
1.17777
1.1785
.5278  .5
238   72
 .4717 or z 
 .4714
.05893
8.48532
.5417  .5
239   72
z
 .7076 or z 
 .7071
.05893
8.48532
z
.5555  .5
240   72
z
 .9418 or z 
 .9428
.05893
8.48532
.5694  .5
241  72
z
 1.1777 or z 
 1.1785
.05893
8.48532
.5833  .5
242   72
 .1.4220 or z 
 1.4142
Version 10
42
.5833 z 
.05893
8.48532
Though these values are all in the ‘do not reject’ zone, note that as x increases, , so does z .
However, it never reaches the ‘reject’ zone.
j. (Extra credit) Use your data to get an approximate 90% 2-sided confidence interval for the
median. Note that z.025  1.645 .
If the index of the lower number is k , the index of the upper number will be n  1  k . The large
sample formula for the index of the lower number in a confidence interval for the median is
n  1  z . 2 n
72  1  1.645 72
k
=29.52, and n  1  k  72  1  29  44 , so use x29 and

2
2
x44 . If we put the numbers in order we have the following limits.
14
252y0612 10/18/06 (Open in ‘Print Layout’ format)
x29
Version
Version
Version
Version
Version
Version
Version
Version
Version
Version
Version
2.
0
1
2
3
4
5
6
7
8
9
10
126.54
126.64
126.74
126.84
126.94
127.04
127.14
127.24
127.34
127.44
127.54
x44
133.00
133.10
133.20
133.30
133.40
133.50
133.60
133.70
133.80
133.90
134.00
Once again, assume that the Normal distribution applies, but assume a population standard
deviation of 16 and that we are testing whether the mean is above 128. (90% 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 a. On the basis of your
p-value, would you reject the null hypothesis? Why? (1)
c. Create a power curve for the test. (6)
Solution: The problem says that   16 , n  72 , H 1 :   128 and   .10
 H :   128
16
16 2

 3.55555  1.8856 and
a. Our hypotheses are  0
and we need  x 
72
72
 H 1 :   128
z  z.10  1.282 .
132 .24  128 

b. Assume that the sample mean is 132.34. The p-value is Px  132 .4   P z 

1.8856 

 Pz  2.25   .5  P0  z  2.25   .5  .4878  .0122 . Since the p-value is below   .10 , we reject the
null hypothesis.
c. First we need a critical value. This is a 1-sided test and since the alternate hypothesis is H 1 :   128 , we
need a critical value above 128. The two-sided formula is xcv   0  z  x , and we will use
2
x cv   0  z  x  128  1.282 1.8856   130 .42 . This is a right-sided test, so we will use points to the
right of 128. Make a diagram: Show a Normal curve with a mean at 128 and a critical value at 130.42.
The ‘reject’ zone is above 130.42. If the population mean is, in fact, above 128, and we do not reject the
null hypothesis because the sample mean is below 130.42, we make a Type II error. The probability of such
130 .42  1 

an error for a population mean of  1 128 is Px  130.42   1   P z 
 . The points that
1.8856 

I will use are 128, 129.21, 130.42, 131.63 and 132.84. I picked these because 129.21 is midway between
128 and the critical value of the sample mean. Because the first three points are apart by 1.21, I picked the
last two points by adding 1.21 to the previous point.
130 .42  128 

Px  130 .42   128   P  z 
 Pz  1.28   .5  .3997  .8997  1  
 1 128
1.8856 

Power  1  .8997  .1003  
130 .42  129 .21 

Px  130 .42   129 .21  P  z 
 1 129.21
  Pz  0.64   .5  .2389  .7389
1.8856


Power  1  .7389  .2611
130 .42  130 .42 

Px  130 .42   130 .42   P  z 
 1 130.42
  Pz  0  .5
1.8856


Power  1  .5  .5 (Power is always .5 at the critical value.)
15
252y0612 10/18/06 (Open in ‘Print Layout’ format)
130 .42  131 .63 

Px  130 .42   131 .63   P  z 
  Pz  0.64   .5  .2389  .2611
1.8856


Power  1  .2611  .7389
130 .42  132 .84 

Px  130 .42   132 .84   P  z 
 1 132.84
  Pz  1.28   .5  .3997  .8997
1.8856


Power  1  .8997  .1003
If we want to try for higher power, add 1.21 to 132.84 and get 134.05
130 .42  134 .05 

 1 134.05
Px  130 .42   134 .05   P  z 
  Pz  1.92   .5  .4726  .9726
1.8856


 1 131.63
3.
(Moore, Notz) A recent survey said that nationwide 73% of all freshman students identified being
well-off financially as an important lifetime goal. You believe that the proportion of freshman
business majors with that goal is higher than the national figure. You take a survey of a random
sample of 200 students and find that 152  a  have being well-off as an important goal, where a
is the second to last digit of your student number. If the second to last digit of your student number
is zero, a  10 . (For example, Seymour Butz’s student number is 976502, so he will add 10 and
say that x  152  10  162 )
a. Formulate your null and alternative hypotheses and do a hypothesis test with a 95% confidence
level. (2)
b. Find a p-value for the null hypothesis. (1)
c. Find the p-value for the null hypothesis if x  162  a (1)
d. (Extra credit) How would your answer to a) change if your sample of 200 came from a
population of 300? (1)
e. (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  .73 . (Up to 6 points)
f. Assume that p  .73 , how large a sample would you need to estimate the proportion above that
have being well-off with an error of .005? (2)
g. Use the proportion that you found in a) to create a 2-sided 95% confidence interval for the
proportion. Does it differ significantly from .73? Why? (2)
[28]
Solution: The problem states that we are testing p  .73 , when n  200 and   .05
In Seymour’s case x  162 or p 
162
 .81 and we can compute
200
p0 q0
.73.27 

 .009855  0.031393 , where p0 comes from our null hypothesis and
n
200
q0  1  p0 .
a. Since p  .73 does not contain an equality, it must be an alternative hypothesis our hypotheses are
p 
 H 0 : p  .73
. This is a right- sided test since values of the observed proportion above .73 are the only

 H 1 : p  .73
values that could lead to a rejection of the null hypotheses. It is most expedient to use a test ratio,
p  p0 .81  .73
z

 2.55 . Make a diagram. Show a normal curve with a mean at zero and an area of
p
.031393
5% above z.05  1.645 . The area above 1.645 is the ‘rejection zone.’ Since our value of z is above 1.645, it
is in the ‘rejection zone,’ and we reject the null hypothesis.
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b. We find a p-value for the null hypothesis.
pvalue  P p  .81  Pz  2.55   .5  P0  z  2.55   .5  .4946  .0054 . Since this is below   .05 , we
reject the null hypothesis.
162  a 152

 .76 . (This should have
200
200
p  p0 .76  .73
been x  152  a , and it should have asked for a conclusion.) z 

 0.96
p
.031393
c. We find the p-value for the null hypothesis if x  162  a . p 
pvalue  P p  .76   Pz  0.96   .5  P0  z  0.96   .5  .3315  .1685 . Since this is not below
  .05 , we do not reject the null hypothesis.
d. We show how our answer to a) would change if our sample of 200 came from a population of 300.
If our sample of 200 came from a population of 300, the standard error would be much smaller.
.73.27 
 .33445 .009855  .5783149 0.031393  .018155
200
p  p0 .81  .73
This would lead to a much larger value of z 

 4.41 . The p-value would be much
p
.018155
p 
N n
N 1
p0 q0
300  200

n
300  1
smaller, but we would still reject the null hypothesis.
e. (Extra credit) We use a critical value of the proportion for testing the null hypothesis and create a power
curve for the test by using the alternate hypothesis and finding the power for values of p1  .73 .
To do this, remember  p 
 H 0 : p  .73
. The critical

 H 1 : p  .73
becomes pcv  p0  z p .73  1.645.031393  .7816
p0 q0
 0.031393 ,   .05 , z.05  1.645 and
n
value is above .73, so that pcv  p0  z  p
2
This is a right-sided test, so we will use points to the right of 0.73. Make a diagram: Show a Normal curve
with a mean at 0.73 and a critical value at .7816. The ‘reject’ zone is above .7816. If the population
proportion is, in fact, above .73, and we do not reject the null hypothesis because the sample mean is below
.7816, we make a Type II error. The probability of such an error for a population proportion of p 1  .73 is

.7816  p1 
p1q1
, where  p 
. The points that I will use are .73, .7558,
P p  .7816 p  p1   P z 


200

p


.7816, .8074 and .8332. I picked these because .7558 is midway between .73 and the critical value of the
sample proportion. Because the first three points are apart by .0258, I picked the last two points by adding
.0258 to the previous point.
p1q1
 .031393
p1  .73 ,  p 
200

.7816  .73 
P z 
 Pz  1.644   .5  .4500  .95  1  


p


Power  1  .95  ..05  
p1  .7558,  p 
.7558 .2442 
 .0009228  .0303781
200

.7816  .7558 
P z 
 Pz  0.85   .5  .3032  .8032


p


Power  1  .8032  .1968
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252y0612 10/18/06 (Open in ‘Print Layout’ format)
.7816 .2184 
 .0008535  .0292148
200
p1  .7816,  p 
Power  1  .5  .5
p1  .8074,  p 

.7816  .7816
P z 

p


  Pz  0  .5


.8074 .1926 
 .00077753  .0278842
200

.7816  ..8074 
P z 
 Pz  0.93   .5  .3238  .1762


p


Power  1  .1762  .8238
p1  .8332,  p 
.8332 .1668 
 .00069489  .026361
200

.7816  .8332 
P z 
 Pz  1.97   .5  .4756  .0244


p


Power  1  .0244  .9756
f. We assume that p  .73 and find how large a sample we need to estimate the proportion above that have
being well-off as a goal with an error of .005.
The formula for sample size for a proportion is in the outline. n 
p  .73 , q  1  p  1  .73  .27 . So n 
.73.27 1.960 2
.005 2
pqz 2
e2
. If   .05 , z .025  1.960 . If
 30287 .1 . Round this up to 30288.
g. We use the proportion that we found in a) to create a 2-sided 95% confidence interval for the proportion
and ask if it differs significantly from .73.
162
 .81 and the formula is p  p  z s p , where
Seymour used p 
2
200
pq
.81.19 

 .0007695  .027740 . So p  .81  1.960 .027740   .81  .0544 or .7556 to
n
200
.8644. Since this interval does not include .73, the proportion differs significantly from .73.
sp 
4.
Standard deviation is often a measure of reliability. A manufacturer is providing a connector with
a mean length of 2.5 mm and is getting complaints that the connector is often too large or too
small for the intended use. The previous standard deviation to the length of the part was 0.025mm,
but the manufacturer introduces a process that should make the standard deviation smaller. A
sample of 25 items is taken which yields a sample standard deviation of 0.030  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 subtract .005 and say that s  0.030  0.005  0.025 . )
a. Formulate the null and alternative hypotheses necessary to see if the goal has been achieved and
test the hypothesis using a 95% 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 (1.5)
d. Do a 95% 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. The number of claims for missing baggage in a large metropolitan airport supposedly follows a
Poisson distribution with a mean of 72 per week. Assume that in a given week 92 are lost. Test
this hypothesis using a test ratio and a 95% confidence level. (2)
h. Find approximate critical values for the number of bags that could be lost in 4g. (2)
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252y0612 10/18/06 (Open in ‘Print Layout’ format)
i. (Extra credit) Find the power of the test in 4g) if the average number of lost bags per week is
really 87. (3)
j. I claim that x is binomially distributed with p  .01 . Test this assertion using a 2-sided test if
there are 4 successes in 10 trials. (2)
[39.5]
Solution: The problem says   2.5 ,   0.025 , n  25 and Seymour uses s  0.025 . This indicated that
I had not played with these results enough to give you a ‘good’ number for s . The manufacturer wants
  0.025 , which is an alternate hypothesis because it does not contain an equality.
a. We formulate the null and alternative hypotheses necessary to see if the goal has been achieved and
test the hypothesis using a 95% confidence level and a test ratio.
 H :   0.025
. DF  n  1  24 . The test ratio for a variance or a standard deviation and a
  .05 and  0
 H 1 :   0.025
small sample is  2 
n  1s 2
 02
. Our test is left sided, so we need a value of  2 with 5% of the
 
24
distribution below it. We therefore say that our ‘reject’ region will be below  2 .95  13.8484 . Make a
diagram. Draw a  2 curve with a mean at 24 and shade the area below 13.8484. Since
2 
24 .025 2
.025 2
 24 , we cannot reject the null hypothesis. Of course, it should be obvious that there is no
improvement.
b. We list assumptions necessary for the test in a). The underlying distribution should be Normal.
c. We try to get a p-value and interpret it. Our table says that the two closest values to 24 are
 
 
. pvalue  Ps  0.025  P
2 24
.90
24
 15.6587 and  2 .10  33.1962 . So all we can say is .10  pvalue  .90 . Actually





 24 and we can say .P  2  15.6587  .10 and .P  2  33.1962  .90
hence .10  pvalue  .90 . This means that we cannot hope to reject the null hypothesis unless we use a
significance level above 10%.
d. We find a 95% confidence interval for the standard deviation. The interval given on Page 1 of the
supplement is
n  1s 2
 2
2
 2 
 
2
2
1
2
24
n  1s 2
 
2
.975  12.4012 and 
24
.025
or
n  1s 2
 .2025
2
 39.3641so we have
 2 
n  1s 2
 .2975
. Table 1 says that
24 .025 2
24 .025 2
or
 2 
39 .3641
12 .4012
.0003811   2  .0012096 or 0.0195    .0348 .
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252y0612 10/18/06 (Open in ‘Print Layout’ format)
e. We find a 95% confidence interval appropriate for testing a). If we use an appropriate confidence
interval, it should be of the type  2  ? or   ? . We want 5% of the distribution above the point that
we pick. The interval that we did in the last section implies that 2.5% of the distribution is above
, so 5% must be above
n  1s 2
 .295
 
24
. The table says  2 .95  13.8484 , so out limit is  2 
n  1s 2
 .2975
24 .025 2
or
13 .8484
 2  0.001083 or   .0329 .
f. We find a critical value for the test in 4a). Table 3 implies that the critical values for a 2-sided hypothesis
 
2
are s cv2 
 
2
.025  0
n 1
 
2
and s cv2 
2
.975  0
n 1
. We need one value below 0.025, so use
13 .8484 0,025 2
 .0003606 . The square root is s cv  .01899 . We reject the null
n 1
24
hypothesis if the sample standard deviation is below .01899.
2
s cv2 
2
.95  0

At this point we are starting a completely new problem. The number of claims for missing baggage in a
large metropolitan airport supposedly follows a Poisson distribution with a mean of 72 per week. Assume
that in a given week 92 are lost.
g. We test the hypothesis that the mean is 72 using a test ratio and a 95% confidence level.
The Poisson distribution with a mean of 72 is approximately the Normal distribution with a mean of 72
72 . We know that x  92 . From what we know about the Normal
 H 0 :   72 Poisson
x   92  72
20
distribution z 
. To do a


 2.36 . Our hypotheses are 

8.48528
72
 H 1 :   72 Poisson
95% test, make a diagram. Show a Normal curve with a mean at zero and ‘reject’ zones below
z.025  1.960 and above z.025  1.960 . Our value of z is in the upper reject zone, so reject the null
hypothesis.
and a standard deviation of
h. We find approximate critical values to do the test in 4g).
95% of the values of z are between -1.960 and 1.960. Use the formula x    z to get
xcv  72  1.960 72  72  16 .63 or 55.37 to 88.37. We reject the null hypothesis if x is below 56 or above
88. (Note that since x  92 , we reject the null hypothesis, but that this was not required.)
i. We are supposed to find the power of the test in 4g) if the average number of lost bags per week is really
87. I’ll try this if you will!
This is another totally new problem.
j. I claim that x is binomially distributed with p  .01 . We test this assertion using a 2-sided test if
there are 4 successes in 10 trials.
 H 0 : p  .01
. n  10 and x  4 Since   np  10.01  0.1 , 4 is above the expected value and , if we

 H 1 : p  .01
use the binomial table, pvalue  2Px  4  21  Px  3  21  1  0 . Since this is below any reasonable
significance level, reject the null hypothesis.
20