252solngr2-06 10/3/06
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Solution to ECO 252 Second Graded Assignment
R. E. Bove
Problem 1: Which of the following could be a null hypothesis? Which could be an alternative hypothesis?
Which could be neither? Why?
(i)   3 , (ii)   3 , (iii)   3 , (iv)   3 , (v) x  3 , (vi) x  3 , (vii) x  3 , (viii) x  3 , (ix) s  5 ,
(x) p  5 , (xi)   3 , (xii)   3 ,(xiii) p  .5.
Solution: Remember the following:
α) Only numbers like , p,  2 ,  and (the population mean, proportion, variance,
standard deviation and median) that are parameters of the population can be in a
hypothesis; x , p, s 2 , s and x.50 (the sample mean, proportion, variance, standard deviation
and median) are statistics computed from sample data and cannot be in a hypothesis because
a hypothesis is a statement about a population;
β) The null hypothesis must contain an equality;
γ) p must be between zero and one;
δ) A variance or standard deviation cannot be negative.
(i)   3 could be H 0 since it contains a parameter and an equality. H 1 would be   3 .
(ii)   3 could be H 1 since it contains a parameter but no equality. H 1 would be   3 .
(iii)   3 could be H 1 since it contains a parameter but no equality. H 1 would be   3 .
(iv)   3 , could be H 0 since it contains a parameter and an equality. H 1 would be   3 .
(v) x  3 can’t be either H 0 or H 1 since the sample mean is not a parameter.
(vi) x  3 can’t be either H 0 or H 1 since the sample mean is not a parameter.
(vii) x  3 can’t be either H 0 or H 1 since the sample mean is not a parameter.
(viii) x  3 can’t be either H 0 or H 1 since the sample mean is not a parameter.
(ix) s  5 can’t be either H 0 or H 1 since the sample standard deviation is not a parameter.
(x) p  5 could not be H 0 or H 1 since p, a parameter, cannot take values below zero or above 1.
(xi)   3 could not be H 0 or H 1 since  , a parameter, cannot take values below zero.
(xii)   3 could be H 0 since it contains a parameter and an equality. H 1 would be   3 .
(xiii) p  .5 can’t be either H 0 or H 1 since the sample proportion is not a parameter.
Learn to make  and call it ‘mu.’ It’s not a ‘u’ and you are too young to be unable
to adjust to using a Greek letter!
Problem 2: A man walks into a bar. He drinks 15 bottles of beer. These bottles are supposed to contain 12
ounces of beer with a population standard deviation of 0.2 ounces. On the basis of the man's condition when
he leaves the bar, we conclude that the sample mean for the bottles was 11.80 ounces. Test the hypothesis
that the population mean for these bottles was 12 ounces. Assume that the confidence level is 95%.
a) State your null and alternative hypotheses.
b) Find critical values for the sample mean and test the hypothesis.
c) Find a confidence interval for the sample mean and test the hypothesis.
d) Use a test ratio for a test of the sample mean
e) Find a p-value for the null hypothesis using the Normal table and use the p-value to test the
hypothesis.
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252solngr2-06 10/3/06
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Solution: a) Since the problem mentions a sample mean and the population standard deviation is given, we
go to table 3 and find the following.
Interval for
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
  x  z 2  x
Mean (
known)
x 
H0 :   0
z
H1 :    0

xcv   0  z 2  x
x  0
x
n
Never use z with s unless degrees of freedom are high! Never use t with  or p !
H :   12
a) What the problem asks is in the following hypotheses:  0
. From the problem
H 1 :   12
statement  0  12, x  11 .80 ,   0.2, n  15 and   .05 . This is a two-sided test of a mean, so we need
x 

0.2
 0.05164 and z  z.025  1.960 .
2
n
15
b) xcv   0  z  x  12  1.9600.05164  12  0.101 or 11.899 to 12.101.
2

Make a diagram. Make a Normal curve with a mean at 12. Shade the rejection zones below 11.899 and
above 12.101. Show that x  11 .80 is in the rejection zone, so reject H 0 .
c)   x  z  x  11.80  1.9600.05164  11.80  0.101 or 11.699 to 11.901.
2
Make a diagram. Make a Normal curve with a mean at 11.80. Shade the confidence interval between
11.899 and 12.101. Since  0  12 is not on this interval, reject H 0 .
d) Make a diagram. Make a Normal curve with a mean at 0. Shade the rejection zones below
x   0 11 .80  12
 z 2   z.025  1.960 and above z 2  z.025  1.960 . Compute z 

 3.873 .
x
0.05164
Show that -3.87 is in the lower rejection zone. reject H 0 .
e) Since this is a two-sided test and the value of z is below zero, p  value
 2Px  11.80   2Pz  3.87   2.5  P 3.87  z  0  2.5  .4999   .0002 . Since
p  value   .05 , reject H 0 .
Problem 3: A psychiatrist is treating a group of aborigines who are suffering from depression. Whether
justifiably or not, she considers this group a random sample of 15 taken from a very large number of
depressed individuals. The numbers below represent the measurement of the sample’s level of depression an
hour after taking the pill using a commonly used (Coolidge Axis II) scale for measuring depression.
Personalize the data as follows: add the digits of your student number to the last six numbers. Example: Ima
Badrisk has the student number 123456; so the last six numbers become {51, 52, 53, 54, 55, 56}.
52
53
58
50
53
58
55
66
53
50
50
50
50
50
50
Compute the sample standard deviation using the computational formula. (If you did this correctly on the
last assignment, just copy your work.) The doctor believes that subjects fed a sugar pill will have an average
score on the same scale of 58.73.
a) Test the validity of the doctor’s hypothesis using a confidence level of 95% and a critical value
for the sample mean. (You cannot test the validity of a hypothesis that you haven’t stated!)
b) Find an approximate p-value for the statement.
c) Will we reject the doctor’s hypothesis at a confidence level of (i) .001? (ii) .01? (iii) .10? Using
the p-value explain why.
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Ima Badrisk did the following on the last assignment.
x
index
x2
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
52
53
58
50
53
58
55
66
53
51
52
53
54
55
56
819
 x  819 ,  x  44931 ,
 x  819  54.60
x
2
2704
2809
3364
2500
2809
3364
3025
4356
2809
2601
2704
2809
2916
3025
3136
44931
n
s x2 
x
n  15
15
2
 nx 2
n 1

44931  1554 .62
14
213 .6
 15 .2571
14
The formula for the sample standard deviation is
in Table 20 of the Supplement.

s x  15 .2571  3.9060 .
s x2
15 .2571

 1.0171  1.0085
n
15
n
a) Since the problem mentions a sample mean and the population standard deviation is not given, we go to
table 3 and find the following.
Interval for
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
sx 
sx

  x  t 2 s x
Mean (
unknown)
H0 :   0
H1 :    0
DF  n 1
t
xcv   0  t  2 s x
x  0
sx
sx 
s
n
 H 0 :   58 .73
The hypotheses are 
xcv   0  t  2 s x and we have found that n  15, x  54.60 and
 H 1 :   58 .73
s  1.0085 .   .05, DF  n  1  14 and t n1  t 14  2.145 . x  58.73  2.145 1.0085 

x
2
cv
.025
 58.73  2.16 . Our critical values are 56.57 and 60.89. Make a diagram. Make an almost Normal curve
with a mean at 58.73. Shade the rejection zones below 56.57 and above 60.89. Since x  54.60 is in the
lower rejection zone, reject H 0 .
b) Compute the test ratio t 
x   0 54 .60  58 .73

 4.095 . This is on the low side of zero, so
sx
1.0085
p  value  2Px  54.60   2Pt  4.095  . Remember that DF  n  1  14 . An excerpt from the t table is
shown below.
Confidence level
df .45
.40
.35
.30
.25
.20
.15
.10
.05 .025
.01 .005 .001
14 0.128 0.258 0.393 0.537 0.692 0.868 1.076 1.345 1.761 2.145 2.624 2.977 3.787





Because of Symmetry, this tells us that P t 14  2.624  .01 , P t 14  2.977  .005 and
P t 14  3.787  .001. From this we can guess that P t 14  4.095  .001 . If we double this probability,



we can say p  value  2Pt  4.095   .002 .
c) Remember the rule on p-value: If the p-value is below the significance level, reject the null
hypothesis. We will reject the doctor’s hypothesis at confidence levels of (ii) .01 and (iii) .10 because the
p-value is below these confidence levels. For (i) .001, we cannot really be sure, so we have to say not to
reject.
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Problem 4 (Moore): 40% of Americans claim that they attend church weekly. You believe that the
proportion is below 40%. Your henchpersons follow around a random sample of 200 people for a week.
They find that x people attend church. (To calculate the number x , add the 2nd to last digit of your student
number to 60. If the second to last digit of your student number is 0, add 10. Example: Ima Badrisk has the
student number 123456; so she says that x  65 .)
a) State your null and alternative hypotheses.
b) Find a test ratio for a test of the proportion.
c) Find a p-value for this ratio and use it to test the hypothesis at a 5% significance level.
Solution: a) Since the problem mentions a proportion, we go to table 3 and find the following.
Interval for
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
Proportion
p  p  z 2 s p
pq
n
q  1 p
H 0 : p  p0
H1 : p  p0
z
p  p0
p
pcv  p0  z 2  p
p0 q0
n
q0  1  p0
a) Since we believe that the proportion is below 40 and this statement does not include an equality,
 H : p  .40
we consider this an alternate hypothesis. Thus our hypotheses are  0
.
 H 1 : p  .40
sp 
b), c) z 
p  p0
and  p 
p 
p0 q0
.40 .60 
. n  200 and p0  .40 .  p 
 .0012  .034641 .
200
n
p
This is a left-sided test since we are worrying about the number of people in the sample who attend church
being too small for the null hypothesis to be true. Thus, if our actual value of z is z1 , the p-value will be
the probability of being below z1 .
61
.305  .40
(i)
x  61, p 
 .305 , z 
 2.74 . pvalue  Px  61  P p  .305 
200
.034641
 Pz  2.74   .5  .4969  .0031 . Since .0031 is below   .05 , reject H 0 .
.310  .40
62
(ii)
 2.60 . pvalue  Px  62   P p  .310 
x  62, p 
 .310 , z 
.034641
200
 Pz  2.60   .5  .4953  .0047 . Since .0047 is below   .05 , reject H 0 .
.315  .40
63
(iii) x  63, p 
 2.45 . pvalue  Px  63  P p  .315 
 .315 , z 
.034641
200
 Pz  2.45   .5  .4929  .0071 . Since .0071 is below   .05 , reject H 0 .
.320  .40
64
(iv) x  64, p 
 2.31 . pvalue  Px  64   P p  .320 
 .320 , z 
.034641
200
 Pz  2.31  .5  .4896  .0104 . Since .0104 is below   .05 , reject H 0 .
65
.325  .40
(v)
 2.17 . pvalue  Px  65   P p  .325 
x  65, p 
 .325 , z 
.034641
200
 Pz  2.17   .5  .4850  .0150 . Since .0150 is below   .05 , reject H 0 .
.330  .40
66
(vi) x  66, p 
 2.02 . pvalue  Px  66   P p  .330 
 .330 , z 
.034641
200
 Pz  2.02   .5  .4783  .0217 . Since .0217 is below   .05 , reject H 0 .
67
.335  .40
(vii) x  67, p 
 1.88 . pvalue  Px  67   P p  .335 
 .335 , z 
.034641
200
 Pz  1.88   .5  .4699  .0301 . Since .0301 is below   .05 , reject H 0 .
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252gras1-06 9/26/06
68
.340  .40
 1.73 . pvalue  Px  68   P p  .340 
 .340 , z 
.034641
200
 Pz  1.73  .5  .4582  .0418 . Since .0418 is below   .05 , reject H 0 .
69
.345  .40
(ix) x  69, p 
 1.59 . pvalue  Px  69   P p  .345 
 .345 , z 
.034641
200
 Pz  1.59   .5  .4441  .0559 . Since .0559 is not below   .05 , do not reject H 0 .
70
.350  40
(x)
 1.44 . pvalue  Px  70   P p  .350 
x  70, p 
 .350 , z 
.034641
200
 Pz  1.44   .5  .4251  .0749 . Since .0749 is not below   .05 , do not reject H 0 .
(viii) x  68, p 
Extra Credit Problem 5:
a) Finish problem 4 by finding an appropriate confidence interval for the proportion and showing
whether it contradicts the null hypothesis.
 H : p  .40
Solution: Our hypotheses are  0
and the two-sided confidence interval formula is
 H 1 : p  .40
pq
. If the confidence interval is to go in the same direction as the alternate
n
hypothesis, it would read p  p  z s p . Since   .05 and this is a one-sided test, use z.05  1.645 . To
p  p  z 2 s p where s p 
show your evidence, make a diagram of a normal curve centered at p . Shade the area below the upper
limit in your confidence interval and show that .40 either is or is not in the interval. Even better, represent
the confidence interval by shading the area below the upper bound and the null hypothesis by shading the
area above .40. If these areas overlap, you cannot reject H 0 . The work below was computer assisted.
.305 .695 
61
 0.0010599  0.0325557 .
 .305 , s p 
200
200
p  .305  1.645 .0325557  .358554 . Since p  .358554 and p  .40 cannot both be true,
reject H 0 .
(i)
x  61, p 
62
.310 .690 
 0.0010695  0.0327032 .
 .310 , s p 
200
200
p  .310  1.645 .0327032  .363797 . Since p  .363797 and p  .40 cannot both be true, reject
H0 .
(ii)
x  62, p 
63
.315 .685 
 0.0010789  0.0328462 .
 .315 , s p 
200
200
p  .315  1.645 .0328462  .369032 . Since p  .358554 and p  .40 cannot both be true, reject
H0 .
(iii)
x  63, p 
64
.320 .680 
 0.010880  0.0329848 .
 .320 , s p 
200
200
p  .320  1.645 .0329848  .374160 . Since p  .374160 and p  .40 cannot both be true, reject
H0 .
(iv)
x  64, p 
65
.325 .675 
 0.0010969  0.0331191 .
 .325 , s p 
200
200
p  .325  1.645 .0331191  .379481 . Since p  .379481 and p  .40 cannot both be true, reject
H0 .
(v)
x  65, p 
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252gras1-06 9/26/06
.330 .670 
66
 0.0011055  0.0332491 .
 .330 , s p 
200
200
p  .330  1.645 .0332491  .384695 . Since p  .384695 and p  .40 cannot both be true, reject
H0 .
(vi)
x  66, p 
.335 .665 
67
 0.0011139  0.0333748 .
 .335 , s p 
200
200
p  .305  1.645 .0333748  .389901 . Since p  .389901 and p  .40 cannot both be true, reject
H0 .
(vii) x  67, p 
.340 .660 
68
 0.0011220  0.0334963 .
 .340 , s p 
200
200
p  .340  1.645 .0334963  .395101 . Since p  .395101 and p  .40 cannot both be true, reject
H0 .
(viii) x  68, p 
.345 .655 
69
 0.0011299  0.0336136 .
 .345 , s p 
200
200
p  .345  1.645 .0336136  .400294 . Since p  .400294 and p  .40 can both be true
(ix)
x  69, p 
(because any value of p between .400000 and .400294 would satisfy both inequalities), do not
reject H 0 .
.350 .650 
70
 0.0011375  0.0337268 .
 .350 , s p 
200
200
p  .350  1.645 .0337268  .405481 . Since p  .405481 and p  .40 can both be true
(x)
x  70, p 
(because any value of p between .400000 and .405481 would satisfy both inequalities), do not
reject H 0 .
b) Use the data in problem 3 to test the hypothesis   50.
Solution:
Interval for
Confidence
Hypotheses
Test Ratio
Interval
VarianceSmall Sample
2 
VarianceLarge Sample
 
n  1s 2
H 0 :  2   02
.25 .5 2 
H1: :  2   02
s 2DF 
H 0 :  2   02
 z 2  2DF 
H1 : 
2
  02
2 
n  1s 2
 02
z 
2  2  2DF   1
Critical Value
2
s cv

s cv 
 .25 .5 2  02
n 1
 2 DF
 z  2  2 DF
Ima Badrisk has already told us that n  15 , x  54.60 , s x2  15.2571 and s x  3.9060 . Since the degrees
of freedom are below 31, we can use the small sample formula above. We have used   .05,
DF  n  1  14 . This is a two-sided test of H 0 :   50 and H 0 :   50 .
The usual way to do this is with a Test Ratio.
n  1s 2  14 15 .2571   0.0854 . To be in the central 95% of values, this ratio should fall between
2 
 02
50 2
14
14
 2.025  26.1190 and  2 .975  5.6287 . Since the chi-squared ratio does not fall between these values,
reject H 0 .
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252gras1-06 9/26/06
If we want a critical value, the formula above is interpreted as
2
s cv

 22  02

 12 2  02
. From
n 1
14
the null hypothesis, we have  02  50 2  2500. . We already know that s x2  15.2571 ,  2 .025  26.1190
14
2
and  2 .975  5.6287 , so that our critical values are scv

n 1
and
2
s cv
5.6287 50 2
 1005 .125 and
14
26 .1190 50 2
 4664 .107 . Since s x2  15.2571 does not fall between these values, reject H 0 .
14
n  1s 2 , which is interpreted as
If we want to use a Confidence Interval, use the formula  2  2
.5 .5 2 
2
s cv

n  1s 2
 2
2
 2 
n  1s 2
12 2
with the two values of chi-squared used with the test ratio. This means
1415.2571 
1415.2571 
. This gives us 8.1179   2  37 .9483 or 2.8597    6.1602 , a
 2 
26.1190
5.6287
comparatively gigantic confidence interval which still does not include H 0 :   50 .
c) Use Minitab to check your answer to problem 4. Do this three ways
First: Enter Minitab. Use the Editor pull-down menu to enable commands. Then enter the
commands below.
Pone 200 x ;
(Replace x with the number you used.)
Test 0.4;
Alter -1;
(Makes H1 ‘less than.’)
useZ.
(Uses normal approx. to binomial)
Second: Use the Stat pull-down menu. Choose ‘Basic Stat’ and then ‘1 proportion.’
Check ‘summarized data’ and enter your n and x . Press Options. Set ‘test proportion’ as 0.4,
alternative hypothesis as ‘less than’ and check ‘Normal distribution.’ Go.
Third: Use the pull-down menu again. But before you start put x yeses and n  x noes in column
1. Uncheck ‘summarized data’ and let Minitab know that the data are in column 1 (C1). Other
options are unchanged.
Solution: See the appendix 252solngr2_06s.
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