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ECO252 QBA2
FIRST HOUR EXAM
October 17-18 2005
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. If you do not use the standard Normal table, explain!
x ~ N 2.5, 6 Make diagrams. For x draw a Normal curve with a vertical line in the center at 2.5. For z
draw a Normal curve with a vertical line in the center at zero.
3  2.5 
  19 .5  2.5
z
1. P19.5  x  3  P 
 P3.67  z  0.08 
6
6 

 P3.67  z  0  P0  z  0.08   .4999  .0319  .5318
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2.5  2.5 
 0  2. 5
z
2. P0  x  2.50   P 
  P0.42  z  0  .1628
6
 6

10 .22  2.5 

3. Px  10 .22   P  z 
  Pz  1.29   Pz  0  P0  z  1.29   .5  .4015  .0985
6


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4. x.06 There is no excuse for not being able to do this at this point, since you should have done it 3 times
in the take-home.
To find z .06 make a Normal diagram for z showing a mean at 0 and 50% above 0, divided into 6% above
z .06 and 44% below z .04 . So P0  z  z.06   .4400 The closest we can come is P0  z  1.55   .4394
or P0  z  1.56   .4406 . So use z .06  1.555 . x.06    z.06  2.5  1.555 6  11.83
For the method of creating the diagrams, go to 252y0551s
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II. (5 points-2 point penalty for not trying part a.) A dealer wishes to test the manufacture’s claim that the
Toyota Caramba gets 28 miles per gallon. The data below is found. (Recomputing what I’ve done for you
is a great way to waste time.)
a. Compute the sample standard deviation, s , of the miles per gallon. Show your work! (2)
b. Is the population mean significantly different (using a 95% confidence level) from 28 mpg?
Show your work! (3)
c. (Extra Credit) Find an approximate p value for your null hypothesis. (2)
Solution: a. Compute the sample
x
Row
x2
standard
deviation, s , of the miles per
1
28.82
830.592
gallon.
Show
your work!(2)
2
23.71
562.164
3
29.25
855.563
x 264 .02
x

 26 .402
4
25.16
633.026
n
10
5
27.60
761.760
6
27.41
751.308
x 2  nx 2 7018 .253  10 26 .402 2
2
s


7
27.23
741.473
n

1
9
8
26.42
698.016
47 .59696
9
21.75
473.063

 5.28855
10
26.67
711.289
9
sum 264.02 7018.253
s  5.28855  2.299685.
Note that excessive rounding can throw this answer way off. Using x  26 , I got


s2 
7018  10 (26 ) 2 7018  6760 258


 28 .667 and s  5.3541 .
9
9
9
b. Is the population mean significantly different (using a 95% confidence level) from 28 mpg?
Show your work! (3)
9
Given: x  26 .402 s  2.2997 , n  7 and   .05 . DF  n 1  9 and t .025
 2.262 .
We are testing H 0 :   28 . Use only one of the following!
5.28855
 0.52855  0.7272 .
10
n
10
  x  t 2 s x  26.402  2.262 0.7272  26.402  1.6449 or 24.757 to 28.047.
Confidence Interval: s x 
s

2.2997

Since 28 is on the confidence interval, it is not significantly different from the sample mean.
5.28855
 0.52855  0.7272 .
10
n
10
   t 2 s x  28  2.262 0.7272  28  1.6449 or 26.36 to 29.65.
Critical Value: s x 
xcv
s

2.2997

Since 26.402 is between the critical values, it is not significantly different from the population
mean.
Test Ratio: s x 
s

2.2997

5.28855
 0.52855  0.7272 .
10
n
10
x   0 26 .402  28
t

 2.197 . The ‘accept’ zone is between -2.262 and 2.262. Since -2.197
sx
0.7272
is between these values, do not reject the null hypothesis.
c. (Extra Credit) Find an approximate p value for your null hypothesis. (2)
9
9
 2.262 and t .05
 1.833 . Since pval  2Pt  2.197  , it is between .05
2.197 is between t .025
and .10.
For Minitab computation of variance and hypothesis test, go to 252y0551s
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III. Do all of the following problems (2 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.
For sample sizes less than 10, the sampling distribution of the mean will be approximately
normally distributed
a) Regardless of the shape of the population.
b) *Only if the shape of the population is symmetrical.
c) Only if the standard deviation of the samples are known.
d) Only if the sample is normally distributed.
2.
The width of a confidence interval estimate for a proportion will be
a) Narrower for 99% confidence than for 95% confidence.
b) Wider for a sample size of 100 than for a sample size of 50.
c) *Narrower for 90% confidence than for 95% confidence.
d) Narrower when the sample proportion is 0.50 than when the sample proportion is 0.20
3.
Which of the following would be an appropriate null hypothesis?
a) *The mean of a population is equal to 55.
b) The mean of a sample is equal to 55.
c) The mean of a population is greater than 55.
d) Only (a) and (c) are true.
4.
A Type II error is committed when
a) We reject a null hypothesis that is true.
b) We don't reject a null hypothesis that is true. Note that b) and c) are not errors.
c) We reject a null hypothesis that is false.
d) *We don't reject a null hypothesis that is false.
5.
6.
If we are performing a two-tailed test of whether  = 100, the power of the test in detecting a shift
of the mean to 105 will be ________ its power detecting a shift of the mean to 94.
a) *Less than
94 is further from 100 than 105.
b) Greater than
c) Equal to
e) Not comparable to
[10]
From a sample of 100 students we find that the mean expenditure for books is $316.40 with a
standard deviation of $43.20. You are asked to test whether the (population) mean expenditure is
above $302 using a 10% significance level.
a) What are the null and alternative hypotheses? (2)
b) What is the ‘rejection zone’ (in terms of x )? (2)
c) What is your conclusion? (2)
Solution: This is basically the revised version of Exercise 9.54 [9.52 in 9th] (9.50 in 8th edition)
that I warned you about.
a) Since the problem asks if the mean is above $302   302  , and this does not contain an equality,
it must be an alternate hypothesis. Our hypotheses are H 0 :   302 , (The average cost of
textbooks per semester at a large university is no more than $302.) and H 1 :   302 (The
average cost of textbooks per semester at a large university is more than $302.) This is a
right-sided test.
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b) Given: 0  302 .00, s  43 .20, n  100 , df  n  1  99, x  316 .40 and   .10 . So
sx 
s

43 .20

 4.320 . Note that tn 1  t ..99
10  1.290 . We need a critical value for x
n
100
above $302. Common sense says that if the sample mean is too far above $302, we will not
believe H 0 :   302 . The formula for a critical value for the sample mean is
x    t n1 s , but we want a single value above 302, so use x    t n1 s
cv
0

2
x
cv
0

x
 302 .00  1.290 4.320   307 .57 . Make a diagram showing an almost Normal curve with
a mean at 302 and a shaded 'reject' zone above 305.57.
c) Since x  316 .40 is in the ‘reject’ zone, we reject the null hypothesis and conclude that the mean is
above $302.
7.
From a sample of 12 students we find that the mean expenditure for books is $316.40 with a
sample standard deviation of $43.20. You are asked to test whether the (population) mean
expenditure is below some number. You compute a t ratio and find that it is 1.645. The p-value is
a) Exactly .05
b) Between 1.60 and 1.80
c) Between .80 and .90
This gets 2 points.
d) Between .10 and .20
e) Between .05 and .10
f) * None of the above – provide a more suitable answer.
[18]
Explanation: For this see the writeup that I announced two weeks ago. If our alternate hypothesis
is that the mean expenditure is below some number, we want Pt  1.645  . There are
11
n  1  11 degrees of freedom. The table says that t.11
10  1.363 and t .05  1.796 . Thus we can see
that .05  Pt  1.645   .10 , and we can conclude that .90  Pt  1.645   .95 . Note that this
answer is worth 3 points.
8.
An employer states that the median wage in a factory is $45000. You believe that it is lower. From
a sample of 300 employees you find that 125 have wages above $45000. The median for the
sample is $40000, the mean is $46273 and the sample standard deviation is $4001. The test should
reflect your beliefs and, if you use p in a hypothesis, you must define it. Your hypotheses are (3)
a)
 H 0 :   45000

 H 1 :   45000
b)
 H 0 :   45000

 H 1 :   45000
c)
 H 0 :   45000

 H 1 :   45000
H 0 :  45000
 H 0 : p .5
d) * 
and 
If p is the proportion above  0
H 1 :  45000
 H 1 : p .5
e)
H 0 :  45000
 H 0 : p .5
and 

H 1 :  45000
 H 1 : p .5
f)
H 0 :  45000
 H 0 : p .5
and 

H 1 :  45000
 H 1 : p .5
H 0 :  45000
 H 0 : p .5
g) * 
and 
If p is the proportion below  0
H
:


45000
 1
 H 1 : p  .5
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h)
i)
H 0 :  45000
 H : p .5
and  0

H 1 :  45000
 H 1 : p  .5
H 0 :  45000
 H : p .5
and  0

H
:


45000
 1
 H 1 : p  .5
[21]
Your belief is that the median is below 45000. This is an alternate hypothesis because it does not
contain an equality. The null hypothesis must be that the median is at least 45000. The correct hypothesis
involving the proportion can be found in the following table from the outline.
Hypotheses about
a median
Hypotheses about a proportion
If p is the proportion
If p is the proportion
above  0
below  0
 H 0 :   0

 H 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 :   0

H 1 :   0
 H 0 : p .5

 H 1 : p  .5
 H 0 : p .5

 H 1 : p  .5
9. (Extra credit) An employer states that the median wage in a factory is $45000. You believe that it
is lower. From a sample of 300 employees you find that 125 have wages above $45000. The median
for the sample is $40000, the mean is $46273 and the sample standard deviation is $4001. The test
should reflect your beliefs and, if you use p in a hypothesis, you must define it. Finish the problem.
(3)
125
 .4167 This is the most natural way to define the
Solution: The proportion above $45000 is p 
300
H 0 :  45000
 H 0 : p  .5
proportion here. So our hypotheses are 
and 
. We also will assume   .05 ,
H 1 :  45000
 H 1 : p  .5
so that z  z.05  1.645 , and we know that p 0  q 0  .5 and n  300 . This is a left-side test. From Table
3, we can use one of the methods..
Interval for
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
Proportion
p  p  z 2 s p
pq
n
q  1 p
sp 
H 0 : p  p0
H1 : p  p0
z
p  p0
p
pcv  p0  z 2  p
p0 q0
n
q0  1  p0
p 
.5.5
.4167 .5833 
 .0008333  .02887 and s p 
 .0008102  .02464
300
300
(i) Critical Value: pcv  p0  z  p  .5 1.645.02887 = .4525. The rejection region is below
We can say  p 
.4525 and the observed proportion, .4167 is in that region, so we reject the null
hypothesis.
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(ii) Test Ratio: z 
p  p0
p

.4167  .5
= -2.8853. The rejection region is below -1.645 and this
.02887
value of z is in that region, so we reject the null hypothesis.
(iii) Confidence Interval: p  p  z s p  .4167  1.645.02464 = .4572. There is no way that the
proportion can be both below .4572 and above .5 as stated in the null hypothesis, so we
reject the null hypothesis.
10. (Extra Credit – Nasty but not that hard – problem due to Prem S. Mann.) Use   .01 . A quality
control inspector is checking machines that are supposed to produce packages of cookies with a mean
weight of 32 oz and a standard deviation of .015 oz. She takes a sample of 25 packages and finds a
sample standard deviation of .029 oz. Do a one-sided test on the variance to see if the machine is out
of control. Do the test by finding a critical value for the sample variance (3) (You can do it in a more
familiar way, but you won’t get as much credit!)
[27]
Solution: H 0 :   .015
H 1:   .015 , n  25 , s  .029 and   .01 . Table 3 says
 .25 .5 2  02
n 1
n 1
, where the  2 , for a 2-sided test would be  2 
or  2  . Since this is a
1
2
n 1
2
right sided test we want a critical value above .015, we take the higher value of the two and change it
n1
24
42.9798 .015 2  0.0004029 . This means that
2
to  2    2 .01  42.9798 . So s cv

24
s cv  0.0004029  .02007 . This is a right-side test, so that the ‘reject’ region is the area under the
curve to the right of 0.02007. Since s  .029 is above this value, reject the null hypothesis and fix the
machine.
2
s cv

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252y0551 10/31/05 (Open in ‘Print Layout’ format)
ECO252 QBA2
FIRST EXAM
October 17-18 2005
TAKE HOME SECTION
Name: _________________________
Student Number and class: _________________________
IV. 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 are not usually acceptable. Neatness
counts!
1. (Prem S. Mann - modified)
Exhibit T1: According to a 1992 survey, 45% of the American population would support higher taxes to
pay for health insurance. A state government is considering offering a health insurance plan and took a
survey of 400 residents and found that 50% would support higher taxes. Use this result to test whether
the results of the 1992 survey apply in the state. Use a 99% confidence level.
Personalize these results as follows. Change 45% by replacing 5 by the last digit of your student number.
We will call your result the ‘proportion of interest.’
(Example: Seymour Butz’s student number is 976512, so he changes 45% to 42%; 42% is his proportion of
interest.)
a. State the null and alternative hypotheses in each case and find a critical value for each case. What is the
‘reject’ region? Compute a test ratio and find a p-value for the hypothesis in each case. (12)
(i) The state government wants to test that the fraction of people who favored higher taxes for health
insurance is below the proportion of interest
(ii) The state government wants to test that the fraction of people who favored higher taxes for health
insurance is above the proportion of interest.
(iii) The state government wants to test that the fraction of people who favored higher taxes for health
insurance is equal to the proportion of interest.
[12]
b. (Extra credit) Find the power of the test if the true proportion is 50% and:
(i) The alternate hypothesis is that the fraction of people is above the proportion of interest. (2)
(ii) The alternate hypothesis is that the fraction of people does not equal the proportion of interest. (2)
c. Assuming that the proportion of interest is correct, is the sample size given above adequate to find the
true proportion within .005 (1/2 of 1%)? (Don’t say yes or no without calculating the size that you actually
need!) (2)
d. If the alternate hypothesis is that the fraction of people is above the proportion of interest, create an
appropriate confidence interval for the hypothesis test. (2)
[20]
Solution: a) From the formula table we have:
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
q  1 p
0
0
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Only the solution for Version 0 is given here. The solutions for the remainder will be in 252y0551s
or 252y0551h. Note that my rule on grading parts like (ii) below is to assume that your alternate
hypothesis was correct and to ask if the critical values agree with it.
p0 q0
.40 .60 

 .0006  .024495
n
400
(i) The state government wants to test that the fraction of people who favored higher taxes for health
insurance is below the proportion of interest
H 0 : p  .40, H 1 : p  .40, n  400 , p  .50 and   .01 . z.01  2.327 .
The critical value must be below .40. pcv  p 0  z  p  .40  2.327.024495 = .3430. The ‘reject’ zone is
a) p 0  .40  p 
Version 0
.50  .40 

 Pz  4.08   1
below .3430. pval  P p  .50   P  z 
.024495 

(ii) The state government wants to test that the fraction of people who favored higher taxes for health
insurance is above the proportion of interest.
H 0 : p  .40, H 1 : p  .40, n  400 , p  .50 and   .01 . z.01  2.327 .
The critical value must be above .40. pcv  p 0  z  p  .40  2.327.024495 = .4570 . The ‘reject’ zone
.50  .40 

 Pz  4.08   0
is above .4570. pval  P p  .50   P  z 
.024495 

(iii) The state government wants to test that the fraction of people who favored higher taxes for health
insurance is equal to the proportion of interest.
[12]
H 0 : p  .40, H 1 : p  .40, n  400 , p  .50 and   .01 . z.005  2.576 .
The critical value must be on either side of .40.
pcv  p 0  z 2  p  .40  2.576.024495  .40  .0631 .
The ‘reject’ zone is below .3369 and above .4631.
.50  .40 

pval  2 P p  .50   2 P  z 
 2 Pz  4.08   0
.024495 


pq
.5.5 .5


 .025 .
n
400
20

b) (i) H 1 : p  .40,   P p  .4570 p1  .50  p 

  Pz 

.4570  .5 
 Pz  1.72   .5  .4573  .0427 Power  1  .0427  .9573
.025 
.4631  .5 
 .3369  .5
z
(ii) H 1 : p  .40,   P .3369  p  .4631 p1  .50  P 
.
025
.025 

 P6.52  z  1.48   .5  .4306  .0694 Power  1  .0694  .9306

c) p 0  .40 n 
pqz 2
e2


.40 .60 2.576 2
.005 2
pq
.5.5 .5


 .025 .
n
400
20
H 0 : p  .40, we must reject it.
d) s p 
 63703 .42 This is above 400, so the sample size is inadequate.
p  p  z s p  .5  2.327.025  .4418 If the null hypothesis is
Minitab Calculation of critical values and test ratios for
all Versions of question 1 of the Take-home exam.
————— 10/17/2005 7:01:49 PM ————————————————————
Available in 252y0551s
10
252y0551 10/31/05 (Open in ‘Print Layout’ format)
2. Customers at a bank complain about long lines and a survey shows a median waiting time of 8 minutes.
Personalize the data as follows: Use the second-to-last digit of your student number – subtract it from the
last digit of every single number. (Example: Seymour Butz’s student number is 976512, so he changes {6.7
6.3 5.7 ….} to {6.6 6.1 5.6 …}) You may drop any number in the data set that you use for this problem
that is exactly equal to 8. Use   .05 . Do not assume that the population is Normal!
Exhibit T2: Changes are made and a new survey of 32 customers is taken. The times are as follows:
6.7
7.6
8.5
6.3
7.6
8.9
5.7
7.7
9.0
5.5
7.7
9.1
4.9
8.0
9.3
7.0
8.0
9.5
7.1
8.1
9.6
7.2
8.2
9.7
7.3
8.4
9.8
7.3
8.4
10.3
7.4
8.4
a. Test that the median is below 8. State your null and alternate hypotheses clearly. (2)
b. (Extra credit) Find a 95% (or slightly more) two-sided confidence interval for the median. (2) [24]
Solution: The data sets for this problem are below. Let n be the count of the sample and x be the number
of numbers below 8.
Row
x1
x2
x3
x4
x5
x6
x7
x8
x9
x10
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
4.8
5.4
5.6
6.2
6.6
6.9
7.0
7.1
7.2
7.2
7.3
7.5
7.5
7.6
7.6
7.9
7.9
8.1
8.3
8.3
8.3
8.4
8.8
8.9
9.0
9.2
9.4
9.5
9.6
9.7
10.2
4.7
5.3
5.5
6.1
6.5
6.8
6.9
7.0
7.1
7.1
7.2
7.4
7.4
7.5
7.5
7.8
7.8
7.9
8.2
8.2
8.2
8.3
8.7
8.8
8.9
9.1
9.3
9.4
9.5
9.6
10.1
n
x
31
17
31
18
4.6
5.2
5.4
6.0
6.4
6.7
6.8
6.9
7.0
7.0
7.1
7.3
7.3
7.4
7.4
7.7
7.7
7.8
7.9
8.1
8.1
8.1
8.2
8.6
8.7
8.8
9.0
9.2
9.3
9.4
9.5
10.0
32
19
4.5
5.1
5.3
5.9
6.3
6.6
6.7
6.8
6.9
6.9
7.0
7.2
7.2
7.3
7.3
7.6
7.6
7.7
7.8
8.1
8.5
8.6
8.7
8.9
9.1
9.2
9.3
9.4
9.9
4.4
5.0
5.2
5.8
6.2
6.5
6.6
6.7
6.8
6.8
6.9
7.1
7.1
7.2
7.2
7.5
7.5
7.6
7.7
7.9
7.9
7.9
8.4
8.5
8.6
8.8
9.0
9.1
9.2
9.3
9.8
29
19
31
22
4.3
4.9
5.1
5.7
6.1
6.4
6.5
6.6
6.7
6.7
6.8
7.0
7.0
7.1
7.1
7.4
7.4
7.5
7.6
7.8
7.8
7.8
7.9
8.3
8.4
8.5
8.7
8.9
9.0
9.1
9.2
9.7
32
23
4.2
4.8
5.0
5.6
6.0
6.3
6.4
6.5
6.6
6.6
6.7
6.9
6.9
7.0
7.0
7.3
7.3
7.4
7.5
7.7
7.7
7.7
7.8
8.2
8.3
8.4
8.6
8.8
8.9
9.0
9.1
9.6
32
23
4.1
4.7
4.9
5.5
5.9
6.2
6.3
6.4
6.5
6.5
6.6
6.8
6.8
6.9
6.9
7.2
7.2
7.3
7.4
7.6
7.6
7.6
7.7
8.1
8.2
8.3
8.5
8.7
8.8
8.9
9.0
9.5
32
23
4.0
4.6
4.8
5.4
5.8
6.1
6.2
6.3
6.4
6.4
6.5
6.7
6.7
6.8
6.8
7.1
7.1
7.2
7.3
7.5
7.5
7.5
7.6
8.1
8.2
8.4
8.6
8.7
8.8
8.9
9.4
4.9
5.5
5.7
6.3
6.7
7.0
7.1
7.2
7.3
7.3
7.4
7.6
7.6
7.7
7.7
8.1
8.2
8.4
8.4
8.4
8.5
8.9
9.0
9.1
9.3
9.5
9.6
9.7
9.8
10.3
31
23
30
15
11
252y0551 10/31/05 (Open in ‘Print Layout’ format)
a) Test that the median is below 8. State your null and alternate hypotheses clearly. (2)
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 0 : p .5
 H 0 : p .5



 H 1 : p  .5
H 1 :   0
 H 1 : p .5
It seems easiest to let p be the proportion below 8. If you defined p as the proportion above .8, the p H :  8
 H : p  .5
values should be identical. Our Hypotheses are  0
and  0
. According to the outline, in
 H 1 :  8
 H 1 : p  .5
the absence of a binomial table we must use the normal approximation to the binomial distribution. If
p  p0
x
p  is our observed proportion, we use z 
. But for the sign test, p  .5 and q  1  .5  .5 .
n
p0 q0
n
So z 
x
n  .5
.25
n

x
n  .5
.5
2x  n
 x  .5n  n  x  .5n


2
.


 n  .5 
n
n
n
(For relatively small values of n , a continuity correction is advisable, so try z 
2x  1  n
, where the +
n
n
n
, and the  applies if x  . ) Values of x and n are repeated below for all ten possible
2
2
n
2x  n
2x 1  n
cases. Both z1 
and the more correct z 2 
(except when x  . ) are computed. Since
2
n
n
the alternative hypothesis is p  .5, the probabilities in the two right columns are p-values. If you used
applies if x 
z1 
p p

x
n  .5
, the values of z 1 below are correct. The exact probabilities were computed by
.25
n
Minitab in the last column.
2x  n
2x 1  n
x
n
z1 
z2 
P  z  z1 
P z  z 2 
1  F x  1
n
n
1
2
3
4
5
6
7
8
9
10
p
17
18
19
19
22
23
23
23
23
15
31
31
32
29
31
32
32
32
31
30
0.53882
0.89803
1.06066
1.67126
2.33487
2.47487
2.47487
2.47487
2.69408
0.00000
0.35921
0.71842
0.88388
1.48556
2.15526
2.29810
2.29810
2.29810
2.51447
0.00000
.5-.2054=.2946
.5-.3133=.1867
.5-.3554=.1445
.5-.4525=.0475*
.5-.4901=.0099*
.5-.4934=.0066*
.5-.4934=.0066*
.5-.4934=.0066*
.5-.4964=.0036*
.5
.5-.1404=.3596
.5-.2642=.2358
.5-.3106=.1894
.5-.4319=.0681
.5-.4846=.0154*
.5-.4893=.0107*
.5-.4893=.0107*
.5-.4893=.0107*
.5-.4940=.0060*
.5
0.360050
0.236565
0.188543
0.068023
0.014725
0.010031
0.010031
0.010031
0.005337
0.572232
Since   .05 , and the starred items are below .05, these are the cases in which we reject the hypothesis
that the median is, at least 8. It makes about as much sense to reject a hypothesis about a median
because the sample median x.50 is above or below  0 with no other information as it does to reject a
hypothesis about a mean because x is above or below  0 without more information.
12
252y0551 10/31/05 (Open in ‘Print Layout’ format)
b) (Extra credit) Find a 95% (or slightly more) two-sided confidence interval for the median. (2) I am going
to assume that you continued to use the data sets above. The outline says that an approximate value for the
index of the lower limit of the interval is k 
n  1  z .2 n
2
. In this formula z  z.025  1.96.
3
We use this formula with the following results.
k
Row n
sqrtn
k * rounded down n  k * 1
1
2
3
4
5
6
7
8
9
10
31
31
32
29
31
32
32
32
31
30
5.56776
5.56776
5.65685
5.38516
5.56776
5.65685
5.65685
5.65685
5.56776
5.47723
10.5436
10.5436
10.9563
9.7225
10.5436
10.9563
10.9563
10.9563
10.5436
10.1323
10
10
10
9
10
10
10
10
10
10
22
22
23
23
22
23
23
23
22
21
interval
7.2    8.4
7.1    8.3
7.0    8.2
6.9    8.6
6.8    8.4
6.7    7.9
6.6    7.8
6.5    7.7
6.4    7.5
7.3    8.5
13
252y0551 10/31/05 (Open in ‘Print Layout’ format)
3. Use the personalized data from Problem 2 (but do not drop any numbers.) test that the mean is below 8.
Minitab found the following the original numbers, which were in C4:
Descriptive Statistics: C4
Variable
N
Mean SE Mean StDev Minimum
C4
32
7.944
0.228 1.291
4.900
Sum of squares (uncorrected) of C4 = 2070.94
Q1
7.225
Median
8.000
Q3
8.975
Maximum
10.300
I do not know the values for your numbers, but the following (copied from last year’s exam) should be
useful:
 x  a    x   na, x  a2   x 2  2a x na2 . Your value of a is negative or
zero. Can you say that the population mean is below 8? Use   .05 .
If you remembered what we learned last semester, you could have done this calculation with little or
no work. The relevant formulas are:
For the mean Ex  a   Ex   a
For the variance Var x  a   Var x 
So that good old Seymour, whose student number is 976512, subtracts 0.1 from every number and gets
Mean  E x  0.1  E x   0.1  7.944  0.1  7.844
Variance  Var x  0.1  Var x   1.291 2
a. State your null and alternative hypotheses (1)
b. Find a critical value for the sample mean and a ‘reject’ zone. Make a diagram! (1)
c. Do you reject the null hypothesis? Use the diagram to show why. (1)
d. Find an approximate p-value for the null hypothesis. Make sure that I know where you got your results.
(1)
e. Test to see if the population standard deviation is 2. (2)
f. (Extra credit) What would the critical values be for this test of the standard deviation? (1)
g. Assume that the population standard deviation is 2, restate your critical value for the mean and create a
power curve for your test. (5)
h. Assume that the population standard deviation is 2 and find a 94% (this does not say 95%!) two sided
confidence interval for the mean. (1)
i. Using a 96% confidence level and assuming that the population standard deviation is 2, test that the mean
is below 8. (1)
j. Using a 96% confidence interval an assuming that the population standard deviation is 2, how large a
sample do you need to have an error in the mean of .005 ? (1)
[39]
Solution:
a. State your null and alternative hypotheses (1)
Since the problem asks if the mean is below   8 , and this does not contain an equality, it must
be an alternate hypothesis. Our hypotheses are
H 0 :   8 , (The average wait is at least 8 minutes.) and
H 1 :   8 (The average wait is less than 8 minutes.) This is a left-sided test.
b. Find a critical value for the sample mean and a ‘reject’ zone. Make a diagram! (1)
Given:  0  8, s  1.291, n  32, df  n  1  31, and   .05 . So
sx 
s

1.291
31
 1.696 . We need a critical value for x below 8.
 0.228 . Note that tn 1  t .10
n
32
Common sense says that if the sample mean is too far below 8, we will not believe H 0 :   8 . The
formula for a critical value for the sample mean is x    t n1 s , but we want a single value
cv
0

2
x
below 8, so use xcv   0  tn1 s x  8  1.696 0.228   7.613 . Make a diagram showing an
almost Normal curve with a mean at 8 and a shaded 'reject' zone below 7.613.
I should not have to say that if you have found a critical value, the ‘reject’ zone should be either
above or below it, depending on whether it is a right or left side test.  0 , the null hypothesis
mean is never in the ‘reject’ zone. If your critical value is x cv and x is in the ‘reject’ zone, you
14
252y0551 10/31/05 (Open in ‘Print Layout’ format)
reject the null hypothesis. Telling me that you reject a hypothesis because x cv is above or
below  0 just doesn’t cut it!
c. Do you reject the null hypothesis? Use the diagram to show why. (1)
x
7.844
7.744
7.644
7.544
7.444
7.344
7.244
7.144
7.044
7.944
Do not
Do not
Do not
Reject
Reject
Reject
Reject
Reject
Reject
Do not
reject
reject
reject
Not below
Not below
Not below
Below cv
Below cv
Below cv
Below cv
Below cv
Below cv
Not below
reject
cv
cv
cv
cv
d. Find an approximate p-value for the null hypothesis. Make sure that I know where you got your results.
n  32 means df  31 . The closest values from the t table to the computed t are shown.
 x  0 
x 8 

The alternative hypothesis, H 1 :   8 , means that the p-value is P t 
  P t 
s
0
.228 

x


Version
Nearest t Values
Approx. p-value
x
t


31
31
1
7.844 -0.68421
p-value is between .20 and .25.
t .25  0.680 t .20  0.853
2
31
31
t .15
 1.054 t .10
 1.309
7.744 -1.12281
3
7.644 -1.56140
4
7.544 -2.00000
31
t .05
 1.309

31
t
 1.309
5
7.444 -2.43860
 2.040
6
7.344 -2.87719
7
7.244 -3.31579
8
7.144 -3.75439
9
7.044 -4.19298
10
.05
31
t .025
31
t .01
31
t .005
31
t .001
31
t .001
31
t .45
7.944 -0.24561
 2.453
 2.744
 2.453
p-value is between .01 and .025
 2.453
p-value is between .005 and .01.
 3.375
p-value is between .001 and .005.
.025
31
t .01
31
t .01
31
t .001
 3.375
p-value is between .025 and .05
p-value is between .025 and .05
p-value is below .005
 3.375
 0.127
p-value is between .10 and .15.
31
t .025
 2.040

31
t
 2.040
p-value is below .005
31
t .40
 0.256
p-value is between .40 and .45.
e. Test to see if the population standard deviation is 2. s  1.291 df  31 . , (2) The outline says
To test H 0 :    0 against H1 :    0
i. Test Ratio:  2 
ii. Critical Value:
n  1s 2
2
s cv
 02

iii. Confidence Interval:
s 2DF 
z 2  2DF 
 
or for large samples z  2  2  2DF   1
 2  02
2
n 1
or
n  1s 2
 22
12 2  02
n 1
2 
or for large samples (from table 3) s cv 
n  1s 2
12 2
 2 DF
 z  2  2 DF
.
or for large samples
s 2DF 
 z 2  2DF 
15
252y0551 10/31/05 (Open in ‘Print Layout’ format)
If we use a test ratio,  2 
n  1s 2
 02

311.291 2
22
 12 .9168 . Since our degrees of freedom are too large
for the table, use z  2  2  2DF   1  212 .9168   231  1  25.8336  61
 5.0827  7.8102  2.7275 . If we are doing a 2-sided 5% test, we do not reject the null hypothesis if z
is between -1.960 and 1.960. It’s not, so we reject the null hypothesis.
If we use a confidence interval,
s 2DF 
z 2  2DF 
 
s 2DF 
 z 2  2DF 
. This becomes
1.2917.8740 
1.2917.8740 
 
or 1.034    1.718 . Since  0 is not in this interval, reject the null
 1.960  7.8740
1.96  7.7840
hypothesis .
f. (Extra credit) What would the critical values be for this test of the standard deviation? (1)
Table 3 says s cv 
 2 DF
 z  2  2 DF

2 231
 1.960  231

27.8740 
. The two critical values are
7.8740  1.960
27.8740 
27.8740 
15 .748
15 .748

 1.6014 and

 2.2663 . Since the standard deviation is
7.8740  1.960
9.834
7.8740  1.960
5.914
not between them, we reject the null hypothesis. pval  2Pz  2.73  2.5  .4968   .0064 .
g. Assume that the population standard deviation is 2, restate your critical value for the mean and create a
power curve for your test. (5). Our hypotheses are H 0 :   8 , (The average wait is at least 8 minutes.) and
H 1 :   8 (The average wait is less than 8 minutes.) This is a left-sided test. If   2 , n  32 ,
x 
2
 0.35355 and   .05 , our critical value is   1.645
2
 8  0.3762  7.4184 . We use the
32
32
following points: 8, 7.709, 7.4184, 7.127 and 6.836. Our results are as follows.
7.4184  8 

Px  7.4184   8  P  z 
 Pz  1.645  = .95
1  8
.35355 

Power  1  .95  .05
7.4184  7.709 

1  7.709
P x  7.4184  7.709  P  z 
  Pz  0.82  =.5 + 0.2939 =.7939
.35355


Power  1  .7939  .2061
7.4184  7.4184 

1  7.4184
P x  7.4184   7.4184  P  z 
  Pz  0   .5
.35355


Power  1  .5  .5
7.4184  7.127 

1  7.127
P x  7.4184   7.127  P  z 
  Pz  0.82  = .5-.2939 = .2061
.35355


Power  1  .2061  .7939
7.4184  6.836 

1  6.836
P x  7.4184   6.836  P  z 
  Pz  1.647  = .5 - .4500 = .0500
.35355


Power  1  .05  .95








h. Assume that the population standard deviation is 2 and find a 94% (this does not say 95%!) two sided
2
 0.35355 . To find
confidence interval for the mean. (1)   .06 . We know   x  z  x .  x 
2
32
z .03 make a Normal diagram for z showing a mean at 0 and 50% above 0, divided into 3% above z .03 and
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252y0551 10/31/05 (Open in ‘Print Layout’ format)
47% below z .03 . So P0  z  z.03   .4700 The closest we can come is P0  z  1.88   .4699 . So
z .03  1.88   x  z  x  x  1.88.35355  x  0.665 . Substitute your value of the sample mean.
2
i. Using a 96% confidence level and assuming that the population standard deviation is 2, test that the mean
is below 8. (1) Our hypotheses are H 0 :   8 , (The average wait is at least 8 minutes.) and H 1 :   8
(The average wait is less than 8 minutes.) This is a left-sided test. We can do this by a critical value, a test
ratio or a confidence interval.
(i)   .04 To find z .04 make a Normal diagram for z showing a mean at 0 and 50% above 0,
divided into 4% above z .04 and 46% below z .04 . So P0  z  z.04   .4600 The closest we can
come is P0  z  1.75   .4579 . So x cv   0  z  x  8  1.750.35355   7.381 If your value of
the sample mean is below 7.381, reject the null hypothesis.
x  0
x 8 
x 8

(ii) z 
. To get a p-value find P z 

 . So, for example, if
0
.
35355 
x
0.35355

7.944  8 

x  7.944 , pval  P z 
  Pz  0.16   .5  .0636  .4364 . If the p-value is below
0.35355 

x  0
x 8

.04 or if the test ratio z 
is below -1.75 , reject the null hypothesis.
x
0.35355
(iii) The one-sided confidence interval is   x  z.04 x  x  1.75.35355   x  0.6187 . If this
value is below 8, reject the null hypothesis.
j. Using a 96% confidence level and assuming that the population standard deviation is 2, how large a
sample do you need to have an error in the mean of .005 ? (1)
  .04 To find z .02 make a Normal diagram for z showing a mean at 0 and 50% above 0, divided into
2% above z .02 and 48% below z .04 . So P0  z  z.02   .4800 The closest we can come is
P0  z  2.08   .4798 From the outline n 
z 22  2
e
2

2.08 2 22
 3461 .12 . Use 3462.
.005
For Minitab Computations for Problem 3, see 252y0551s..
What follows below is excerpted from the solution to the first exam from Spring
2005.
Possible Rubric for Statistics Exams.
I have been hearing a lot about rubrics lately, and have taken a while to be assured that they are
not the materials that the third pig built his house out of. My first attempt at this came to me in a recent
assessment meeting. It is slightly expanded here.
1.
2.
3.
4.
5.
Did the student make a good effort to understand the question? This would include asking
the instructor and consulting notes and texts if he/she did not understand what was
desired.
Was the method used to solve the problem the best and most appropriate for the problem?
Was the method used correctly?
Did the student present the solution in such a way that the instructor can understand how
the student got the answers presented? This should include all formulas, equations and
tables used. Is it evident from the way the work is presented that the student understood
what he/she was doing? Is it legible?
Was the conclusion stated clearly? Was the null hypotheses rejected or not rejected? Was
a valid null and alternate hypothesis clearly stated at the beginning? What were the
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252y0551 10/31/05 (Open in ‘Print Layout’ format)
6.
implications of the conclusion for a relevant goal, for example the decision to buy a new
product?
Did the student demonstrate knowledge of the difference between sample statistics and
population parameters? Was a statistical test using sample statistics used to evaluate a
null hypothesis containing population parameters and an equality?
In view of what was said here, it is incredible that, on every exam I give, students give me confidence
intervals and tests for means when I ask for confidence intervals and tests for medians, variances and even
proportions. Check the wording on the questions that you misunderstood. Can you identify what wording in
the question made you think it was about a mean? Can you tell me what it was? It is also remarkable that
there are any people out there who do not know that proportions, probabilities and p-values (which are
probabilities) must be between one and zero. It is also amazing to me that that so many of you cannot
express the difference between t and z . In the most practical sense a value of t  comes from the t table
and must be used with s , the sample standard deviation in confidence intervals and tests for the population
mean. There are only a few other cases where we use t  and they will be discussed later in the course. On
the other hand z  , which comes only from the bottom line of the t table, but can be calculated using the
table of the standardized Normal distribution, must be used with  , the population standard deviation, in
confidence intervals and tests for the mean. z  is also used in large sample tests for the population
proportion, population mean, population standard deviation, population median and the means of the
Poisson and Binomial distribution if the correct formulas are used, but don’t push it. The Normal
distribution should not be used if more accurate methods are available. In any case, look at Things You
Should Never Do on and Exam or Anywhere Else before you do another assignment and frequently
thereafter.
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