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```252y0753 10/19/07 (Open in ‘Print Layout’ format)
ECO252 QBA2
FIRST EXAM
October 4 and 8, 2007
Version 3
Name:_KEY___________
Class hour: _____________
Student number: __________
Show your work! Make Diagrams! Include a vertical line in the middle! Exam is normed on 50
points. Answers without reasons are not usually acceptable.
I. (8 points) Do all the following.
x ~ N 4, 11
0  4

1. Px  0  P  z 
 Pz  0.36   0.36  z  0  Pz  0 = .1406 + .5 = .6406
11 

2  4
  33  4
z
 P 3.36  z  0.18   P3.36  z  0  P0.18  z  0
2. P33  x  2  P 
11 
 11
=.4996 - .0714 = .4282
1
252y0753 10/19/07 (Open in ‘Print Layout’ format)
4  4
 4  4
z
 P 0.73  z  0 = .2673
3. P4  x  4  P 
11 
 11
4. x.075 (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 .075 is the value of z with 7.5% of the
distribution above it. Since 100 – 7.5 = 92.5, it is also the .925 fractile. Since 50% of the standardized
Normal distribution is below zero, your diagram should show that the probability between z .075 and zero
is 92.5% - 50% = 42.5% or P0  z  z.075   .4250 . If we check this against the Normal table, the closest
we can come to .4250 is P0  z  1.44   .4251 . So z .075  1.44 . This is the value of z that you need for
a 85% confidence interval. To get from z .075 to x.075 , use the formula x    z , which is the
x
. x  4  1.4411  19 .84 . If you wish, make a completely separate diagram for x .

Draw a Normal curve with a mean at 4. Show that 50% of the distribution is below the mean (4). If 7.5% of
the distribution is above x.075 , it must be above the mean and have 42.5% of the distribution between it and
the mean.
19 .84  4 

 Pz  1.44   Pz  0  P0  z  1.44 
Check: Px  19 .84   P  z 
11 

 .5  .4251  .0749  .075 . This is identical to the way you normally get a p-value for a right-sided
test.
opposite of z 
2
252y0753 10/19/07 (Open in ‘Print Layout’ format)
II. (9 points-2 point penalty for not trying part a.)
Monthly incomes (in thousands) of 6 randomly picked individuals in the little town of Rough Corners are
shown below.
2.5 7.3 3.1 2.6 2.4 3.0
a. Compute the sample standard deviation, s , of expenditures. Show your work! (2)
b. Assuming that the underlying distribution is Normal, compute a 99% confidence interval for the
mean. (2)
c. Redo b) when you find out that there were only 50 people living in Rough Corners. (2)
d. Assume that the population standard deviation is 2 and create an 85% two-sided confidence
interval for the mean. (2)
e. Use your results in a) to test the hypothesis that the mean income is above 2.3(thousand) at the
99% level. (3) State your hypotheses clearly!
f. (Extra Credit) Given the data, test the hypothesis that the population standard deviation is below
2. (3)
Solution: a) Compute the sample standard deviation, s , of expenditures.
The first two columns are needed for the
x
Row
x2
xx
x  x 2
computational (shortcut) method. The first, third
1 2.5
6.25 -0.98333
0.9669
and fourth are needed for the definitional
2 7.3 53.29
3.81667 14.5669
method. Using (both methods or) the
3 3.1
9.61 -0.38333
0.1469
definitional method wastes time.
4 2.6
6.76 -0.88333
0.7803
5 2.4
5.76 -1.08333
1.1736
x  x   0 (a
x  20 .9 ,
x 2  90 .67 ,
6
x
3.0
20.9
9.00
90.67
-0.48333
0.0
 x  20.9  3.4833 s   x
n
2
x
6

0.2336
17.8682
2
 nx 2
n 1
check),



 x  x 
2
 544 and
n  6.
90 .67  63.4833 2 17 .8697

 3.5739
5
5
s x  3.5739  1.8905 . If you used the definitional method, you would have gotten 1.8904. There seems
to be a lot of potential for rounding error here. Note that the x  x column, even though it carries an extra
place, does not quite add to the expected zero but to .00002.
b) Assuming that the underlying distribution is Normal, compute a 99% confidence interval for the mean.
(2)   x  tn1 s x  3.4833  4.032 0.77178   3.483  3.112 or 0.371 to 6.595.
2
1.8905
3.5739
5


 0.59565  0.77178
t n1  t.005
 4.032
2
6
n
6
c) Redo b) when you find out that there were only 50 people living in Rough Corners. (2)
  x  tn1 s x 136 .00  4.604 4.3304   136 .00  19.98 or 116.02 to 155.98
sx 
sx
2
N  n 1.8905 50  6
3.5739  44 
9


   0.59565    0.41237  0.6422
6
49
N

1
50

1
 
 13 
n
6
d) Assume that the population standard deviation is 2 and create an 85% two-sided confidence interval for
the mean. (2) (2) We found z.075  1.44 on the last page. We have   2 , n  6 , x  3.4833 and
sx 
x 
sx
x

2

n
6
2.3075 to 4.6591.
4
 0.66667  0.8165   x  z 3  x1  3.4833 1.440.8165  3.4833 1.1758 or
6
e) Use your results in a) to test the hypothesis that the mean income is above 2.3(thousand) at the 99%
level. (3) State your hypotheses clearly! The statement that the mean is above 2.3 does not contain an
equality, so it must be an alternate hypothesis. We have the following information.   .01 , x  3.4833 ,
s
5
 3.365 .
n  6 and s x  x  0.77178 . Since this is a one-sided hypothesis we will use tn 1  t .01
n
Needless to say, because of the small sample size, we are assuming that the parent distribution is Normal.
3
252y0753 10/19/07 (Open in ‘Print Layout’ format)
H :   2.3
Our hypotheses are  0
so  0  2.3 . Since we are worrying about the mean being too large, this
H 1 :   2.3
is a right-sided test.
There are three ways to do this. Do only one of them.
x   0 3.4833  2.3
(i) Test Ratio: t 

 1.5332 . This is a right-sided test - the larger the sample mean
sx
0.77178
is, the more positive will be this ratio. We will reject the null hypothesis if the ratio is larger than
5
tn 1  t .01
 3.365 . Make a diagram showing a Normal curve with a mean at 0 and a shaded 'reject' zone
above 3.365. Since the test ratio is below 3.365, we cannot reject H 0 .
If you wish to find a p-value for your hypothesis, note that the t-ratio is 1.5332. The p-value will be the
probability that t is above 3.365. The line of the t table for 5 degrees of freedom is below. {ttable}
df .45 .40 .35 .30 .25 .20 .15 .10 .05 .025 .01 .005 .001
5 0.132 0.267 0.408 0.559 0.727 0.920 1.156 1.476 2.015 2.571 3.365 4.032 5.893
What this tells us, among other things, is that Pt  1.476   .10 and Pt  2.015   .05 . Since 1.5332 lies
between 1.476 and 2.015, the probability that t lies above 1.5332 must be between .05 and .10.
.05  p  value  .10 . This is above our significance level of .01, so we will not reject the null hypothesis.
(ii) Critical value: We need a critical value for x above 2.3. Common sense says that if the sample mean is
too far above 2.3, we will not believe H 0 :   2.3 . The formula for a critical value for the sample mean is
x    t n1 s , but we want a single value above 2.3, so use x    t n1 s
cv
0

2
x
cv
0

x
 2.3  3.365 0.77178   2.3  2.5970  4.8970 . Make a diagram showing an almost Normal curve with
a mean at 2.3 and a shaded 'reject' zone above 4.8970. Since x  3.4833 is not above 4.8970, 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 :   2.3 , the confidence interval is   x  tn1 s x or
  3.4833  3.365 0.77178   0.8863 . Make a diagram showing an almost Normal curve with a mean at
x  3.4833 and, to represent the confidence interval, shade the area above 0.8863 in one direction. Then,
on the same diagram, to represent the null hypothesis, H 0 :   2.3 , shade the area below 2.3 in the
opposite direction. Notice that these overlap. What the diagram is telling you is that it is possible for
  0.8863 and H 0 :   2.3 to both be true. (If you follow my more recent suggestions, it is actually
enough to show that 2.3 is on the confidence interval.) So we do not reject H 0 .
f.) (Extra Credit) Given the data, test the hypothesis that the population standard deviation is below 2. (3)
This is an alternate hypothesis, H 1 :   2 . The null hypothesis is H 0 :   2 Remember n  6 ,
  .01 s x2  3.5739 . Table 3 says that the test ratio is  2 
n  1s 2
 02

53.5739 
22
 4.4674 .
4
252y0753 10/19/07 (Open in ‘Print Layout’ format)
Recall df  n  1  5. The first paragraph of the chi-squared table appears below. If we look at the 5
4
column, we see that the lower 1% of values of chi-squared are cut off by  2 .99  0.2971, so that the reject
region is below 0.5543.
Degrees of Freedom

0.005
0.010
0.025
0.050
0.100
0.900
0.950
0.975
0.990
0.995
1
2
3
4
5
6
7.87946 10.5966 12.8382 14.8603 16.7496 18.5476
6.63491 9.2103 11.3449 13.2767 15.0863 16.8119
5.02389 7.3778 9.3484 11.1433 12.8325 14.4494
3.84146 5.9915 7.8147 9.4877 11.0705 12.5916
2.70554 4.6052 6.2514 7.7794 9.2364 10.6446
0.01579 0.2107 0.5844 1.0636 1.6103 2.2041
0.00393 0.1026 0.3518 0.7107 1.1455 1.6354
0.00098 0.0506 0.2158 0.4844 0.8312 1.2373
0.00016 0.0201 0.1148 0.2971 0.5543 0.8721
0.00004 0.0100 0.0717 0.2070 0.4117 0.6757
7
20.2778
18.4753
16.0128
14.0671
12.0170
2.8331
2.1674
1.6899
1.2390
0.9893
8
21.9550
20.0902
17.5346
15.5073
13.3616
3.4895
2.7326
2.1797
1.6465
1.344
9
23.5893
21.6660
19.0228
16.9190
14.6837
4.1682
3.3251
2.7004
2.0879
1.7349
Computer output for parts b) d) e) and f) follows
————— 10/5/2007 1:39:55 PM ————————————————————
Welcome to Minitab, press F1 for help.
MTB > WOpen "C:\Documents and Settings\RBOVE\My Documents\Minitab\252x075123.MTW".
Retrieving worksheet from file: 'C:\Documents and Settings\RBOVE\My
Documents\Minitab\252x0751-23.MTW'
Worksheet was saved on Wed Oct 03 2007
Results for: 252x0751-23.MTW
MTB > describe x
Descriptive Statistics: x
Variable
x
N
6
N*
0
Mean
3.483
SE Mean
0.772
MTB > onet c1;
SUBC> conf 99.
StDev
1.890
Minimum
2.400
Q1
2.475
Median
2.800
Q3
4.150
Maximum
7.300
Part b)
One-Sample T: x
Variable
x
N
6
Mean
3.48333
StDev
1.89041
MTB > onez c1;
SUBC> sigma 2;
SUBC> conf 85.
SE Mean
0.77176
99% CI
(0.37149, 6.59517)
Part d)
One-Sample Z: x
The assumed standard deviation = 2
Variable N
Mean
StDev SE Mean
x
6 3.48333 1.89041 0.81650
MTB > Onet c1;
SUBC>
Test 2.3;
SUBC>
Confidence 99;
SUBC>
Alternative 1.
85% CI
(2.30796, 4.65871)
Part e)
One-Sample T: x
Test of mu = 2.3 vs > 2.3
Variable
x
N
6
Mean
3.48333
StDev
1.89041
SE Mean
0.77176
99%
Lower
Bound
0.88642
T
1.53
P
0.093
MTB > Save "C:\Documents and Settings\RBOVE\My Documents\Minitab\252x075123.MTW";
SUBC>
Replace.
5
252y0753 10/19/07 (Open in ‘Print Layout’ format)
Saving file as: 'C:\Documents and Settings\RBOVE\My
Documents\Minitab\252x0751-23.MTW'
Existing file replaced.
Part f)
MTB > %sigtest c1 4
Tests data in column 1 for variance of 4. Packaged
Minitab Macro. This is a 2-sided test.
Executing from file: sigtest.MAC
The value of the test statistic is
4.4671.
If the test statistic is less than
0.8312 or greater
than 12.8325 then there is statistical evidence indicating
that your variance does not equal to
4.0000, at alpha =
0.0500.
MTB > %sigtest c1 4;
Since I didn’t have time to input a 1sided test. I ran two-sided tests with
a confidence level of 98% and 90%
because their lower critical values
would be the same as for 1-sided tests
with confidence levels of 99% and 95%.
SUBC> alpha 98.
Executing from file: sigtest.MAC
The value of the test statistic is
4.4671.
If the test statistic is less than
4.2789 or greater
than
4.4249 then there is statistical evidence indicating
that your variance does not equal to
4.0000, at alpha =
0.9800.
MTB > %sigtest c1 4;
SUBC> alpha 90.
Executing from file: sigtest.MAC
The value of the test statistic is
4.4671.
If the test statistic is less than
3.9959 or greater
than
4.7278 then there is statistical evidence indicating
that your variance does not equal to
4.0000, at alpha =
0.9000.
6
252y0753 10/19/07 (Open in ‘Print Layout’ format)
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 if possible.
1) If I want to test to see if the mean of x is smaller than the given population mean  0 my null
hypothesis is:
i)    0
ii)    0
iii) *    0
iv)    0
vi) None of the above
Explanation:    0 is our alternate hypothesis since it doesn’t contain an equality.    0 is the
opposite, so it must be the null hypothesis.
2) Assuming that you have a sample mean of 100 based on a sample of 36 taken from a population of 300
with a sample standard deviation of 80, the 99% confidence interval for the population mean is
 80 

a) 100  2.576 
 36 
 300  36 80 

b) 100  2.576 
 300  1 36 


 80 

c) 100  2.576 
 300 
 80 

d) 100  2.724 
 300 
 300  36 80 

e) * 100  2.724 
 300  1 36 


 80 

f) 100  2.724 
 36 
 80 

g) 100  2.438 
 300 
 300  36 80 

h) 100  2.438 
 300  1 36 


 80 

i) 100  2.438 
 36 
 300  36 80 

j) 100  2.438 
 300  1 300 


g) None of the above. Fill in a correct answer.
Explanation: The formula for a confidence interval when the variance is known when the sample is more
than 20% of the population was given in the solution to problem A2 as   x  t n1 x , where x  100,
n  36, N  300, s  80 , 1    99% and   .01 .
2
7
252y0753 10/19/07 (Open in ‘Print Layout’ format)
Here s x 
80
N n

N 1
36
sx
n
 80
 36

  100  2.724 
300  36
300  1
300  36
35
 2.724 . So
and t n 1  t .005
2
300  1




3) Which of the following is a Type 2 error?
a) Rejecting the null hypothesis when the null hypothesis is true.
b) Not rejecting the null hypothesis when the null hypothesis is true.
c) *Not rejecting the null hypothesis when the null hypothesis is false.
d) Rejecting the null hypothesis when the null hypothesis is false.
e) All of the above
f) None of the above.
4) If a random sample is gathered to get information about a population proportion, what do we mean by a
p-value?
a) P-value is the probability that, if the null hypothesis was false, that, if we were to repeat the
experiment many times, we would get a sample proportion as extreme as or more extreme
than the sample proportion actually observed.
b) *P-value is the probability that, if the null hypothesis was true, that, if we were to repeat
the experiment many times, we would get a sample proportion as extreme as or more extreme
than the sample proportion actually observed.
c) P-value is the population proportion in the null hypothesis.
d) P-value is the population proportion in the alternate hypothesis.
e) P-value is the probability of a type 2 error.
f) P-value is the probability that the alternate hypothesis is true, given the sample proportion
actually observed.
g) None of the above is true.
5) If a difference in proportions (in a business-related problem) is called statistically significant at the 1%
significance level, this means that
a) *If the null hypothesis is true, the difference in proportions is surprisingly large.
b) There is a 99% chance that the null hypothesis is true.
c) The difference in proportions is large enough so that we must take account of it in our
d) All of the above
6) (Wonnacott & Wonnacott) When an industrial process is in control, it produces bolts with a hardness
that has a mean of at least 80 and a (population) standard deviation of 8. If the hardness is too far below 80,
you must shut down the process. Every hour you take a sample of 16 bolts. How low must the average
hardness of these bolts be before we shut the process down? (Use a 10% significance level, and don’t forget
Solution: We have  0  80 ,   8 , n  16 , H 1 :   80 , H 0 :   80 ,   .10 , z .10  1.282 ,
x 

n

8
16
 2 , xcv   0  z  x  80  1.282 2  77.436
8
252y0753 10/19/07 (Open in ‘Print Layout’ format)
7) Your boss, who doesn’t know any statistics, tells you to shut down the process in 6) if the hardness level
from a sample of 16 bolts is 79 or lower. . It is known that the population standard deviation is 8. If we
assume that the process is producing bolts with an average hardness of 80, what is the probability that it
will be shut down? (Think p-value?) (2)
[15]
79  80 

Solution: Px  79   P  z 
 Pz  0.5  .5  .1915  .3085
2 

8) A 1989 Gallup poll revealed that 59% of women believed that the Republican Party were more likely
than the Democratic Party to keep the country prosperous (My, how things change!). We were already sure
that 68% of men believed that this was true.
a) How many women had to be polled before we could state that the proportion for women is
.59  .03 ? (Use a 5% significance level.) (2)
Solution: The outline says “The usually suggested formula is n 
pqz 2
…..This is the formula everyone
e2
forgets that we covered.” So, we have p  .59 , q  1  p  1  .59  .41 , z  1.960 , e  .03 and
n
.59 .411.960 2
.032
 1032 .54 . So use a sample of at least 1033.
b) If we wished to test our belief that women were less likely to believe that the Republicans were
more likely than the Democratic Party to keep the country prosperous and we were already sure that 68%
of men believed that this was true, let p represent the proportion of women. What are our null and
alternative hypotheses? (1)
Solution: The statement implied in the problem p  .68 does not contain an equality, so the null
hypothesis is the opposite p  .68 .
c) The actual poll covered a sample of 750 women. Using a 95% confidence level and assuming
that your hypothesis in b) is correct, test the hypothesis. (2)
[20]
Solution: From Table 3.
Interval for
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
Proportion
p  p0
H 0 : p  p0
p  p  z 2 s p
pcv  p0  z 2  p
z

H
:
p

p
p
1
0
pq
p0 q0
sp 
p 
n
n
q  1 p
q0  1  p0
  .05 , z  z.05  1.645 , p 0  .68 , q 0  1  p 0  1  .68  .32 , p  .59 and n  750 . First, for the test
ratio or critical value we need  p 
p0 q0
.68.32 

 .000290  .0170333 . For the confidence
750
n
pq
.59 .41

 .000323  .01796 .
n
750
Use one of the following 3 methods. H 0 : p  .68 , H 1 : p  .68
interval we need p  .59 , q  1  p  1  .59  .41 . s p 
Critical Value Method: Since we have H 1 : p  .68 , this is a left-sided test. We use a critical value for the
proportion of pcv  p0  z  p  .68  1.645 .0170333   .6520 . Make a diagram showing a normal curve
with a center at p 0  .68 and a rejection region below .6528. Since p  .59 , is below .6520, we reject the
null hypothesis.
p  p0
.59  .68

 5.284 . Make a diagram showing a normal curve with a
Test Ratio Method: z 
.0170333
p
center at zero and a rejection region below -1.645. Since z  5.284 , is below -1.645, we reject the null
hypothesis. The p-value would be P p  .59   Pz  5.284   .5  .5000 =.0.
9
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This would lead to rejection of the null hypothesis for most values of  , since the p-value would be below
the significance level.
Confidence interval method: Since we have H 1 : p  .68 , we need a one-sided ‘  ’ interval. This would be
p  p  z p  .59  1.645 0.01796   .6195 The null hypothesis H 0 : p  .68 is contradicted by the
confidence interval . p  .6195 .
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ECO252 QBA2
FIRST EXAM
October 8, 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. Failing to be transparent about which section of which
problem you are doing can lose you 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).
A group of 30 employees are interviewed to determine the minimum amount that they will take to give up a
vacation day. After careful interviewing, a psychologist repots the following amounts.
479
616
627
648
488
622
522
557
512
595
621
631
547
628
657
511
578
634
539
625
My calculations say that the sum of these 30 numbers is
x
2
553
612
520
509
499
633
606
616
612
598
 x  17395 and that the sum of squares is
 10171575 . This is a sample of 30.
Personalize these data as follows. Take the second to last digit of your student number and multiply it by 5.
Add this quantity to each of the 30 numbers. If the second to last digit of your student number is 0, add 50.
Label your exam by version number as follows. If the second to last digit of your student number is 1, you
are doing Version 1. If the second to last digit is 2, you are doing Version 2. Etc. If the second to last digit
is zero you are doing version 10. Last term's exam said the following.
If you add a quantity a to a column of numbers,
 x  a   x na,
 x  a   x  2a x na . For example, if a  60 ,
 x  60    x 3060 ,  17395 + 1800 = ? and
 x  60   x  260 x 3060  1017157512017395  303600.
2
2
2
2
2
2
Test the following
Problem 1: Count the number of people in your sample that demand more than \$602.50 and make it into a
sample proportion. Test the following 3 hypotheses: I) that 60% demand more than \$602.50, II) that more
than 60% demand more than \$602.50 and III) that less than 60% demand more than \$602.50, using a 98%
confidence level.
For each of these three tests a) state your null and alternative hypotheses (2), b) test each one using
a test ratio or a critical value for the proportion (2) and c) find a p-value for the null hypotheses (3). Label
each part clearly so that I know which is I, II and III and a), b) c). Make sure that I know where the ‘reject’
zone is.
d) Using the proportion you found above, how large a sample would you need to estimate a 2-sided 98%
confidence interval for the proportion with and error of at most .001? Assume that your sample is of that
size and show that the confidence interval has an error of at most .001. (3)
[10]
e) (Extra credit) Assume that you are testing the hypothesis that (II) more than 60% demand over \$602.50,
find the power of the test if you use a sample of 30 the true proportion is 70% (3)
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Problem 2: Assume that the underlying data for problem 1 is not Normal and using the data for problem 1
test the following three hypotheses: I) that the median demand is \$602.50, II) that median demand is more
than \$602.50 and III) that the median demand is less than \$602.50, using a 98% confidence level. a) state
your null and alternative hypotheses and the hypotheses that you will actually test for each of the 3 tests
(3), b) test each one using a test ratio or a critical value (3), c) find a p-value for the 2-sided test and explain
whether and why it would lead to a rejection of the null hypothesis at the 95% confidence level (1), d)
(extra credit) Show explicitly what the conclusion in c) would be if the sample of 30 came from a
population of 60. (1) e) (extra credit) Find a two sided confidence interval for the median (2)
[17]
Problem 3: a) Find the sample mean and sample standard deviation for the data in Problem 1 (1)
b) Test the hypothesis that the mean is 602.50 using critical values for the sample mean, first stating your
hypotheses clearly. Use a 98% confidence level (2)
c) Test the hypothesis in b) using a test ratio. Find an approximate p-value and state and explain whether
this will lead to a rejection of the null hypothesis if we continue to use a 98% confidence level. (2)
d) Using the test ratio you found in c) find a p-value for the null hypothesis that the mean is at most 602.50
(1)
e) Using the test ratio you found in c) find a p-value for the null hypothesis that the mean is at least 602.50
(1)
f) Test the null hypothesis that the mean is at most 602.50 using an appropriate confidence interval (1)
g) Test the null hypothesis that the mean is at least 602.50 using an appropriate confidence interval (1)
[26]
Problem 4: Assume that the population standard deviation is known to be 30 but that we are still working
with a problem like Problem 3. (98% confidence level, sample of 30.) Do either Problem 4.1 or Problem
4.2. Make sure that I know which one!
Problem 4.1. a) Find a critical value for the sample mean if we are testing whether the population
mean is below 30. Clearly state your null and alternative hypotheses (2)
b) Assume that the sample mean is 30 minus the second to last digit of your student number. (Use 10 if this
digit is zero.) Find a p-value for your null hypothesis. (1)
c) Create a power curve for the test (6)
Problem 4.2. a) Find critical values for the sample mean if we are testing whether the population
mean is 30. Clearly state your null and alternative hypotheses (2)
b) Assume that the sample mean is 30 minus the second to last digit of your student number. (Use 10 if this
digit is zero.) find a p-value for your null hypothesis. (1)
c) Create a power curve for the test (8)
[37]
Problem 5: In problem 4 we assumed that the population standard deviation is 30.
a) Do a 98% confidence interval for the mean using the mean that you found in Problem 3 and assuming
that our sample of 30 came from a population of 300. (2)
b) How large a sample would we need if we wanted to make the error term no more than 1 and the
sample came from an infinite population? (2)
c) Using a 98% confidence level and a sample size of 30 create a confidence interval for the population
standard deviation using your sample variance or standard deviation from Problem 3. (2)
d) Repeat c) assuming that you had a sample of 300. (2)
e) Can we say that the standard deviation is significantly different from 30 on the basis of c) and d)? (1)
f) Using the data and sample size from problem 3 can we say that the standard deviation is above 30? State
your hypotheses and do an appropriate hypothesis test. (3)
[49]
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252y0753 10/19/07 (Open in ‘Print Layout’ format)
Problem 1: Count the number of people in your sample that demand more than \$602.50 and make it into a
sample proportion. Test the following 3 hypotheses: I) that 60% demand more than \$602.50, II) that more
than 60% demand more than \$602.50 and III) that less than 60% demand more than \$602.50, using a 98%
confidence level.
For each of these three tests a) state your null and alternative hypotheses (2), b) test each one using
a test ratio or a critical value for the proportion (2) and c) find a p-value for the null hypotheses (3). Label
each part clearly so that I know which is I, II and III and a), b) c). Make sure that I know where the ‘reject’
zone is.
d) Using the proportion you found above, how large a sample would you need to estimate a 2-sided 98%
confidence interval for the proportion with and error of at most .001? Assume that your sample is of that
size and show that the confidence interval has an error of at most .001. (3)
[10]
e) (Extra credit) Assume that you are testing the hypothesis that (II) more than 60% demand over \$602.50,
find the power of the test if you use a sample of 30 the true proportion is 70% (3)
Solution: The data sets that you had are presented in order. A line divides the numbers above \$602.50 from
those below. xb is the number below 602.50. x  30  x b is the number below 602.50.
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
xb
x
V1
484
493
504
514
516
517
525
527
544
552
558
562
583
600
603
611
617
617
621
621
626
627
630
632
633
636
638
639
653
662
14
16
V2
489
498
509
519
521
522
530
532
549
557
563
567
588
605
608
616
622
622
626
626
631
632
635
637
638
641
643
644
658
667
13
17
V3
494
503
514
524
526
527
535
537
554
562
568
572
593
610
613
621
627
627
631
631
636
637
640
642
643
646
648
649
663
672
13
17
V4
499
508
519
529
531
532
540
542
559
567
573
577
598
615
618
626
632
632
636
636
641
642
645
647
648
651
653
654
668
677
13
17
V5
504
513
524
534
536
537
545
547
564
572
578
582
603
620
623
631
637
637
641
641
646
647
650
652
653
656
658
659
673
682
12
18
V6
509
518
529
539
541
542
550
552
569
577
583
587
608
625
628
636
642
642
646
646
651
652
655
657
658
661
663
664
678
687
12
18
V7
514
523
534
544
546
547
555
557
574
582
588
592
613
630
633
641
647
647
651
651
656
657
660
662
663
666
668
669
683
692
12
18
V8
519
528
539
549
551
552
560
562
579
587
593
597
618
635
638
646
652
652
656
656
661
662
665
667
668
671
673
674
688
697
12
18
V9
524
533
544
554
556
557
565
567
584
592
598
602
623
640
643
651
657
657
661
661
666
667
670
672
673
676
678
679
693
702
12
18
V10
529
538
549
559
561
562
570
572
589
597
603
607
628
645
648
656
662
662
666
666
671
672
675
677
678
681
683
684
698
707
10
20
Let p be the proportion that demand more than \$602.50.
a) State your null and alternative hypotheses (2)
 H 0 : p  .6
I) 60% demand more than \$602.50 
 H 1 : p  .6
 H 0 : p  .6
II) More than 60% demand more than \$602.50 
 H 1 : p  .6
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 H : p  .6
III) Less than 60% demand more than \$602.50  0
 H 1 : p  .6
b) Test each hypothesis using a test ratio or a critical value for the proportion (2)
The relevant formulas are in Table 3. n  30 ,   .02 , z   z.02  2.054 was found in Grass1 and the t
table says z  z.01  2.327  p 
2
x
V1
16
V2
17
V3
17
p0 q0
.6.4 

= .008  .08944
30
n
V4
17
V5
18
V6
18
V7
18
V8
18
V9
18
V10
20
x
x

.5333 .5667 .5667 .5667 .6000 .6000 .6000 .6000 .6000 .6667
n 30
Here is the slice of Table 3 for proportions.
Interval for
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
Proportion
p  p0
p  p  z 2 s p
pcv  p0  z 2  p
H 0 : p  p0
z
p
H1 : p  p0
pq
p0 q0
sp 
p 
n
n
q  1 p
q0  1  p0
p
 H : p  .6
I) 60% demand more than \$602.50  0
Critical Value: pcv  p0  z  p
2
 H 1 : p  .6
 .6  2.327 .08966   .6  .2086 Make a diagram. Draw a Normal curve centered at .6 with
rejection zones below .3914 and above .8086. None of the values of p falls into the rejection
region.
p  p0
p  .6

Test Ratio: z 
. We reject the null hypothesis unless z falls between
.08944
p
.5333  .6
 0.7458 , V2-4:
.08944
.5667  .6
.6000  .6
.6667  .6
z
 0.3723 , V5-9: z 
 0 , V10: z 
 0.7458 None of these
.08944
.08944
.08944
falls in the ‘reject’ region.
 H 0 : p  .6
II) More than 60% demand more than \$602.50 
Critical Value: pcv  p0  z  p
 H 1 : p  .6
 .6  2.054 .08966   .6  0.1842  .7842 . Make a diagram. . Draw a Normal curve centered at .6
with a rejection zone above .7842. None of our values of p falls into the rejection zone.
 z 2   z.01  2.327 and z 2  z.01  2.327 . V1: z 
Test Ratio: z 
p  p0
p

p  .6
. We reject the null hypothesis if z falls above z.02  2.054 .
.08944
None of our values of z falls into the rejection zone.
 H 0 : p  .6
III) Less than 60% demand more than \$602.50 
Critical Value: pcv  p0  z  p
 H 1 : p  .6
 .6  2.054 .08966   .6  0.1842  .4158 . Make a diagram. Draw a Normal curve centered at .6
with a rejection zone below .4158. None of our values of p falls into the rejection zone.
p  p0
p  .6

Test Ratio: z 
. We reject the null hypothesis if z falls below z .02  2.054 .
.08944
p
None of our values of z falls into the rejection zone.
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c) Find a p-value for the null hypotheses
In response to a student inquiry, I wrote the following paragraph about p-value.
We could say that to compute a value for z or t, if it is a left sided test, find the probability below your
value of z using what you know about finding Normal probabilities (if it is t approximate the probability
using the t table.) If it is a right sided test find the probability above your value of z. If it is a 2-sided test
and z is negative, proceed as you would in a left sided test and double the probability. If it is a 2 sided test
and z is positive, proceed as you would in a right sided test and double the probability.
 H : p  .6
I) 60% demand more than \$602.50  0
 H 1 : p  .6
V1: z  0.7458 , p  value  2Pz  0.7458   2.5  .2734   .4532
V2-4: z  0.3723 , p  value  2Pz  0.3723   2.5  .1443   .7114
V5-9: z  0 , p  value  2Pz  0  2.5  1.0000
V10: z  0.7458 , p  value  2Pz  0.7458   2.5  .2734   .4532
 H : p  .6
II) More than 60% demand more than \$602.50  0
 H 1 : p  .6
V1: z  0.7458 , p  value  Pz  0.7458   .5  .2734   .7734
V2-4: z  0.3723 , p  value  Pz  0.3723   .5  .1443   .6443
V5-9: z  0 , p  value  Pz  0  .5
V10: z  0.7458 , p  value  Pz  0.7458   .5  .2734   .2266
 H : p  .6
III) Less than 60% demand more than \$602.50  0
 H 1 : p  .6
V1: z  0.7458 , p  value  Pz  0.7458   .5  .2734   .2266
V2-4: z  0.3723 , p  value  Pz  0.3723   .5  .1443   .3557
V5-9: z  0 , p  value  Pz  0  .5
V10: z  0.7458 , p  value  Pz  0.7458   .5  .2734   .7734
d) Using the proportion you found above, how large a sample would you need to estimate a 2-sided 98%
confidence interval for the proportion with and error of at most .001? Assume that your sample is of that
size and show that the confidence interval has an error of at most .001. (3)
[10]
n
pqz 2
. I have not worked this out for all versions, and it is up to you to decide what
e2
confidence level you will use. A solution for Version 1 with   .01 is probably the largest result
that you could get. n 
.5333 .6667 2.328 2
.0012
 1203 .3 . The sample should be, at least 1204.
e) (Extra credit) Assume that you are testing the hypothesis that (II) more than 60% demand over \$602.50,
find the power of the test if you use a sample of 30 the true proportion is 70% (3)




.7842  .7. 

Find P  p  .7842 p  .7   P z 
 . If you answer was close to this and I didn’t give
.7.3 



30 

you credit, complain.
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Problem 2: Assume that the underlying data for problem 1 is not Normal and using the data for problem 1
test the following three hypotheses: I) that the median demand is \$602.50, II) that median demand is more
than \$602.50 and III) that the median demand is less than \$602.50, using a 98% confidence level. a) state
your null and alternative hypotheses and the hypotheses that you will actually test for each of the 3 tests
(3), b) test each one using a test ratio or a critical value (3), c) find a p-value for the 2-sided test and explain
whether and why it would lead to a rejection of the null hypothesis at the 95% confidence level (1), d)
(extra credit) Show explicitly what the conclusion in c) would be if the sample of 30 came from a
population of 60. (1) e) (extra credit) find a two sided confidence interval for the median (2)
[17]
Let p be the proportion that demand more than \$602.50. The data has been exhibited in Problem 1. We
have calculated the following.
V1
16
x
V2
17
V3
17
V4
17
V5
18
V6
18
V7
18
V8
18
V9
18
V10
20
x
x

.5333 .5667 .5667 .5667 .6000 .6000 .6000 .6000 .6000 .6667
n 30
The relevant formulas are in Table 3. n  30 ,   .02 , z   z.02  2.054 was found in Grass1 and the t
p
table says z  z.01  2.327  p 
2
p0 q0
.5.5

= .00833  .091287 .
30
n
If we check the table in the outline {252ones}, we have the correspondences below. We will use the
hypotheses about a proportion on the left.
A median
proportion
If p is the
If p is the
proportion
proportion
above  0
below  0
 H 0 :   0

 H 1 :   0
 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
 H 0 : p .5

 H 1 : p  .5
 H 0 : p .5

 H 1 : p  .5
a) State your null and alternative hypotheses and the hypotheses that you will actually test for each of the 3
tests (3)
b) Test each one using a test ratio or a critical value (3)
 H 0 :  602 .50  H 0 : p  .5
I) The median demand is \$602.50 

 H 1 :  602 .50  H 1 : p  .5
Critical Value: pcv  p0  z  p  .5  2.327 .091287   .5  .2124 . Make a diagram. Draw a
2
Normal curve centered at .5 with rejection zones below .2878 and above .7124. None of the
values of p fall into the rejection region.
Test Ratio: z 
p  p0
p

p  .5
. We reject the null hypothesis unless z falls between
.091287
.5333  .5
 0.3648 , V2-4:
.091287
.5667  .5
.6000  .5
.6667  .5
z
 0.7307 , V5-9: z 
 1.0954 , V10: z 
 1.8261 None of
.091287
.091287
.091287
these falls in the ‘reject’ region.
 z 2   z.01  2.327 and z 2  z.01  2.327 . V1: z 
16
252y0753 10/19/07 (Open in ‘Print Layout’ format)
 H :  602 .50
II) The median demand is more than \$602.50  0
 H 1 :  602 .50
Critical Value: This is a right-sided test so our critical value is
H 0 : p  .5

H 1 : p  .5
pcv  p0  z  p
 .5  2.054 .091287   .5  .1875  .6875 . Make a diagram. Draw a Normal curve centered at .5
with a rejection zone above .6875. None of the values of p fall into the rejection region.
p  p0
p  .5

Test Ratio: z 
. We reject the null hypothesis if z falls above z  
.091287
p
z.02  2.054 . V1: z  0.3648 , V2-4: z  0.7307 , V5-9: z  1.0954 , V10: z  1.8261 None of
these falls in the ‘reject’ region.
 H :  602 .50 H 0 : p  .5
III) The median demand is less than \$602.50  0

 H 1 :  602 .50 H 1 : p  .5
Critical Value: This is a right-sided test so our critical value is pcv  p0  z  p
 .5  2.054 .091287   .5  .1875  .3125 . Make a diagram. Draw a Normal curve centered at .5
with a rejection zone below .3125. None of the values of p fall into the rejection region.
p  p0
p  .5

Test Ratio: z 
. We reject the null hypothesis if z falls below  z  
.091287
p
z .02  2.054 . V1: z  0.3648 , V2-4: z  0.7307 , V5-9: z  1.0954 , V10: z  1.8261 None
of these falls in the ‘reject’ region.
Alternate formulas for this section include those below.
i. Test Ratio: With continuity correction z 

same as testing z against   z 


2

p  .5 n  p 0
p
, p 
p0 q0
2x 1  n
or z 
. This is the
n
n
2x  n
.5 
. Without continuity correction z 
. To allow for a finite
n p 
n
N n
.
N 1
 p0  1 2n  z 2 p . To make a finite population correction, multiply  p by
population, divide these by
ii. Critical Value: pcv
N n
.
N 1


 2n  z
iii. Confidence Interval: p  p  1
by

2

s p . To make a finite population correction, multiply s p
N n
.
N 1
c) Find a p-value for the 2-sided test and explain whether and why it would lead to a rejection of the null
hypothesis at the 95% confidence level (1) . See problem 1 for an explanation of p-value.
 H 0 :  602 .50  H 0 : p  .5
I) The median demand is \$602.50 

 H 1 :  602 .50  H 1 : p  .5
V1: z  0.3648 , p  value  2Pz  0.3648   2.5  .1406   .7188
V2-4: z  0.7307 , p  value  2Pz  0.7307   2.5  .2673   .4654
V5-9: z  1.0954 , p  value  2Pz  1.0954   2.5  .3621   .2758
V10: z  1.8261 p  value  2Pz  1.8261   2.5  .4664   .0672
17
252y0753 10/19/07 (Open in ‘Print Layout’ format)
Since none of these are below the significance level of 5%, none of these lead to a rejection of the
null hypothesis at a 95% confidence level.
 H :  602 .50 H 0 : p  .5
II) The median demand is more than \$602.50  0

 H 1 :  602 .50 H 1 : p  .5
V1: z  0.3648 , p  value  Pz  0.3648   .5  .1406   .4406
V2-4: z  0.7307 , p  value  Pz  0.7307   .5  .2673   .2327
V5-9: z  1.0954 , p  value  Pz  1.0954   .5  .3621   .1379
V10: z  1.8261 p  value  Pz  1.8261   .5  .4664   .0336
Since only the p-value for Version 10 is below the significance level of 5%, only in version 10 do
we reject the null hypothesis at a 95% confidence level.
 H :  602 .50 H 0 : p  .5
III) The median demand is less than \$602.50  0

 H 1 :  602 .50 H 1 : p  .5
V1: z  0.3648 , p  value  Pz  0.3648   .5  .1406   .3594
V2-4: z  0.7307 , p  value  Pz  0.7307   .5  .2673   .7673
V5-9: z  1.0954 , p  value  Pz  1.0954   .5  .3621   .8621
V10: z  1.8261 p  value  Pz  1.8261   .5  .4664   .9664
Since none of these are below the significance level of 5%, none of these lead to a rejection of the
null hypothesis at a 95% confidence level.
d) (Extra credit) Show explicitly what the conclusion in c) would be if the sample of 30 came from a
population of 60. (1)
p0 q0
60  30 .5.5

= 0.50847 .00833  .0042373  .065094
60  1
30
n
p  p0
p  .5

The test ratio is z 
and is now larger in absolute value than it was in c). We can put the
.065094
p
p 
N n
N 1
p-values for the one-sided hypotheses under the hypotheses in the table below.
Version
z-score
.5333  .5
 0.5116
.065094
.5667  .5
 0.7307
V2-4: z 
.091287
.6000  .5
 1.5362
V5-9: z 
.065094
.6667  .5
 2.5609
V10: z 
.065094
V1: z 
 H 0 :  602 .50

 H 1 :  602 .50
H 0 : p  .5

H 1 : p  .5
 H 0 :  602 .50

 H 1 :  602 .50
H 0 : p  .5

H 1 : p  .5
Pz  0.5116   .5  .1950  .3050 Pz  0.5116   .5  .1950  .6950
Pz  0.7307   .5  .2673  .7673 Pz  0.7307   .5  .2673  .2327
Pz  1.5362   .5  .4382  .9382 Pz  1.5362   .5  .4382  .0618
Pz  2.5609   .5  .4948  .9948 Pz  2.5609   .5  .4948  .0052 .
If we look at these, mentally double the smaller of the two probabilities and compare the p-values with
  .05 , we see that, though some of these p-values have fallen, the only change to our results is that for
 H 0 :  602 .50  H 0 : p  .5
Version 10 we will now reject the null hypothesis for 
as well as for the

 H 1 :  602 .50  H 1 : p  .5
right-sided test.
e) (Extra credit) Find a two sided confidence interval for the median (2) . At this point I’m unsure if there is
any significance level implied. The easiest way for me to do this is to copy out the first part of the binomial
18
252y0753 10/19/07 (Open in ‘Print Layout’ format)
table for n  30 . {bin} Recall that if we take the k th number from both the top and the bottom as our
interval. We get a significance level of   2Px  k  1 , when p  .5 . For example if n  50
  2Px  19  1  203245   .06490 ,   2Px  18  1  2.01642   .03264
  2Px  17  1  2.00767   .01534 and   2Px  16  1  200330   .00660 . If the confidence
level is to be at least 1   , the significance level must be at most  . So, if we want a 95% confidence
interval we need k  18 (and 50  k  1  33 ), for a 98% confidence level we need k  17 (and 32) and for a
99% confidence level we need k  16 (and 33). Unfortunately, we do not have a binomial table for n  30
n  1  z .2 n
30  1  1.960 30
 10 .13 , for 98% this
2
2
30  1  2.327 30
30  1  2.576 30
would be k 
 9.13 and for 99% this would be k 
 8.44 . These
2
2
are all rounded down and are paired with the number with index 30  k  1 , which takes the values 21, 22
and 23.
The confidence intervals are given on the following table.
so we must try k 
. For 95% this would be k 
95% confidence interval for the median
Index V1
10
552
to
21
626
V2
557
V3
562
V4
567
V5
572
V6
577
V7
582
V8
587
V9
592
V10
597
631
636
641
646
651
656
661
666
671
98% confidence interval for the median
Index V1
9 544
to
22 627
V2
549
V3
554
V4
559
V5
564
V6
569
V7
574
V8
579
V9
584
V10
589
632
637
642
647
652
657
662
667
672
99% confidence interval for the median
Index V1
8 527
to
23 630
V2
532
V3
537
V4
542
V5
547
V6
552
V7
557
V8
562
V9
567
V10
572
635
640
645
650
655
660
665
670
675
19
252y0753 10/19/07 (Open in ‘Print Layout’ format)
Problem 3: a) Find the sample mean and sample standard deviation for the data in Problem 1 (1)
b) Test the hypothesis that the mean is 602.50 using critical values for the sample mean, first stating your
hypotheses clearly. Use a 98% confidence level (2)
c) Test the hypothesis in b) using a test ratio. Find an approximate p-value and state and explain whether
this will lead to a rejection of the null hypothesis if we continue to use a 98% confidence level. (2)
d) Using the test ratio you found in c) find a p-value for the null hypothesis that the mean is at most 602.50
(1)
e) Using the test ratio you found in c) find a p-value for the null hypothesis that the mean is at least 602.50
(1)
f) Test the null hypothesis that the mean is at most 602.50 using an appropriate confidence interval (1)
g) Test the null hypothesis that the mean is at least 602.50 using an appropriate confidence interval (1)
[26]
The data in use is as below.
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
V1
484
653
527
600
552
662
583
544
558
525
504
611
617
621
493
562
626
633
516
639
630
617
514
638
621
603
632
627
517
636
V2
489
658
532
605
557
667
588
549
563
530
509
616
622
626
498
567
631
638
521
644
635
622
519
643
626
608
637
632
522
641
V3
494
663
537
610
562
672
593
554
568
535
514
621
627
631
503
572
636
643
526
649
640
627
524
648
631
613
642
637
527
646
V4
499
668
542
615
567
677
598
559
573
540
519
626
632
636
508
577
641
648
531
654
645
632
529
653
636
618
647
642
532
651
V5
504
673
547
620
572
682
603
564
578
545
524
631
637
641
513
582
646
653
536
659
650
637
534
658
641
623
652
647
537
656
V6
509
678
552
625
577
687
608
569
583
550
529
636
642
646
518
587
651
658
541
664
655
642
539
663
646
628
657
652
542
661
V7
514
683
557
630
582
692
613
574
588
555
534
641
647
651
523
592
656
663
546
669
660
647
544
668
651
633
662
657
547
666
V8
519
688
562
635
587
697
618
579
593
560
539
646
652
656
528
597
661
668
551
674
665
652
549
673
656
638
667
662
552
671
V9
524
693
567
640
592
702
623
584
598
565
544
651
657
661
533
602
666
673
556
679
670
657
554
678
661
643
672
667
557
676
V10
529
698
572
645
597
707
628
589
603
570
549
656
662
666
538
607
671
678
561
684
675
662
559
683
666
648
677
672
562
681
Minitab offers the summary statistics below.
Version
1
2
3
4
5
6
7
8
9
10
n
30
30
30
30
30
30
30
30
30
30
x
584.83
589.83
594.83
599.83
604.83
609.83
614.83
619.83
624.83
629.83
sx
9.91
9.91
9.91
9.91
9.91
9.91
9.91
9.91
9.91
9.91
sx
Q1
Median
Q3
54.26
54.26
54.26
54.26
54.26
54.26
54.26
54.26
54.26
54.26
526.50
531.50
536.50
541.50
546.50
551.50
556.50
561.50
566.50
571.50
607.00
612.00
617.00
622.00
627.00
632.00
637.00
642.00
647.00
652.00
630.50
635.50
640.50
645.50
650.50
655.50
660.50
665.50
670.50
675.50
x x
17545
17695
17845
17995
18145
18295
18445
18595
18745
18895
2
10346275
10522475
10700175
10879375
11060075
11242275
11425975
11611175
11797875
11986075
20
252y0753 10/19/07 (Open in ‘Print Layout’ format)
a) Find the sample mean and sample standard deviation for the data in Problem 1 (1)
Solution: There isn’t a good reason to repeat the calculations here for more than one example.
So I will stick to Version 1
Index
x
1
484
234256
2
653
426409
3
527
277729
4
600
360000
5
552
304704
6
662
438244
7
583
339889
8
544
295936
9
558
311364
10
525
275625
11
504
254016
12
611
373321
13
617
380689
14
621
385641
15
493
243049
16
562
315844
17
626
391876
18
633
400689
19
516
266256
20
639
408321
21
630
396900
22
617
380689
23
514
264196
24
638
407044
25
621
385641
26
603
363609
27
632
399424
28
627
393129
29
517
267289
30
636
404496
Sum 17545 10346275
For these numbers
x
 x  17545,  x
2
 x  x 2
xx
x2
-100.833
68.167
-57.833
15.167
-32.833
77.167
-1.833
-40.833
-26.833
-59.833
-80.833
26.167
32.167
36.167
-91.833
-22.833
41.167
48.167
-68.833
54.167
45.167
32.167
-70.833
53.167
36.167
18.167
47.167
42.167
-67.833
51.167
0.000
10167.4
4646.7
3344.7
230.0
1078.0
5954.7
3.4
1667.4
720.0
3580.0
6534.0
684.7
1034.7
1308.0
8433.4
521.4
1694.7
2320.0
4738.0
2934.0
2040.0
1034.7
5017.4
2826.7
1308.0
330.0
2224.7
1778.0
4601.4
2618.0
85374.2
 10346275 and n  30 . This means that
 x  17545  584 .3333 . If we subtract this mean from all 30 numbers in the first column, we get
n
30
the 3 column which has the sum
rd
 x  x 
2
 x  x   0 . If we square the 3
rd
column, we get
 85374.2 .
Using the computational or definitional formula, we get the following.
s x2
x

2
 nx 2
n 1
10346275  30 584 .8333 2


29
 x  x 
n 1
2

85374 .17
 2943 .9367
29
2
s
2943 .9367

 98 .1312  9.9061 .
n
30
You could have gotten this using the shortcut at the beginning of the Takehome document as follows.
So s x  2943 .9367  54.25806 . s x 
 x  a    x na,  17395  305  17545
 1071575  2517545  330 52  10346275 .
x  a   x  2a x na
2
2
2
21
252y0753 10/19/07 (Open in ‘Print Layout’ format)
b) Test the hypothesis that the mean is 602.50 using critical values for the sample mean, first stating your
hypotheses clearly. Use a 98% confidence level (2)
Solution: As usual, we go back to Table 3.
Interval for
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
Mean (
  x  t 2 s x
xcv    t 2 s x
x  0
H0 :   0
t
unknown)
sx
H1 :    0
DF  n 1
s
sx 
n

29 

n 1
Note that   .02 and n  30 , so that t
=2.462. Recall that s  9.9061 and
t

2
x
.01
H 0 :   602 .50
. xcv    t s x  602 .50  2.462 9.9061   602 .50  24.39 or 578.11 to 626.89

2
H 1 :   602 .50
Make a diagram. Show an approximately Normal curve with a mean at 602.50 and shaded ‘reject’ zones
above 626.89 and below 578.11. None of the means below will fall into a ‘reject’ zone except the sample
mean for Version 10.
Version
x
1
2
3
4
5
6
7
8
9
10
584.83
589.83
594.83
599.83
604.83
609.83
614.83
619.83
624.83
629.83
c) Test the hypothesis in b) using a test ratio. Find an approximate p-value and state and explain whether
this will lead to a rejection of the null hypothesis if we continue to use a 98% confidence level. (2)
x   0 x  602 .50
t

If you wish, make an approximately Normal curve with a mean at zero and
sx
9.9061
29 
‘reject’ zones above t .01
= 2.462 and below -2.462 and compare your value of t with the ratios computed
below.
In order to find the p-value, we look at the t table to find the following for 29 degrees of freedom.
df .45 .40 .35 .30 .25 .20 .15 .10 .05 .025 .01 .005 .001
29 0.127 0.256 0.389 0.530 0.683 0.854 1.055 1.311 1.699 2.045 2.462 2.756 3.396
Values of t appear in the table below. If we compare t  1.78375 , the ratio for Version 1, with the table
29 
29 
29 
 1.699< 1.78375 < t .025
 2.045. Since t .05
 1.699, means Pt  1.699   .05 , we can
values we find t .05
conclude that .025  Pt  1.78735   .05 or, by symmetry, .025  Pt  1.78735   .05 . For a two-sided
test the p-value is the probability of getting something as extreme as or more extreme than x  584 .83 is
twice the probability Pt  1.78735  , so we can say that .05  p  value  .10 . The rest are shown on the
table below.
t comp
Location of t comp
Version
x
1
584.83
-1.78375 t .05  1.699< t comp < t .025  2.045 .05  p  value  .10
2
589.83
-1.27901 t .15  1.055< t comp < t .10  1.311 .20  p  value  .30
3
594.83
-0.77427 t .25  0.683< t comp < t .20  0.854 .40  p  value  .50
4
599.83
-0.26953 t .40  0.256< t comp < t .35  0.389 .70  p  value  .80
5
604.83
0.23521 t .45  0.127< t comp < t .40  0.256 .80  p  value  .90
Approximate p-value
29
29
29
29
29
29
29
29
29
29
22
252y0753 10/19/07 (Open in ‘Print Layout’ format)
29
29
29
29
29
29
29
29
29
29
6
609.83
0.73995 t .25  0.683< t comp < t .20  0.854 .40  p  value  .50
7
614.83
1.24469 t .15  1.055< t comp < t .10  1.311 .20  p  value  .30
8
619.83
1.74943 t .05  1.699< t comp < t .025  2.045 .05  p  value  .10
9
624.83
2.25417 t .025  2.045< t comp < t .01  2.462 .02  p  value  .05
10
629.83
2.75891 t .005  2.756< t comp < t .001  3.396 .002  p  value  .01
If   .02 , only Version 10 would give a rejection of the null hypothesis.
d) Using the test ratio you found in c) find a p-value for the null hypothesis that the mean is at most 602.50
H :   602 .50
(1) We are now testing  0
. This is a right-sided test.
H 1 :   602 .50
Values of t are repeated in the table below. If we compare t  1.78375 , the ratio for Version 1, with the
29 
29 
29 
table values we find t .05
 1.699< 1.78375 < t .025
 2.045. Since t .05
 1.699, means Pt  1.699   .05 ,
we can conclude that .025  Pt  1.78735   .05 or, by symmetry, .025  Pt  1.78735   .05 . For a
right-sided test the p-value is the probability of getting something as high as or higher than x  584 .83 is
the probability Pt  1.78735  , so we can say that .95  p  value  .975 . The rest are shown on the table
below. Note that if the significance level is .02, we will definitely reject the null hypothesis in Version 10
and probably in Version 9.
Version
Location of t comp
Approximate p-value
x
t comp
29
29
29
29
29
29
29
29
1
584.83
-1.78375 t .05  1.699< t comp < t .025  2.045 .95  p  value  .975
2
589.83
-1.27901 t .15  1.055< t comp < t .10  1.311 .85  p  value  .90
3
594.83
-0.77427 t .25  0.683< t comp < t .20  0.854 .75  p  value  .90
4
599.83
-0.26953 t .40  0.256< t comp < t .35  0.389 .60  p  value  .75
5
604.83
0.23521 t .45  0.127< t comp < t .40  0.256 .40  p  value  .45
6
609.83
0.73995 t .25  0.683< t comp < t .20  0.854 .20  p  value  .25
7
614.83
1.24469 t .15  1.055< t comp < t .10  1.311 .10  p  value  .15
8
619.83
1.74943 t .05  1.699< t comp < t .025  2.045 .025  p  value  .05
9
624.83
2.25417 t .025  2.045< t comp < t .01  2.462 .025  p  value  .01
10
629.83
2.75891 t .005  2.756< t comp < t .001  3.396 .001  p  value  .005
29
29
29
29
29
29
29
29
29
29
29
29
e) Using the test ratio you found in c) find a p-value for the null hypothesis that the mean is at least 602.50
(1)
H 0 :   602 .50
We are now testing 
. This is a left -sided test.
H 1 :   602 .50
Values of t are repeated in the table below. If we compare t  1.78375 , the ratio for Version 1, with the
29 
29 
29 
 1.699< 1.78375 < t .025
 2.045. Since t .05
 1.699, means Pt  1.699   .05 ,
table values we find t .05
we can conclude that .025  Pt  1.78735   .05 or, by symmetry, .025  Pt  1.78735   .05 . For a
right-sided test the p-value is the probability of getting something as low as or lower than x  584 .83 is the
probability Pt  1.78735  , so we can say that .025  p  value  .05 . The rest are shown on the table
below. Note that your p-values for d) and e) should add to 1.
23
252y0753 10/19/07 (Open in ‘Print Layout’ format)
t comp
Location of t comp
Version
x
1
584.83
-1.78375 t .05  1.699< t comp < t .025  2.045 .025  p  value  .05
2
589.83
-1.27901 t .15  1.055< t comp < t .10  1.311 .10  p  value  .15
3
594.83
-0.77427 t .25  0.683< t comp < t .20  0.854 .20  p  value  .25
4
599.83
-0.26953 t .40  0.256< t comp < t .35  0.389 .35  p  value  .40
5
604.83
0.23521 t .45  0.127< t comp < t .40  0.256 .55  p  value  .60
6
609.83
0.73995 t .25  0.683< t comp < t .20  0.854 .75  p  value  .80
7
614.83
1.24469 t .15  1.055< t comp < t .10  1.311 .85  p  value  .90
8
619.83
1.74943 t .05  1.699< t comp < t .025  2.045 .95  p  value  .975
9
624.83
2.25417 t .025  2.045< t comp < t .01  2.462 .975  p  value  .99
10
629.83
2.75891 t .005  2.756< t comp < t .001  3.396 .995  p  value  .999
Approximate p-value
29
29
29
29
29
29
29
29
29
29
29
29
29
29
29
29
29
29
29
29
Note that if the significance level is .02, we will never reject the null hypothesis.
f) Test the null hypothesis that the mean is at most 602.50 using an appropriate confidence interval (1)
29 
I’m surprised that no one called me on this. To do this correctly you need t .02
 2.150, which is not on any
H :   602 .50
of your tables. Here you get points for thinking, so I’ll see what you did.  0
. Recall
H 1 :   602 .50
  .02 , n  30 , s x  9.9061 and the two-sided formula is   x  t s x , which becomes   x  t s x
 x  2.150 9.9061   x  21 .30 . For the results see the table after g).
2
g) Test the null hypothesis that the mean is at least 602.50 using an appropriate confidence interval (1)
H 0 :   602 .50
29 
 2.150, n  30 , s x  9.9061 and the two-sided formula is
Recall   .02 , t .02

H
:


602
.
50
 1
  x  t 2 s x , which becomes   x  t s x  x  2.150 9.9061   x  21.30 . The intervals in both f) and
g) contain the sample mean.
H 0 :   602 .50 H 0 :   602 .50
  x  t s x
  x  t s x
Version x
1 584.83
= 563.53
= 606.13
2 589.83
= 568.53
= 611.13
3 594.83
= 573.53
= 616.13
4 599.83
= 578.53
= 621.13
5 604.83
= 583.53
= 626.13
6 609.83
= 588.53
= 631.13
7 614.83
= 593.53
= 636.13
8 619.83
= 598.53
= 641.13
9 624.83
= 603.53 *
= 646.13
10 629.83
= 608.53 *
= 651.13
Note that the two starred confidence intervals contradict the null hypothesis and thus imply rejection.
24
252y0753 10/19/07 (Open in ‘Print Layout’ format)
Problem 4: Assume that the population standard deviation is known to be 30 but that we are still working
with a problem like Problem 3. (98% confidence level, sample of 30.) Do either Problem 4.1 or Problem
4.2. Make sure that I know which one!

30
Let’s start with Table 3.   .02 , n  30 ,   30 and  x 

 30  5.4772
n
30
Interval for
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
Mean (
  x  z 2  x
xcv   0  z 2  x
x  0
H0 :   0
z
known)
x
H1 :    0

x 
n
Problem 4.1. a) Find a critical value for the sample mean if we are testing whether the population
mean is below 30. Clearly state your null and alternative hypotheses (2)
b) Assume that the sample mean is 30 minus the second to last digit of your student number. (Use 10 if this
digit is zero.) Find a p-value for your null hypothesis. (1)
c) Create a power curve for the test (6)
Solution: a) We find a critical value for the sample mean if we are testing whether the population mean is
 H :   30
below 30. We state our null and alternative hypotheses (2)  0
z.02  2.054 We need a critical
 H 1 :   30
value for the mean that is below 30. We use xcv   0  z  x  30  2.054 5.4772   30  11.250  18.750
b) We assume that the sample mean is 30 minus the second to last digit of our student number. (Use 10 if
x  0
x  30

this digit is zero.) We find a p-value for our null hypothesis. z calc 
will be our test ratio
x
5.4772
and we will calculate p  value  Pz  z calc  . For example if the mean is 29, we compute
29  30
 0.18 . Using the Normal table we find p  value  Pz  0.18   .5  .0714  .4286 . The
5.4772
values below were computer generated. Yours should be close.
Version x
z calc Pz  z calc 
z calc 
1
2
3
4
5
6
7
8
9
10
29
28
27
26
25
24
23
22
21
20
-0.18258
-0.36515
-0.54773
-0.73030
-0.91288
-1.09545
-1.27803
-1.46060
-1.64318
-1.82575
0.427566
0.357500
0.291940
0.232603
0.180654
0.136660
0.100620
0.072063
0.050173
0.033944
c) We create a power curve for the test. We do not reject the null hypothesis if our sample mean is above
 H 0 :   30
and that we need a power curve for all
x cv  18 .750 . Remember that our hypotheses are 
 H 1 :   30
possible values of  that are below 30. The distance between 30 and the critical value is 30 – 18.750 =
11.25, half of that is 5.62, which we can round to 6. Let’s try using 30, 24, 18.75, 12 and 6 as 1 . We will
18 .750  1 

 . Make a diagram. Show a Normal curve with a mean of
compute Px  18 .750 1   P z 
5.4772 

30 and shade a ‘reject’ zone below 18.750. On the same diagram make a second Normal curve of the same
size as the first one with a mean at a value of 1 and shade a ‘do not reject’ zone that includes the entire
25
252y0753 10/19/07 (Open in ‘Print Layout’ format)


area under the second curve above 18.750. For 1  29 this becomes   P x  18.750 1  24
18 .750  24 

 P z 
  Pz  0.96   .5  .3315  .8315 . If we let the computer do the dirty work, we get
5.4772 

the following.
18 .750  1
z calc 
Point 1
  Pz  z calc  power 1  
5.4772
1
2
3
4
5
30.00
24.00
18.75
12.00
6.00
-2.05397
-0.95852
0.00000
1.23238
2.32783
0.980011
0.831099
0.500000
0.108903
0.009961
0.019989
0.168901
0.500000
0.891097
0.990039
As was explained in class, you do not need to do the calculations for points 1 and 3 since the power for
1   0 is always equal to the significance level and the power at 1  xcv is always .5. Graph the power
Problem 4.2. a) Find critical values for the sample mean if we are testing whether the population
mean is 30. Clearly state your null and alternative hypotheses (2)
b) Assume that the sample mean is 30 minus the second to last digit of your student number. (Use 10 if this
digit is zero.) find a p-value for your null hypothesis. (1)
c) Create a power curve for the test (8)
[37]
Solution: a) We find critical values for the sample mean if we are testing whether the population mean is
 H 0 :   30
30. We state our null and alternative hypotheses 
 x  5.4772 ,   .02 and z.01  2.327 .
 H 1 :   30
We need critical values for the sample mean that are both above and below 30. These are xcv   0  z  x
2
 30  2.327 5.4772   30  12.745 or 17.255 and 42.745. We reject the null hypothesis if the sample mean
does not fall between these values.
b) We assume that the sample mean is 30 minus the second to last digit of our student number. (We use 10
x  0
x  30

if this digit is zero.) and find a p-value for our null hypothesis. z calc 
will be our test
x
5.4772
ratio and we will calculate p  value  2Pz  z calc  (since all our values of x are to the left of the alleged
29  30
 0.18 . Using the Normal table we
5.4772
find p  value  2Pz  0.18   2(.5  .0714 )  2(.4286 )  .8572 . If we let the computer do the work, we
get the table below. Your results should be similar.
Version x
z calc 2Pz  z calc 
mean.). For example if the mean is 29, we compute z calc 
1
2
3
4
5
6
7
8
9
10
29
28
27
26
25
24
23
22
21
20
-0.18258
-0.36515
-0.54773
-0.73030
-0.91288
-1.09545
-1.27803
-1.46060
-1.64318
-1.82575
0.855131
0.714999
0.583881
0.465207
0.361308
0.273319
0.201241
0.144125
0.100347
0.067888
c) We create a power curve for the test . We do not reject the null hypothesis if the sample mean lies
between 17.255 and 42.745. The alleged mean is 30 and the distance between 30 and the critical values is
12.745, half of which we can round to 6.5. We need the power for every value of the mean. Let’s try using
30, 36.5, 42.745, 49.5 and 56 for 1 on the top side of 30 and 30, 23.5, 17.255, 10.5 and 4 on the bottom
26
252y0753 10/19/07 (Open in ‘Print Layout’ format)
42 .745  1 
 17 .255  1
z
 . For example
side of 30. We will compute P17 .255  x  42 .745 1   P
5.4772 
 5.4772
42 .745  36 .5 
 17 .255  36 .5
z
if 1  36 .5 , we find   P17 .255  x  42 .745 1  36 .5  P

5.4772
 5.4772

 P3.51  1.14   .4998  .3729  .8727 .
The table that follows is computer generated. Because of rounding error in the standard deviation only the
first four significant figures of the operating characteristic   and the power columns should be taken
seriously, but your results should be very close to these.
17 .255  1
42 .745  1
z calc1 
z calc2 
Point 1
  Pz calc1  z  z calc2  power 1  
5.4772
5.4772
1
2
3
4
5
6
7
8
9
4.000
10.500
17.255
23.500
30.000
36.500
42.745
49.500
56.000
2.42003
1.23329
0.00000
-1.14018
-2.32692
-3.51366
-4.65384
-5.88713
-7.07387
7.07387
5.88713
4.65384
3.51366
2.32692
1.14018
0.00000
-1.23329
-2.42003
0.007760
0.108733
0.499998
0.872674
0.980030
0.872674
0.499998
0.108733
0.007760
0.992240
0.891267
0.500002
0.127326
0.019970
0.127326
0.500002
0.891267
0.992240
Of course, this is much less work than it looks like. Only points 1, 2 and 4 need to be computed. Note that
points 3 and 7 are at critical values and give powers of .5 and that point 5 is the null hypothesis mean and
gives a power equal to the significance level (2%). Also the power for the points 9 through 6 is identical to
the power for points 1 through 4, so that only three computations are necessary to compute the operating
characteristic curve.
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252y0753 10/19/07 (Open in ‘Print Layout’ format)
Problem 5: In problem 4 we assumed that the population standard deviation is 30.
a) Do a 98% confidence interval for the mean using the mean that you found in Problem 3 and assuming
that our sample of 30 came from a population of 300. (2)
b) How large a sample would we need if we wanted to make the error term no more than 1 and the
sample came from an infinite population? (2)
c) Using a 98% confidence level and a sample size of 30 create a confidence interval for the population
standard deviation using your sample variance or standard deviation from Problem 3. (2)
d) Repeat c) assuming that you had a sample of 300. (2)
e) Can we say that the standard deviation is significantly different from 30 on the basis of c) and d)? (1)
f) Using the data and sample size from problem 3 can we say that the standard deviation is above 30? State
your hypotheses and do an appropriate hypothesis test. (3)
[49]
Solution:
Interval for
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
Mean (
x  0
  x  t 2 s x
xcv    t 2 s x
H0 :   0
t
unknown)
s
H
:



DF  n 1
x
s
1
0
sx 
n
a) Do a 98% confidence interval for the mean using the mean that you found in Problem 3 and assuming
that our sample of 30 came from a population of 300. (2)
29 
In problem 3 we found t .01
 2.462, s x2  2943.9367 , s x  54.25806 and used   x  t s x . With the
2
finite population correction, we have the following. s x 

N n
(9.9061 )
N 1
270
2943 .9367   2658 .4044  51 .5597 , so that   x  2.462 51.5597   x  126 .94 .
299
If we use the population variance at the beginning of this problem, z.01  2.327 ,  x2  900 ,  x  30 and
  x  z x . With the finite population correction we have the following.  x 
N n
(5.4772 )
N 1
270
900   812 .7090  28 .5081 , so that   x  2.327 28.5081   x  66.34 .
299
Using the means for the various versions, we can get our intervals easily.

x
Version
1
2
3
4
5
6
7
8
9
10
584.83
589.83
594.83
599.83
604.83
609.83
614.83
619.83
624.83
629.83
  x  126.94
457.89
462.89
467.89
472.89
477.89
482.89
487.89
492.89
497.89
502.89
to
to
to
to
to
to
to
to
to
to
711.77
716.77
721.77
726.77
731.77
736.77
741.77
746.77
751.77
756.77
  x  66.34
518.49
523.49
528.49
533.49
538.49
543.49
548.49
553.49
558.49
563.49
to
to
to
to
to
to
to
to
to
to
651.17
656.17
661.17
666.17
671.17
676.17
681.17
686.17
691.17
696.17
b) How large a sample would we need if we wanted to make the error term no more than 1 and the
sample came from an infinite population? (2)
Solution: n 
z 2 2
e2
. Depending on what we believe, we can use  2  900 or s x2  2943.9367 in the
 2 slot. If the confidence level is 98%, we will use z.01  2.327 and since e 2  1 , we can leave it out of
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252y0753 10/19/07 (Open in ‘Print Layout’ format)
the equation. We have either n  2.327 2 (900) = 4873.44 or n  2.3272 (2943.9367)  15941.21 and use
4874 or 15942
Problems c-f concern the variance and standard deviation and use formulas from Table 3.
Interval for
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
VarianceH 0 :  2   02
n  1s 2
n  1s 2
 .25 .5 2  02
2
2 
2  2
Small Sample
s

cv
.5 .5 2 
 02
n 1
H1: :  2   02
VarianceLarge Sample
 
s 2DF 
 z 2  2DF 
H 0 :  2   02
H1 : 
2
  02
z 
2  2  2DF   1
s cv 
 2 DF
 z  2  2 DF
c) Using a 98% confidence level and a sample size of 30 create a confidence interval for the population
standard deviation using your sample variance or standard deviation from Problem 3. (2)
Solution: Recall the following. n  30 s x2  2943.9367 or s x  54.25806 . For (30 – 1) degrees of
29
29
n  1s 2   2  n  1s 2 , and
freedom,  2 .99  14.2565 and  2 .01  49.5881, take the formula
 2
2 
substitute s x2  2943.9367 to get
29 2943 .9367
49 .5881
Or, if we take square roots, 41.49    77.38.
29 2943 .9367
 2 
14 .2565
2
1
2
or 1721.6663   2  59844.4379
d) Repeat c) assuming that you had a sample of 300. (2) For the appropriate value of z and the square root
of twice the degrees of freedom z.01  2.327 ,
s 2 DF
z   2 DF
 
2
s 2 DF
 z   2 DF
2 DF  2299   24 .45404 , take the formula
and substitute s x  54.25806 to get , for   .02 ,
2
54 .25806 24 .45404 
54 .25806 24 .45404 
 
or 49.54    59.96
2.327  24 .45404
 2.327  24 .45404
e) Can we say that the standard deviation is significantly different from 30 on the basis of c) and d)? (1)
It is enough to check our results from the confidence intervals, though a more formal test of H 0 :   30
could be done using  2 
n  1s 2
 02
and/or setting z  2 2  2DF  1 could be done. Simply put,
since 30 falls on neither interval, there is a significant difference between 30 and the standard deviation
from our sample.
f) Using the data and sample size from problem 3 can we say that the standard deviation is above 30? State
 H 0 :   30
your hypotheses and do an appropriate hypothesis test. (3) The pair of hypotheses 
are
 H 1 :   30
 H 0 :  2  900
equivalent to 
. If n  30 , so there are 29 degrees of freedom, we can use the test ratio
 H 1 :  2  900
29 2943 .9367 
 94 .8602 . If we maintain a 98% confidence level, our ‘reject’ zone will
900
29
29
be the area above  2 .02 . Our table does not give us this value, but we can say that  2 .01  49.5881
2 
n  1s 2
 02

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252y0753 10/19/07 (Open in ‘Print Layout’ format)
29
29
and  2 .025  45.7224 , so that  2 .02 must lie between them and 94.8602 must be in the ‘reject’ zone. A p29
value approach would observe that the largest number in the df = 29 column is  2 .005  52.3360 and, since


94.8602 is above this p  value  P  2  94.8602  .005    .02 . So we reject H 0 .
30
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