252y0551s 10/28/05 (Open in ‘Print Layout’ format)
ECO252 QBA2
FIRST HOUR EXAM
October 17-18 2005
Name _Combined Key______
Hour of class registered _____
Class attended if different ____
This key is extremely long and shows most of what I did to put together the exam.
Show your work! Make Diagrams! Exam is normed on 50 points. Answers without reasons are not
usually acceptable.
Version 1
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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252y0551s 10/28/05 (Open in ‘Print Layout’ format)
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


2
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
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
Version 2
I. (8 points) Do all the following. If you do not use the standard Normal table, explain!
x ~ N 4.5, 6 Make diagrams. For x draw a Normal curve with a vertical line in the center at 4.5. For z
draw a Normal curve with a vertical line in the center at zero.
3  4.5 
  17 .5  4.5
z
1. P17.5  x  3  P 
 P3.67  z  0.25 
6
6 

 P3.67  z  0  P0.25  z  0  .4999  .0987  .4012
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252y0551s 10/28/05 (Open in ‘Print Layout’ format)
7  4. 5 
 4.5  4.5
z
2. P4.5  x  7.00   P 
 P0  z  0.42   .1628
6
6 

10 .22  4.5 

3. Px  10 .22   P  z 
  Pz  0.95   Pz  0  P0  z  0.95   .5  .3289  .1711
6


4
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
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  4.5  1.555 6  13.83
Generation of solutions and graphs for Normal distribution using Minitab
————— 10/18/2005 7:35:09 PM ————————————————————
Welcome to Minitab, press F1 for help.
MTB > WOpen "C:\Documents and Settings\rbove\My Documents\Minitab\252x05054.MTW".
Retrieving worksheet from file: 'C:\Documents and Settings\rbove\My
Documents\Minitab\252x0505-4.MTW'
Worksheet was saved on Tue Oct 18 2005
Results for: 252x0505-4.MTW
MTB
MTB
MTB
MTB
MTB
>
>
>
>
>
let c6 = c3-c1
let c7 = c4-c1
let c6 = c6/c2
let c7 = c7/c2
print c1-c5
#computing values of z.
#original data
Data Display
Row
1
2
3
4
5
6
mn
2.5
2.5
2.5
4.5
4.5
4.5
sd
6
6
6
6
6
6
ll
-19.50
0.00
10.22
-17.50
4.50
10.22
ul
3.0
2.5
100.0
3.0
7.0
100.0
MTB > print c3 c4 c6 c7
C5
0
0
2
0
0
2
#values of x and z
Data Display
Row
1
2
3
4
5
6
MTB
MTB
MTB
MTB
MTB
MTB
MTB
ll
-19.50
0.00
10.22
-17.50
4.50
10.22
>
>
>
>
>
>
>
ul
3.0
2.5
100.0
3.0
7.0
100.0
let c6 =
round c6
let c7 =
round c7
let c6 =
let c7 =
print c3
zl
-3.66667
-0.41667
1.28667
-3.66667
0.00000
0.95333
100*c6
c6
100*c7
c7
c6/100
c7/100
c4 c6 c7
zu
0.0833
0.0000
16.2500
-0.2500
0.4167
15.9167
#rounding z to 2 places to right of decimal point
#value of x and z
Data Display
Row
1
2
3
4
5
ll
-19.50
0.00
10.22
-17.50
4.50
ul
3.0
2.5
100.0
3.0
7.0
zl
-3.67
-0.42
1.29
-3.67
0.00
zu
0.08
0.00
16.25
-0.25
0.42
5
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
6
MTB
MTB
MTB
MTB
10.22
>
>
>
>
stack
stack
stack
erase
100.0
c3
c4
c5
c6
0.95
c6 c3
c7 c4
c5 c5
c7
15.92
#putting values of z in c3 and c4
MTB > print c1-c5
#Setup for normarea6c
Data Display
Row
1
2
3
4
5
6
7
8
9
10
11
12
mn
2.5
2.5
2.5
4.5
4.5
4.5
0.0
0.0
0.0
0.0
0.0
0.0
sd
6
6
6
6
6
6
1
1
1
1
1
1
ll
-19.50
0.00
10.22
-17.50
4.50
10.22
-3.67
-0.42
1.29
-3.67
0.00
0.95
ul
3.00
2.50
100.00
3.00
7.00
100.00
0.08
0.00
16.25
-0.25
0.42
15.92
C5
0
0
2
0
0
2
0
0
2
0
0
2
MTB > %normarea6c
#call to macro
Executing from file: normarea6c.MAC
Executing from file: NormArea6.MAC
...working...
#Graph stored in jpg format as graph 20505-401 and copied
#from storage into document.
Normal Curve Area
...working...
#402
Normal Curve Area
6
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
...working...
#403
Normal Curve Area
...working...
#404
Normal Curve Area
...working...
#405
Normal Curve Area
7
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
...working...
#406
Normal Curve Area
...working...
#407
Normal Curve Area
...working...
#408
Normal Curve Area
8
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
...working...
#409
Normal Curve Area
...working...
#410
Normal Curve Area
...working...
#411
Normal Curve Area
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252y0551s 10/28/05 (Open in ‘Print Layout’ format)
...working...
#412
Normal Curve Area
MTB >
10
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
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)
Version 1
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


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 .
s2 
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.
————— 10/12/2005 6:39:53 PM ————————————————————
Welcome to Minitab, press F1 for help.
11
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
MTB > WOpen "C:\Documents and Settings\rbove\My Documents\Minitab\252x0505-2.MTW".
Retrieving worksheet from file: 'C:\Documents and Settings\rbove\My
Documents\Minitab\252x0505-2.MTW'
Worksheet was saved on Wed Oct 12 2005
Version 1
Results for: 252x0505-2.MTW
MTB > sum x1
Sum of x1
Sum of x1 = 264.02
MTB > ssq x1
Sum of Squares of x1
Sum of squares (uncorrected) of x1 = 7018.253
MTB > print c1 c3
Data Display
Row
x1
x1sq
1
28.82
830.592
2
23.71
562.164
3
29.25
855.563
4
25.16
633.026
5
27.60
761.760
6
27.41
751.308
7
27.23
741.473
8
26.42
698.016
9
21.75
473.063
10
26.67
711.289
264.02 7018.253
MTB > describe c1
Descriptive Statistics: x1
Variable N N* Mean SE Mean StDev Minimum
Q1 Median Q3 Maximum
x1
10 0 26.402 0.727
2.300 21.750
24.798 26.950 27.905 29.250
MTB > Onet c1.
One-Sample T: x1
Variable N Mean StDev SE Mean
95% CI
x1
10 26.4020 2.2997 0.7272 (24.7569, 28.0471)
Version 2
x
Row
x2
1
23.56
555.074
2
26.09
680.688
3
27.55
759.003
4
26.92
724.686
5
29.20
852.640
6
29.02
842.160
7
22.48
505.350
8
27.45
753.503
9
25.90
670.810
10
25.81
666.156
sum 263.98 7010.070
a. Compute the sample standard deviation, s , of the miles per gallon. Show your work! (2)
12
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
x
 x  263 .98  26.398
s2 
x
2
 nx 2

7010 .07  10 26 .398 2
9
n
10
n 1
41 .52596

 4.61400 s  4.61400  2.148021 Note that excessive rounding can throw this
9
answer way off. Using x  26 , I got s 2 
7010  10 (26 ) 2 7010  6760 250


 27 .778 and
9
9
9
s  5.3541 .
b. Is the population mean significantly different (using a 95% confidence level) from 28 mpg? Show your
work! (3)
9
Given: x  26 .398 s  2.14802 , 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!
4.61400
 0.461400  0.6793 .
10
n
10
  x  t 2 s x  26.398  2.262 0.6793  26.398  1.5366 or 21.861 to 27.935.
Confidence Interval: s x 
s

2.14802

Since 28 is not on the confidence interval, it is significantly different from the sample mean.
4.61400
 0.461400  0.6793 .
10
n
10
   t 2 s x  28  2.262 0.6793  28  1.5366 or 26.46 to 29.53.
Critical Value: s x 
xcv
s

2.14802

Since 26.398 is not between the critical values, it is significantly different from the population
mean.
Test Ratio: s x 
s

2.14802

4.61400
 0.461400  0.6793 .
10
n
10
x   0 26 .398  28
t

 2.358 . The ‘accept’ zone is between -2.262 and 2.262. Since -2.197
sx
0.6793
is not between these values, reject the null hypothesis.
c. (Extra Credit) Find an approximate p value for your null hypothesis. (2)
9
9
 2.262 and t .01
 2.821 . Since pval  2Pt  2.358  , it is between .02
2.358 is between t .025
and .05.
MTB > sum c2
Sum of x2
Sum of x2 = 263.98
MTB > ssq c2
Sum of Squares of x2
Sum of squares (uncorrected) of x2 = 7010.07
MTB > describe c2
Descriptive Statistics: x2
Variable N N* Mean SE Mean StDev Minimum
Q1 Median Q3 Maximum
x2
10 0 26.398 0.679
2.148 22.480
25.248 26.505 27.918 29.200
MTB > Onet c2.
One-Sample T: x2
Variable N Mean StDev SE Mean
95% CI
x2
10 26.3980 2.1480 0.6793 (24.8614, 27.9346)
13
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
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.
Version 1
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.
c) We reject a null hypothesis that is false.
d) *We don't reject a null hypothesis that is false.
5.
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]
Version 2
1. For sample sizes greater than 100, 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 population is normally distributed.
2.
When determining the sample size for a proportion for a given level of confidence and sampling
error, the closer to 0.50 that p is estimated to be the __________ the sample size required.
a) Smaller
b) *Larger
c) Sample size is not affected.
d) The effect cannot be determined from the information given .
3.
Which of the following would be an appropriate null hypothesis?
a) The population proportion is less than 0.65.
14
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
b) The sample proportion is less than 0.65.
c) *The population proportion is at most 0.65.
d) The sample proportion is at most 0.65.
4.
5.
A Type I error is committed when
a) *We reject a null hypothesis that is true.
b) We don't reject a null hypothesis that is true.
c) We reject a null hypothesis that is false.
d) We don't reject a null hypothesis that is false
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 96.
a) *Less than
96 is closer than 105 to 100.
b) Greater than
c) Equal to
d) Not comparable to
Version 1
6. 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.
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.
Version 2
6. From a sample of 100 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 above $314 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)
[16]
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.
15
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
a) Since the problem asks if the mean is above $314   314  , and this does not contain an equality,
it must be an alternate hypothesis. Our hypotheses are H 0 :   314 , (The average cost of
textbooks per semester at a large university is no more than $314.) and H 1 :   314 (The
average cost of textbooks per semester at a large university is more than $314.) This is a
right-sided test.
b) Given:  0  314 .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 $314. Common sense says that if the sample mean is too far above $314, we will not
believe H 0 :   314 . The formula for a critical value for the sample mean is
x    t n1 s , but we want a single value above 314, so use x    t n1 s
cv
0

2
x
cv
0

x
 314 .00  1.290 4.320   319 .57 . Make a diagram showing an almost Normal curve with
a mean at 314 and a shaded 'reject' zone above 319.57.
c) Since x  316 .40 is not in the ‘reject’ zone, we do not reject the null hypothesis and cannot
conclude that the mean is above $314.
Version 1
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
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 .
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 you 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
d)
H 0 :  45000
 H 0 : p .5
and 

H 1 :  45000
 H 1 : p .5
e)
H 0 :  45000
 H 0 : p .5
and 

H 1 :  45000
 H 1 : p .5
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252y0551s 10/28/05 (Open in ‘Print Layout’ format)
f)
H 0 :  45000
 H : p .5
and  0

H 1 :  45000
 H 1 : p .5
g)
H 0 :  45000
 H : p .5
and  0

H
:


45000
 1
 H 1 : p  .5
h)
i)
From the outline.
Hypotheses about
a median
 H 0 :   0

 H 1 :   0
 H 0 :   0

H 1 :   0
 H 0 :   0

H 1 :   0
H 0 :  45000
 H : p .5
and  0

H 1 :  45000
 H 1 : p  .5
H 0 :  45000
 H : p .5
and  0

H 1 :  45000
 H 1 : p  .5
[21]
Hypotheses about a proportion
If p is the proportion
If p is the proportion
above  0
below  0
 H 0 : p .5

 H 1 : p .5
 H 0 : p .5

 H 1 : p .5
 H 0 : p .5

 H 1 : p .5
 H 0 : p .5

 H 1 : p  .5
 H 0 : p .5

 H 1 : p  .5
 H 0 : p .5

 H 1 : p  .5
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 you beliefs and, if you use p in a hypothesis, you must define it. Finish the problem. (3)
Version 2:
7. From a sample of 10 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 1.960. The p-value is
a) Between .025 and .05
b) Between .05 and .10
c) Between .95 and .975
d) Between 1.90 and 1.95
e) Exactly .025
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  1.960  . There are
n  1  9 degrees of freedom. The table says that t.9025  2.262 and t.905  1.833 . Thus we can see
that .025  Pt  1.960   .05 , and we can conclude that .95  Pt  1.645   .975 .
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 170 have wages below $45000. The median for the
sample is $41000, the mean is $47273 and the sample standard deviation is $4051. The test should
reflect you beliefs and, if you use p in a hypothesis, you must define it. Your hypotheses are (3)
a)
 H 0 :   45000

 H 1 :   45000
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252y0551s 10/28/05 (Open in ‘Print Layout’ format)
b)
 H 0 :   45000

 H 1 :   45000
c)
 H 0 :   45000

 H 1 :   45000
d)
H 0 :  45000
 H : p .5
and  0

H 1 :  45000
 H 1 : p .5
e)
H 0 :  45000
 H : p .5
and  0

H 1 :  45000
 H 1 : p .5
f)
H 0 :  45000
 H : p .5
and  0

H 1 :  45000
 H 1 : p .5
g)
H 0 :  45000
 H 0 : p .5
and 

H
:


45000
 1
 H 1 : p  .5
h)
i)
H 0 :  45000
 H 0 : p .5
and 

H 1 :  45000
 H 1 : p  .5
H 0 :  45000
 H 0 : p .5
and 

H 1 :  45000
 H 1 : p  .5
From the outline.
Hypotheses about
a median
 H 0 :   0

 H 1 :   0
 H 0 :   0

H 1 :   0
 H 0 :   0

H 1 :   0
[21]
Hypotheses about a proportion
If p is the proportion
If p is the proportion
above  0
below  0
 H 0 : p .5

 H 1 : p .5
 H 0 : p .5

 H 1 : p .5
 H 0 : p .5

 H 1 : p .5
 H 0 : p .5

 H 1 : p  .5
 H 0 : p .5

 H 1 : p  .5
 H 0 : p .5

 H 1 : p  .5
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 170 have wages below $45000. The median
for the sample is $41000, the mean is $47273 and the sample standard deviation is $4051. The test
should reflect your beliefs and, if you use p in a hypothesis, you must define it. Finish the problem.
(3)
Version 1
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
2
s cv

18
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
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.
Version 2
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 40 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  40 , s  .029 and   .01 . Table 3 says
s cv 
 2 DF
 z  2  2 DF
. If we are doing a right sided test we want a critical value above .015, we use
.015 8.84176 
0.13222

 .0203 .
 2.327  8.84176
6.5148
This is a right-side test, so that the ‘reject’ region is the area under the curve to the right of 0.0203.
Since s  .029 is above this value, reject the null hypothesis and fix the machine.
 z   z.01  2.372 .
2 DF  239   78  8.84176 s cv 
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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)
[14]
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
q0  1  p0
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252y0551s 10/28/05 (Open in ‘Print Layout’ format)
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
p0 q0
.41.59 

 .0006048  .0245917
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  .41, H 1 : p  .41, n  400 , p  .50 and   .01 . z.01  2.327 .
The critical value must be below .40. pcv  p 0  z  p  .41 2.327.0245917 = .3528. The ‘reject’ zone is
Version 1
a) p 0  .41  p 
.50  .41 

 Pz  3.65   .5  .4999  .9999
below .3528. pval  P p  .50   P  z 
.0245917 

(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.
21
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
H 0 : p  .41, H 1 : p  .41, n  400 , p  .50 and   .01 . z.01  2.327 .
The critical value must be above .41. pcv  p 0  z  p  .41 2.327.0245917 = .4672 . The ‘reject’
.50  .41 

 Pz  3.66   .5  .4999  .0001
zone is above .467270. pval  P p  .50   P  z 
.0245917 

(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]
n

400
,
H 0 : p  .41, H 1 : p  .41,
p  .50 and   .01 . z.005  2.576 .
The critical value must be on either side of .40.
pcv  p 0  z 2  p  .41 2.576.0245917  .41 .0633 .
The ‘reject’ zone is below .3467 and above .4733.
.50  .41 

pval  2 P p  .50   2 P  z 
  2 Pz  3.66   2.5  .4999   .0002
.
0245917




b) (i) H 1 : p  .41,   P p  .4672 p1  .50  p 

  Pz 
pq
.5.5 .5


 .025 .
n
400
20
.4672  .5 
 Pz  1.31  .5  .4049  .0951 Power  1  .0951  .9049
.025 

(ii) H 1 : p  .41,
.4733  .5 
 .3467  .5
z
 P 6.13  z  1.07 
.
025
.025 

 .5  .3577  .1423 Power  1  .1423  .8577
  P.3467  p  .4733 p1  .50  P 
c) p 0  .40 n 
pqz 2
e
2

.41.59 2.576 2
.005 2
pq
.5.5 .5


 .025 .
n
400
20
H 0 : p  .41, we must reject it.
d) s p 
 64207 .8 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
p0 q0
.42.58 

 .0006090  .0246779
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  .42, H 1 : p  .42, n  400 , p  .50 and   .01 . z.01  2.327 .
The critical value must be below .42. pcv  p 0  z  p  .42  2.327.0246779 = .3626. The ‘reject’ zone
Version 2
a) p 0  .42  p 
.50  .42 

 Pz  3.24   .5  .4994  .9994
is below .3626. pval  P p  .50   P  z 
.0246779 

(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  .42, H 1 : p  .42, n  400 , p  .50 and   .01 . z.01  2.327 .
The critical value must be above .42. pcv  p 0  z  p  .42  2.327.02446779 = .4774 . The ‘reject’
.50  .42 

 Pz  3.24   .5  .4994  .0006
zone is above .4774. pval  P p  .50   P  z 
.
0246779 

(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 .
22
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
The critical value must be on either side of .42.
pcv  p 0  z 2  p  .42  2.576.0246779  .40  .0636 .
The ‘reject’ zone is below .3564 and above .4836.
.50  .40 

pval  2 P p  .50   2 P  z 
 2 Pz  3.24   .0012
.0246779 



b) (i) H 1 : p  .42,   P p  .4774 p1  .50  p 

  Pz 
pq
.5.5 .5


 .025 .
n
400
20
.4774  .5 
 Pz  0.90   .5  .3159  .1841 Power  1  .1841  .8159
.025 

(ii) H 1 : p  .42,
.4836  .5 
 .3564  .5
z
 P 5.74  z  0.66 
.025 
 .025
 .5  .2454  .2546 Power  1  .2546  .7454
  P.3564  p  .4836 p1  .50  P 
c) p 0  .40 n 
pqz 2
e2

.42 .58 2.576 2
.005 2
pq
.5.5 .5


 .025 .
n
400
20
H 0 : p  .42, we must reject it.
d) s p 
 64659 .0 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
p0 q0
.43.57 

 .0006128  .0247538
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  .43, H 1 : p  .43, n  400 , p  .50 and   .01 . z.01  2.327 .
The critical value must be below .43. pcv  p 0  z  p  .43  2.327.0247538 = .3724. The ‘reject’ zone is
Version 3
a) p 0  .43  p 
.50  .43 

 Pz  2.83   .5  .4977  .9977
below .3724. pval  P p  .50   P  z 
.0247538 

(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  .43, H 1 : p  .43, n  400 , p  .50 and   .01 . z.01  2.327 .
The critical value must be above .43. pcv  p 0  z  p  .43  2.327.027538 = .4876. The ‘reject’ zone is
.50  .43 

 Pz  2.83   .5  .4977  .0023
above .4876. pval  P p  .50   P  z 
.
0247538 

(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  .43, H 1 : p  .43, 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  .43  2.576.0247538  .43  .0638 .
The ‘reject’ zone is below .3662 and above .4938.
.50  .43 

pval  2 P p  .50   2 P  z 
 2 Pz  2.83   2.0023   .0046
.0247538 



b) (i) H 1 : p  .43,   P p  .4876 p1  .50  p 

  Pz 

pq
.5.5 .5


 .025 .
n
400
20
.4876  .5 
 Pz  0.50   .5  .1915  .3085 Power  1  .3085  .6915
.025 
23
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
(ii) H 1 : p  .43,
.4938  .5 
 .3662  .5
z
 P 5.35  z  0.24 
.025 
 .025
 .5  .0948  .4052 Power  1  .4052  .5948
  P.3662  p  .4938 p1  .50  P 
c) p 0  .43 n 
pqz 2
e2

.43.57 2.576 2
 65057 .1 This is above 400, so the sample size is inadequate.
.005
pq
.5.5 .5


 .025 .
n
400
20
H 0 : p  .43, we must reject it.
d) s p 
p  p  z s p  .5  2.327.025  .4418 If the null hypothesis is
p0 q0
.44.56 

 .0006160  .0248193
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  .44, H 1 : p  .44, n  400 , p  .50 and   .01 . z.01  2.327 .
The critical value must be below .44. pcv  p 0  z  p  .44  2.327.0248193 = .3822. The ‘reject’ zone is
a) p 0  .44  p 
Version 4
.50  .44 

 Pz  2.42   .5  .4922  .9922
below .3822. pval  P p  .50   P  z 
.
0248193 

(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  .44, H 1 : p  .44, n  400 , p  .50 and   .01 . z.01  2.327 .
The critical value must be above .44. pcv  p 0  z  p  .44  2.327.0248193 = .4978. The ‘reject’ zone
.50  .44 

 Pz  2.42   .5  .4922 = .0078
is above .4978. pval  P p  .50   P  z 
.0248193 

(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  .44, H 1 : p  .44, n  400 , p  .50 and   .01 . z.005  2.576 .
The critical value must be on either side of .44.
pcv  p 0  z 2  p  .44  2.576.0248193  .44  .0639 .
The ‘reject’ zone is below .3761 and above .5039.
.50  .40 

pval  2 P p  .50   2 P  z 
 2 Pz  2.42   2.0078   .0156
.028193 


pq
.5.5 .5


 .025 .
n
400
20

b) (i) H 1 : p  .44,   P p  .4977 p1  .50  p 

  Pz 

.4977  .5 
 Pz  0.90   .5  .3159  .1841 Power  1  .1841  .8159
.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  .44 n 
pqz 2
e2


.44 .60 2.576 2
.005 2
pq
.5.5 .5


 .025 .
n
400
20
H 0 : p  .44, we must reject it.
d) s p 
 65402 .2 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
24
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
p0 q0
.45.55 

 .0006188  .0248747
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  .45, H 1 : p  .45, n  400 , p  .50 and   .01 . z.01  2.327 .
The critical value must be below .45. pcv  p 0  z  p  .45  2.327.0248747 = .3921. The ‘reject’ zone
a) p 0  .45  p 
Version 5
.50  .45 

 Pz  2.01  .5  .4778  .9778
is below .3921. pval  P p  .50   P  z 
.0248747 

(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  .45, H 1 : p  .45, n  400 , p  .50 and   .01 . z.01  2.327 .
The critical value must be above .40. pcv  p 0  z  p  .45  2.327.0248747 = .5079 . The ‘reject’
.50  .45 

 Pz  2.01  .5  .4778  .0452
zone is above .5029. pval  P p  .50   P  z 
.0248747 

(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  .45, H 1 : p  .45, n  400 , p  .50 and   .01 . z.005  2.576 .
The critical value must be on either side of .45.
pcv  p 0  z 2  p  .45  2.576.0248747  .45  .0641 .
The ‘reject’ zone is below .3859 and above .5141.
.50  .45 

pval  2 P p  .50   2 P  z 
 2 Pz  2.01  2.0452   .0904
.0248747 



b) (i) H 1 : p  .45,   P p  .5079 p1  .50  p 

  Pz 
pq
.5.5 .5


 .025 .
n
400
20
.5079  .5 
 Pz  0.32   .5  .1255  .6255 power  1  .6255  .3745
.025 

(ii) H 1 : p  .45,
.5141  .5 
 .3859  .5
z
 P 4.56  z  0.56 
.025 
 .025
 .5  .2123  .2877 Power  1  .2877  .7123
  P.3859  p  .5141 p1  .50  P 
c) p 0  .45 n 
pqz 2
e2

.45 .55 2.576 2
.005 2
 65694 .2 This is above 400, so the sample size is inadequate.
pq
.5.5 .5


 .025 . p  p  z s p  .5  2.327.025  .4418. If the null hypothesis is
n
400
20
H 0 : p  .45, p could be between .4418 and .45 so we must not reject the null hypothesis.
d) s p 
p0 q0
.46.54 

 .0006210  .0249199
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  .46, H 1 : p  .46, n  400 , p  .50 and   .01 . z.01  2.327 .
The critical value must be below .46. pcv  p 0  z  p  .46  2.327.0249199 = .4020. The ‘reject’ zone
Version 6
a) p 0  .46  p 
.50  .46 

 Pz  1.60   .5  .4452  .9452
is below .4020. pval  P p  .50   P  z 
.0249199 

25
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
(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  .46, H 1 : p  .46, n  400 , p  .50 and   .01 . z.01  2.327 .
The critical value must be above .46. pcv  p 0  z  p  .46  2.327.0249199 = .5180 . The ‘reject’
.50  .46 

 Pz  1.60   .5  .4452  .0548
zone is above .5180. pval  P p  .50   P  z 
.0249199 

(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  .46, H 1 : p  .46, n  400 , p  .50 and   .01 . z.005  2.576 .
The critical value must be on either side of .46.
pcv  p 0  z 2  p  .46  2.576.0249199  .46  .0642 .
The ‘reject’ zone is below .3958 and above .5242.
.50  .46 

pval  2 P p  .50   2 P  z 
 2 Pz  1.60   2.0548   .1096
.0249199 



b) (i) H 1 : p  .46,   P p  .5180 p1  .50  p 

  Pz 
pq
.5.5 .5


 .025 .
n
400
20
.5180  .5 
 Pz  0.72   .5  .2642  .7642 Power  1  .7642  .2358
.025 

(ii) H 1 : p  .46,
.5242  .5 
 .3958  .5
z
 P 4.17  z  0.97 
.025 
 .025
 .5  .3340  .8340 Power  1  .8340  .1660
  P.3958  p  .5242 p1  .50  P 
c) p 0  .46 n 
pqz 2
e2

.46 .54 2.576 2
.005 2
 65933 .1 This is above 400, so the sample size is inadequate.
pq
.5.5 .5


 .025 . p  p  z s p  .5  2.327.025  .4418 If the null hypothesis is
n
400
20
H 0 : p  .46, p could be between .4418 and .46, so we must not reject the null hypothesis.
d) s p 
p0 q0
.47 .53 

 .0006228  .0249550
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  .47, H 1 : p  .47 , n  400 , p  .50 and   .01 . z.01  2.327 .
The critical value must be below .47. pcv  p 0  z  p  .47  2.327.0249550 = .4119. The ‘reject’ zone
Version 7
a) p 0  .47  p 
.50  .47 

 Pz  1.20   .5  .3849  .8849
is below .4119. pval  P p  .50   P  z 
.0249550 

(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  .47, H 1 : p  .47 , n  400 , p  .50 and   .01 . z.01  2.327 .
The critical value must be above .40. pcv  p 0  z  p  .47  2.327.0249550 = .5281 . The ‘reject’
.50  .46 

 Pz  1.20   .5  .3849 = .1151
zone is above .5281. pval  P p  .50   P  z 
.0249550 

(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  .47, H 1 : p  .47 , n  400 , p  .50 and   .01 . z.005  2.576 .
26
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
The critical value must be on either side of .47. pcv  p 0  z  p  .47  2.576.0249550  .40  .0643 .
2
The ‘reject’ zone is below .4057 and above .5343.
.50  .47 

pval  2 P p  .50   2 P  z 
 2 Pz  1.20   2.1151   .2302
.024495 



b) (i) H 1 : p  .47 ,   P p  .5281 p1  .50  p 

  Pz 
pq
.5.5 .5


 .025 .
n
400
20
.5281  .5 
 Pz  1.12   .5  .3708  .8708 Power  1  .8708  .1292
.025 

(ii) H 1 : p  .47 ,
.5343  .5 
 .4057  .5
z
 P 3.77  z  1.37 
.025 
 .025
 .4999  .4147  .9146 Power  1  .9146  .0854
  P.4057  p  .5343 p1  .50  P 
c) p 0  .47 n 
pqz 2
e2

.47 .53 2.576 2
.005 2
 66118 .9 This is above 400, so the sample size is inadequate.
pq
.5.5 .5


 .025 . p  p  z s p  .5  2.327.025  .4418 If the null hypothesis is
n
400
20
H 0 : p  .47, p could be between .4418 and .47, so we must not reject the null hypothesis.
d) s p 
p0 q0
.48.52 

 .0006240  .0249800
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  .48, H 1 : p  .48, n  400 , p  .50 and   .01 . z.01  2.327 .
The critical value must be below .40. pcv  p 0  z  p  .48  2.327.0249800 = .4219. The ‘reject’ zone is
Version 8
a) p 0  .48  p 
.50  .48 

 Pz  0.80   .5  .2881  .7881
below .4219. pval  P p  .50   P  z 
.0249800 

(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  .48, H 1 : p  .48, n  400 , p  .50 and   .01 . z.01  2.327 .
The critical value must be above .40. pcv  p 0  z  p  .48  2.327.0249800 = .5381 . The ‘reject’
.50  .48 

 Pz  0.80   .5  .2119  .2881
zone is above .5381. pval  P p  .50   P  z 
.
0249800 

(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  .48, H 1 : p  .48, n  400 , p  .50 and   .01 . z.005  2.576 .
The critical value must be on either side of .48.
pcv  p 0  z 2  p  .48  2.576.0249800  .48  .0643 .
The ‘reject’ zone is below .4157 and above .5443.
.50  .48 

pval  2 P p  .50   2 P  z 
 2 Pz  0.80   .5762
.0249800 



b) (i) H 1 : p  .48,   P p  .5381 p1  .50  p 

  Pz 

pq
.5.5 .5


 .025 .
n
400
20
.5381  .5 
 Pz  1.53   .5  .4370  .9370 Power  1  .9370  .6300
.025 
27
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
(ii) H 1 : p  .48,
  P.4156  p  .5443 p1  .50
.5443  .5 
 .4156  .5
 P
z
 P 3.37  z  1.77   .4996  .4616  .9612
.025 
 .025
 P6.52  z  1.48   .5  .4306  .0694 Power  1  .9612  .0388
c) p 0  .48 n 
pqz 2
e2

.48 .52 2.576 2
.005 2
 66251 .6 This is above 400, so the sample size is inadequate.
pq
.5.5 .5


 .025 . p  p  z s p  .5  2.327.025  .4418. If the null hypothesis is
n
400
20
H 0 : p  .48, p could be between .4418 and .48, so we must not reject the null hypothesis.
d) s p 
p0 q0
.49.51

 .0006248  .0249950
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  .49, H 1 : p  .49, n  400 , p  .50 and   .01 . z.01  2.327 .
The critical value must be below .49. pcv  p 0  z  p  .49  2.327.0249950 = .4318. The ‘reject’ zone
a) p 0  .49  p 
Version 9
.50  .49 

 Pz  0.40   .5  .1554  .6554
is below .4318. pval  P p  .50   P  z 
.
0249950 

(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  .49, H 1 : p  .49, n  400 , p  .50 and   .01 . z.01  2.327 .
The critical value must be above .49. pcv  p 0  z  p  .49  2.327.0249950 = .5481 . The ‘reject’
.50  .49 

 Pz  0.40   .5  .1554  .3446
zone is above .5481. pval  P p  .50   P  z 
.0249950 

(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  .49, H 1 : p  .49, n  400 , p  .50 and   .01 . z.005  2.576 .
The critical value must be on either side of .49.
p  p 0  z 2  p  .49  2.576.0249950  .49  .0643 .
The ‘reject’ zone is below .4256 and above .5544.
.50  .49 

pval  2 P p  .50   2 P  z 
 2 Pz  0.40   2.3446   .6892
.0249950 



b) (i) H 1 : p  .49,   P p  .5481 p1  .50  p 

  Pz 
pq
.5.5 .5


 .025 .
n
400
20
..5481  .5 
 Pz  1.93   .5  .4732  .9732 Power  1  .9732  .0268
.025 

(ii) H 1 : p  .40,
.5544  .5 
 .4256  .5
z
 P 2.98  z  2.18 
.025 
 .025
 .4986  .4854  .9840 Power  1  .9840  .0160
  P.4256  p  .5544 p1  .50  P 
c) p 0  .49 n 
pqz 2
e2

.49 .512.576 2
.005 2
 66331 .2 . This is above 400, so the sample size is inadequate.
28
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
pq
.5.5 .5


 .025 . p  p  z s p  .5  2.327.025  .4418 If the null hypothesis is
n
400
20
H 0 : p  .49, p could be between .4418 and .49, so we must not reject the null hypothesis.
d) s p 
————— 10/17/2005 7:01:49 PM ————————————————————
Welcome to Minitab, press F1 for help.
Results for: 252x0505-3.MTW
MTB > WSave "C:\Documents and Settings\rbove\My Documents\Minitab\252x05053.MTW";
SUBC>
Replace.
Saving file as: 'C:\Documents and Settings\rbove\My
Documents\Minitab\252x0505-3.MTW'
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let c2=1-c1
let c3 c1*c2
let c3=c1*c2
let c3=c3/400
let c4 = sqrt(c3)
let c5 = c1*c2
let c5=6.635776 * c5
let c5 = c5/.005
let c5 = c5/.005
let c6 = c1-2.327*c3
let c6 = c1 - 2.327 * c4
let c7 = c1 + 2.327 * c4
let c8 = c1 + 2.576* c4
let c9 = c1 - 2.576 * c4
let c10= .5-c1
let c10 = c10/c4
let c11 = let c11 = c8
let c11 = c8
let c8 = c9
let c9 = c11
let c11 = c7 - .5
let c11 = c11/.025
let c12 = c8 - .5
let c12 = c12/.025
let c13 = c9 - .5
let c13 = c13/.025
print c1-c7
Data Display
p 0  .40
Row
1
2
3
4
5
6
7
8
9
10
p
0.40
0.41
0.42
0.43
0.44
0.45
0.46
0.47
0.48
0.49
q  1  p  2p 
q
0.60
0.59
0.58
0.57
0.56
0.55
0.54
0.53
0.52
0.51
p0 qo
p 
n
var
0.0006000
0.0006048
0.0006090
0.0006128
0.0006160
0.0006188
0.0006210
0.0006228
0.0006240
0.0006248
p0 q0
pqz 2
n 2
n
e
sqrt
0.0244949
0.0245917
0.0246779
0.0247538
0.0248193
0.0248747
0.0249199
0.0249550
0.0249800
0.0249950
n
63703.4
64207.8
64659.0
65057.1
65402.2
65694.2
65933.1
66118.9
66251.6
66331.2
p cv  p 0  z  p p cv  p 0  z  p
cv1
0.343000
0.352775
0.362574
0.372398
0.382245
0.392117
0.402011
0.411930
0.421872
0.431837
cv2
0.457000
0.467225
0.477426
0.487602
0.497755
0.507883
0.517989
0.528070
0.538128
0.548163
MTB > print c8-c13
Data Display
29
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
pcv  p 0  z 2  p pcv  p 0  z 2  p z 
Row
1
2
3
4
5
6
7
8
9
10
cv3
0.336901
0.346652
0.356430
0.366234
0.376065
0.385923
0.395806
0.405716
0.415652
0.425613
cv4
0.463099
0.473348
0.483570
0.493766
0.503935
0.514077
0.524194
0.534284
0.544348
0.554387
.50  p 0
p
tr1
4.08248
3.65978
3.24176
2.82785
2.41747
2.01008
1.60514
1.20217
0.80064
0.40008
z
cv 2  .5
cv3  .5
cv 4  .5
z
z
.025
.025
.025
tr2
-1.72001
-1.31101
-0.90298
-0.49592
-0.08982
0.31534
0.71954
1.12281
1.52514
1.92653
tr3
-6.52395
-6.13393
-5.74281
-5.35063
-4.95739
-4.56309
-4.16774
-3.77136
-3.37394
-2.97548
tr4
-1.47605
-1.06607
-0.65719
-0.24937
0.15739
0.56309
0.96774
1.37136
1.77394
2.17548
MTB > let c14=2.576*c4
MTB > print c1 c14
Data Display
Row
1
2
3
4
5
6
7
8
9
10
p0
z .005 x
p
0.40
0.41
0.42
0.43
0.44
0.45
0.46
0.47
0.48
0.49
er
0.0630989
0.0633481
0.0635703
0.0637658
0.0639346
0.0640772
0.0641936
0.0642840
0.0643485
0.0643871
MTB > print c7 c11
Data Display
Row
cv2
tr2
p cv  p 0  z  p z 
1
2
3
4
5
6
7
8
9
10
0.457000
0.467225
0.477426
0.487602
0.497755
0.507883
0.517989
0.528070
0.538128
0.548163
cv 2  .5
.025
-1.72001
-1.31101
-0.90298
-0.49592
-0.08982
0.31534
0.71954
1.12281
1.52514
1.92653
MTB > print c8 c9 c12 c13
Data Display
pcv  p 0  z 2  p pcv  p 0  z 2  p z 
Row
1
2
cv3
0.336901
0.346652
cv4
0.463099
0.473348
cv3  .5
cv 4  .5
z
.025
.025
tr3
-6.52395
-6.13393
tr4
-1.47605
-1.06607
30
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
3
4
5
6
7
8
9
10
0.356430
0.366234
0.376065
0.385923
0.395806
0.405716
0.415652
0.425613
0.483570
0.493766
0.503935
0.514077
0.524194
0.534284
0.544348
0.554387
-5.74281
-5.35063
-4.95739
-4.56309
-4.16774
-3.77136
-3.37394
-2.97548
-0.65719
-0.24937
0.15739
0.56309
0.96774
1.37136
1.77394
2.17548
MTB > Save "C:\Documents and Settings\rbove\My Documents\Minitab\252x05053.MTW";
SUBC>
Replace.
Saving file as: 'C:\Documents and Settings\rbove\My
Documents\Minitab\252x0505-3.MTW'
Existing file replaced.
31
252y0551s 10/28/05 (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)
[16]
b. (Extra credit) Find a 95% (or slightly more) two-sided confidence interval for the median. (2)
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
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
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
n
31
31
29
31
31
30
x
17
18
19
22
23
15
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
 H 0 :   0

H 1 :   0
above  0
below  0
 H 0 : p .5

 H 1 : p .5
 H 0 : p .5

 H 1 : p  .5
32
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
It seems easiest to let p be the proportion below 8. If you defined p as the proportion above .8, the p-
 H : p  .5
values should be identical. Our Hypotheses are  0
. According to the outline, in the absence of a
 H 1 : p  .5
x
binomial table we must use the normal approximation to the binomial distribution. If p  is our
n
p  p0
observed proportion, we use z 
. But for the sign test, p  .5 and q  1  .5  .5 . So
p0 q0
n
z
x
n  .5
x

n  .5
.5
.25
n
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.
applies if x 
x
1
2
3
4
5
6
7
8
9
10
17
18
19
19
22
23
23
23
23
15
n
31
31
32
29
31
32
32
32
31
30
z1 
2x  n
n
0.53882
0.89803
1.06066
1.67126
2.33487
2.47487
2.47487
2.47487
2.69408
0.00000
z2 
2x 1  n
P  z  z1 
n
0.71842
1.07763
1.23744
1.85695
2.51447
2.65165
2.65165
2.65165
2.87368
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
P z  z 2 
.5-.2642=.2358
.5-.3599=.1401
.5-.3925=.1075
.5-.4686=.0314*
.5-.4940=.0060*
.5-.4960=.0040*
.5-.4960=.0040*
.5-.4960=.0040*
.5-.4979=.0021*
.5
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. Note that z2 is wrong!!! See 252y0551h for correct values.
Computation of exact probabilities for this Problem
————— 10/31/2005 4:26:58 PM ————————————————————
Welcome to Minitab, press F1 for help.
MTB > WOpen "C:\Documents and Settings\rbove\My Documents\Minitab\252x05051c.MTW".
Retrieving worksheet from file: 'C:\Documents and Settings\rbove\My
Documents\Minitab\252x0505-1c.MTW'
Worksheet was saved on Mon Oct 31 2005
Results for: 252x0505-1c.MTW
MTB >
MTB >
SUBC>
MTB >
let c3 = c1-1
cdf c3 c4;
binomial 29 .5.
cdf c3 c5;
#Computes F x  1 for n  29
33
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
SUBC>
MTB >
SUBC>
MTB >
SUBC>
MTB >
MTB >
MTB >
MTB >
MTB >
binomial 30 .5.
cdf c3 c6;
binomial 31 .5.
cdf c3 c7;
binomial 32 .5.
let c4=1-c4 #so if x is 17, we now have Px  17 
let c5=1-c5
let c6 = 1-c6
let c7 = 1-c7
print c1-c7 #Note that correct answers are in ‘()’
Data Display
Row
1
2
3
4
5
6
7
8
9
x
17
18
19
19
22
23
23
23
15
n
31
31
32
29
31
32
32
31
30
x-1
16
17
18
18
21
22
22
22
14
n29
n30
0.229129
0.292332
0.132465
0.180797
0.068023
0.100244
(0.068023) 0.100244
0.004065
0.008062
0.001158
0.002611
0.001158
0.002611
0.001158
0.002611
0.500000 (0.572232)
n31
n32
(0.360050) 0.430025
(0.236565) 0.298307
0.140521 (0.188543)
0.140521
0.188543
(0.014725) 0.025051
0.005337 (0.010031)
0.005337 (0.010031)
(0.005337) 0.010031
0.639950
0.701693
MTB >
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
————— 10/17/2005 9:38:18 PM ————————————————————
Welcome to Minitab, press F1 for help.
MTB
MTB
MTB
MTB
MTB
MTB
MTB
>
>
>
>
>
>
>
let
let
let
let
let
let
let
c3 = 2*c1
c3=c3-c2
c5 = sqrt(c2)
c4 = c3+1
c3=c3/c5
c4=c4/c6
c4 = c4/c5
MTB > print c1-c5
34
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
Data Display
Row
1
2
3
4
5
6
7
8
9
10
x
17
18
19
19
22
23
23
23
23
15
n
31
31
32
29
31
32
32
32
31
30
z1
0.53882
0.89803
1.06066
1.67126
2.33487
2.47487
2.47487
2.47487
2.69408
0.00000
z2
0.71842
1.07763
1.23744
1.85695
2.51447
2.65165
2.65165
2.65165
2.87368
0.18257
sqrtn
5.56776
5.56776
5.65685
5.38516
5.56776
5.65685
5.65685
5.65685
5.56776
5.47723
MTB > print c1-c4
Data Display
Row
1
2
3
4
5
6
7
8
9
10
x
17
18
19
19
22
23
23
23
23
15
n
31
31
32
29
31
32
32
32
31
30
z1
0.53882
0.89803
1.06066
1.67126
2.33487
2.47487
2.47487
2.47487
2.69408
0.00000
z2
0.71842
1.07763
1.23744
1.85695
2.51447
2.65165
2.65165
2.65165
2.87368
0.18257
MTB > MTB > let c6 = c2+1-1.960*c5
MTB > let c6 = c6/2
MTB > print c2 c5 c6
Data Display
Row
1
2
3
4
5
6
7
8
9
10
n
31
31
32
29
31
32
32
32
31
30
sqrtn
5.56776
5.56776
5.65685
5.38516
5.56776
5.65685
5.65685
5.65685
5.56776
5.47723
k
10.5436
10.5436
10.9563
9.7225
10.5436
10.9563
10.9563
10.9563
10.5436
10.1323
35
252y0551s 10/28/05 (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
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.
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)
[30]
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
 0.228 . Note that tn 1  t .10
 1.696 . We need a critical value for x below 8.
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.
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
reject
Not below
Not below
Not below
Below cv
Below cv
Below cv
Below cv
Below cv
Below cv
Not below
cv
cv
cv
cv
36
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
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 

sx 
 0.228 

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
t 31  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
7.944 -0.24561
.05
31
t .025
31
t .01
31
t .005
31
t .001
31
t .001
31
t .45
 2.453
 2.744
p-value is between .10 and .15.
31
t .025
 2.040
t 31  2.040
 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
p-value is between .025 and .05
p-value is between .025 and .05
 3.375
p-value is below .005
 3.375
p-value is below .005
31
 0.127 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
 02

2
s cv
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
12 2  02
n 1
2 
 22
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 
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.
f. (Extra credit) What would the critical values be for this test of the standard deviation? (1)
2 231
27.8740 
Table 3 says s cv   2 DF
. The two critical values are


 1.960  231 7.8740  1.960
 z   2 DF
2
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.
37
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
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  7.812
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  7.6238
P x  7.4184   7.4184  P  z 
  Pz  0   .5
.35355


Power  1  .5  .5
7.4184  7.127 

1  7.436
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  7.248.
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
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 2  x  x  1.88.35355  x  0.665 . Substitute your value of the sample mean.
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 

.04, reject the null hypothesis.
(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.
38
252y0551s 10/28/05 (Open in ‘Print Layout’ format)
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
e2

2.08 2 22
 3461 .12 . Use 3462.
.005
————— 10/17/2005 11:10:20 PM ————————————————————
Welcome to Minitab, press F1 for help.
MTB > WOpen "C:\Documents and Settings\rbove\My Documents\Minitab\252x05051.MTW".
Retrieving worksheet from file: 'C:\Documents and Settings\rbove\My
Documents\Minitab\252x0505-1.MTW'
Worksheet was saved on Tue Oct 11 2005
Results for: 252x0505-1.MTW
MTB > describe c1-c10
Descriptive Statistics: C1, C2, C3, C4, C5, C6, C7, C8, C9, C10
Variable
C1
C2
C3
C4
C5
C6
C7
C8
C9
C10
MTB
MTB
MTB
MTB
MTB
>
>
>
>
>
N
32
32
32
32
32
32
32
32
32
32
N*
0
0
0
0
0
0
0
0
0
0
Mean
7.844
7.744
7.644
7.544
7.444
7.344
7.244
7.144
7.044
7.944
SE Mean
0.228
0.228
0.228
0.228
0.228
0.228
0.228
0.228
0.228
0.228
StDev
1.291
1.291
1.291
1.291
1.291
1.291
1.291
1.291
1.291
1.291
Minimum
4.800
4.700
4.600
4.500
4.400
4.300
4.200
4.100
4.000
4.900
Q1
7.125
7.025
6.925
6.825
6.725
6.625
6.525
6.425
6.325
7.225
Median
7.900
7.800
7.700
7.600
7.500
7.400
7.300
7.200
7.100
8.000
Q3
8.875
8.775
8.675
8.575
8.475
8.375
8.275
8.175
8.075
8.975
Maximum
10.200
10.100
10.000
9.900
9.800
9.700
9.600
9.500
9.400
10.300
let c13=c12-8
let c13=c13/0.288
let c14=c12-8
let c14=c14/.3536
print c12-c14
Data Display
1
2
3
4
5
6
7
8
9
10
x 8
sx
x
t
7.844
7.744
7.644
7.544
7.444
7.344
7.244
7.144
7.044
7.944
-0.54167
-0.88889
-1.23611
-1.58333
-1.93056
-2.27778
-2.62500
-2.97222
-3.31944
-0.19444
Row
z
x 8
x
-0.44118
-0.72398
-1.00679
-1.28959
-1.57240
-1.85520
-2.13801
-2.42081
-2.70362
-0.15837
MTB >
39