252y0551h 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
[20] appropriate confidence interval for the hypothesis test. (2)
Solution: a) From the formula table we have:
Interval for
Proportion
Confidence
Interval p
 p
 z

2 s p
Hypotheses
H
0
H
1
:
: p p

 p
0 p
0 s p
 p q q
 p n

1
Test Ratio z
 p

 p p
0
Critical Value p
 q
0 cv
 p
0
 z

2
 p p
 p
0

1
 p
0 n q
0
1

252y0551h 10/31/05 (Open in ‘Print Layout’ format)
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.
Version 0 a) p
0

.
40
 p
 p
0 n q
0 
.
40
 
400

.
0006

.
024495
(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
The critical value must be below .40. p cv
 p
and
0
 z



 p
.
01

.
.
40 z

.
01

2 .
327
2 .
327
.

.
024495

= .3430. The ‘reject’ zone is below .3430. pval

P
 p

.
50


P

 z

.
50
.

.
40
024495



P
 z

4 .
08


1
(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
The critical value must be above .40. p cv
 p
0
 z



 p
.
01

.
.
40 z
.
01


2 .
327
2 .
327
.

.
024495

= .4570 . The ‘reject’ zone is above .4570. pval

P
 p

.
50


P

 z

.
50

.
40
.
024495



P
 z

4 .
08


0
(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.
H
0
: p

.
40 , H
1
: p

.
40 , n

400 , p

.
50 and
The critical value must be on either side of .40.
 p cv

.
01
 p
0
.
 z z

.
005

2
The ‘reject’ zone is below .3369 and above .4631. pval

2 P
 p

.
50


2 P

 z

.
50
.

.
40
024495



2 P
 z

4 .
08


0 p

2 .
576

.
40

.
2 .
576

.
024495

[12]

.
40

.
0631 . b) (i) H
1
: p

.
40 ,
 
P
 p

.
4570 p
1

.
50

 p
 pq n

.
5
400

.
5
20

.
025 .
 
P

 z

.
4570
.
025

.
5



P
 z
 
1 .
72


.
5

.
4573

.
0427 Power

1

.
0427

.
9573
(ii)
 c) d)
H
1
P


6 .
52 p
0 s p


.
40
:
 p p q n z n

.
40 ,



 
P

.
3369
 p

.
4631 p
1

1 .
48


.
5

.
4306

.
0694 pqz
2
2 e
.
5
 


400
.
40 .
 
2 .
576

2
.
5
20
.
005

2
.
025 .
p



Power
63703 p
 z

.
50


P


.
3369
.
025

.
5

1

.
0694
.
42 s p

.
5


.
9306
2 .
327
 
 z

.
4631
.
025

.
5


This is above 400, so the sample size is inadequate.

.
4418 If the null hypothesis is
H
0
: p

.
40 , we must reject it.
2

252y0551h 10/31/05 (Open in ‘Print Layout’ format)
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.
Version 1 a) p
0

.
41
 p
 p
0 q
0 n

.
41
 
400

.
0006048

.
0245917
(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
The critical value must be below .40. p cv
 p
0
 z



 p
.
01

.
.
41 z

.
01

2 .
327
2 .
327
.

.
0245917

= .3528. The ‘reject’ zone is below .3528. pval

P
 p

.
50


P

 z

.
50

.
41
.
0245917



P
 z

3 .
65


.
5

.
4999

.
9999
(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

.
41 , H
1
: p

.
41 , n

400 , p

.
50 and
The critical value must be above .41. p cv
 p
0
 z



 p
.
01

. z
.
41
.
01


2 .
327
2 .
327
.

.
0245917

= .4672 . The ‘reject’ zone is above .467270. pval

P
 p

.
50


P

 z

.
50

.
41
.
0245917



P
 z

3 .
66


.
5

.
4999

.
0001
(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.
H
0
: p

.
41 , H
1
: p

.
41 , n

400 , p

.
50 and
The critical value must be on either side of .40.
The ‘reject’ zone is below .3467 and above .4733.
 p cv

.
01
 p
0
.
 z z
.
005

2

 p pval

2 P
 p

.
50


2 P

 z

.
50

.
41
.
0245917



2 P
 z

3 .
66
2 .
576

.
5

.
4999
.
41



.
.
2 .
576

.
0245917

0002
[12]

.
41

.
0633 . b) (i) H
1
: p

.
41 ,
 
P
 p

.
4672 p
1

.
50

 p
 pq n

.
5
 
400

.
5
20

.
025 .
 
P

 z
(ii) H
1

:
.
4672
.
025

.
5

 p

.
41 ,

P
 z
 
1 .
31


.
5

.
4049

.
0951 Power

1

.
0951

.
9049

 c) d)
 p
0 s
P
.
5
 p

.
3467
 p

.
4733 p
1

.
50


P


.
3467
.
025

.
5
 z

.
4733
.
025

.
5



P


6 .
13
 z
 
1 .
07

.
3577


.
40 p q n
 n
.
1423


2
Power pqz
2 e
.
5
 

400
.
41

.
5

20
.
1
59


.
1423
2 .
576
.
005
2
.
025 .


2 p
.
8577


64207 p
 z

.
8 s p
This is above 400, so the sample size is inadequate.

.
5

2 .
327
 

.
4418 If the null hypothesis is
H
0
: p

.
41 , we must reject it.
3

252y0551h 10/31/05 (Open in ‘Print Layout’ format)
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.
Version 2 a) p
0

.
42
 p
 p
0 q
0 n

.
42

.
0006090

.
0246779
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
The critical value must be below .42. p cv
 p
and
0
 z



 p
.
01

.
.
42 z

.
01

2 .
327
2 .
327
.

.
0246779

= .3626. The ‘reject’ zone is below .3626. pval

P
 p

.
50


P

 z

.
50

.
42
.
0246779



P
 z

3 .
24


.
5

.
4994

.
9994
(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
The critical value must be above .42. p cv
 p
0
 z



 p
.
01

.
.
42 z
.
01


2 .
327
2 .
327
.

.
02446779

= .4774 . The ‘reject’ zone is above .4774. pval

P
 p

.
50


P

 z

.
50

.
42
.
0246779



P
 z

3 .
24


.
5

.
4994

.
0006
(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.
H
0
: p

.
40 , H
1
: p

.
40 , n

400 , p

.
50 and
The critical value must be on either side of .42.
 p cv

.
01
 p
0
.
 z z

.
005

2
The ‘reject’ zone is below .3564 and above .4836. pval

2 P
 p

.
50


2 P

 z

.
50

.
42
.
0246779



2 P
 z

3 .
24


.
0012 p

2 .
576

.
42

.
2 .
576

.
0246779

[12]

.
40

.
0636 . b) (i) H
1
: p

.
42 ,
 
P
 p

.
4774 p
1

.
50

 p
 pq n

.
5
400

.
5
20

.
025 .
 
P

 z
(ii) H
1

:
.
4774
.
025

.
5

 p

.
42 ,

P
 z
 
0 .
90


.
5

.
3159

.
1841 Power

1

.
1841

.
8159

 c) d)

P

.
3564
.
5

.
2454 p
0 s p


.
40
 p q n
 n
 p


.
2546
.
4836 pqz
2
Power
2 e
.
5
 
400

.
42
 p
1

.
5
20
.
50
.
005



2
.
025 .
P


.
3564
.
025


.
1
58

.
2546
2 .
576


2 p
.
7454


64659 p

.
5 z


.
0 s z p

.
4836
.
025
.
5


.
5



P



This is above 400, so the sample size is inadequate.

2 .
327
 

5 .
74
 z
 
0 .
66
.
4418 . If the null hypothesis is
H
0
: p

.
42 , we must reject it.
4

252y0551h 10/31/05 (Open in ‘Print Layout’ format)
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.
Version 3 a) p
0

.
43
 p
 p
0 n q
0 
.
43
 
400

.
0006128

.
0247538
(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
The critical value must be below .43. p cv
 p
and
0
 z



 p
.
01

.
.
43 z
.
01


2 .
327
2 .
327
.

.
0247538

= .3724. The ‘reject’ zone is below .3724. pval

P
 p

.
50


P

 z

.
50

.
43
.
0247538



P
 z

2 .
83


.
5

.
4977

.
9977
(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
The critical value must be above .43. p cv
 p
0
 z



 p
.
01

.
.
43 z

.
01

2 .
327
2 .
327
.

.
027538

= .4876. The ‘reject’ zone is above .4876. pval

P
 p

.
50


P

 z

.
50

.
43
.
0247538



P
 z

2 .
83


.
5

.
4977

.
0023
(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.
H
0
: p

.
43 , H
1
: p

.
43 , n

400 , p

.
50 and
The critical value must be on either side of .40.
 p cv

.
01

. p
0
 z z

.
005

2
The ‘reject’ zone is below .3662 and above .4938. pval

2 P
 p

.
50


2 P

 z

.
50

.
43
.
0247538



2 P
 z

2 .
83
 
.
0023


.
0046 p

2 .
576

.
43

.
2 .
576

.
0247538

[12]

.
43

.
0638 . b) (i) H
1
: p

.
43 ,
 
P
 p

.
4876 p
1

.
50

 p
 pq n

.
5
400

.
5
20

.
025 .
 
P

 z
(ii) H
1

:
.
4876
.
025

.
5

 p

.
43 ,

P
 z
 
0 .
50


.
5

.
1915

.
3085 Power

1

.
3085

.
6915
 
P

.
3662
 p

.
4938 p
1

.
50


P


.
3662
.
025

.
5
 c)
.
5
 d) p
0 s p
.
0948


.
43
 p q n n
.
4052


2
Power pqz
2 e
.
5
 

400
.
43

.
5

20
.
1
57


.
4052
2 .
576
.
005
.
025 .


2 p
.
5948


65057 p
 z

 z

.
4938
.
025
.
5


.
5



P

 
0 .
24

.
1 This is above 400, so the sample size is inadequate. s p

2 .
327
 

5 .
35
 z

.
4418 If the null hypothesis is
H
0
: p

.
43 , we must reject it.
5

252y0551h 10/31/05 (Open in ‘Print Layout’ format)
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.
Version 4 a) p
0

.
44
 p
 p
0 q
0 n

.
44

.
0006160

.
0248193
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
The critical value must be below .44. p cv
 p
and
0
 z



 p
.
01

.
.
44 z

.
01

2 .
327
2 .
327
.

.
0248193

= .3822. The ‘reject’ zone is below .3822. pval

P
 p

.
50


P

 z

.
50

.
44
.
0248193



P
 z

2 .
42


.
5

.
4922

.
9922
(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
The critical value must be above .44. p cv
 p
0
 z



 p
.
01

.
.
44 z
.
01


2 .
327
2 .
327
.

.
0248193

= .4978. The ‘reject’ zone is above .4978. pval

P
 p

.
50


P

 z

.
50

.
44
.
0248193



P
 z

2 .
42


.
5

.
4922 = .0078
(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.
H
0
: p

.
44 , H
1
: p

.
44 , n

400 , p

.
50 and
The critical value must be on either side of .44.
 p cv

.
01
 p
0
.
 z z

.
005

2
The ‘reject’ zone is below .3761 and above .5039. pval

2 P
 p

.
50


2 P

 z

.
50
.

.
44
028193



2 P
 z

2 .
42
 
.
0078


.
0156 p

2 .
576

.
44

.
2 .
576

.
0248193

[12]

.
44

.
0639 . b) (i) H
1
: p

.
44 ,
 
P
 p

.
4977 p
1

.
50

 p
 pq n

.
5
400

.
5
20

.
025 .
 
P

 z

.
4977
.
025

.
5



P
 z
 
0 .
90


.
5

.
3159

.
1841 Power

1

.
1841

.
8159
(ii)
 c) d)
H
1
P


6 .
52 p
0 s p


.
44
:
 p p q n z n

.
40 ,



 
P

.
3369
 p

.
4631 p
1

1 .
48


.
5

.
4306

.
0694 pqz
2
2 e
.
5
 


400
.
44 .
 
2 .
576

2
.
5
20
.
005
2

.
025 .
p



Power
.
50



.
5

 z

65402 .
2 This is above 400, so the sample size is inadequate. p
 z


1

.
0694 s p
P


.
3369
.
025

.
5

.
9306
2 .
327
 
.
4631
.
025

.
5



.
4418 If the null hypothesis is
H
0
: p

.
44 , we must reject it.
6

252y0551h 10/31/05 (Open in ‘Print Layout’ format)
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.
Version 5 a) p
0

.
45
 p
 p
0 q
0 n

.
45

.
0006188

.
0248747
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
The critical value must be below .45. p cv
 p
and
0
 z



 p
.
01

.
.
45 z

.
01

2 .
327
2 .
327
.

.
0248747

= .3921. The ‘reject’ zone is below .3921. pval

P
 p

.
50


P

 z

.
50

.
45
.
0248747



P
 z

2 .
01


.
5

.
4778

.
9778
(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
The critical value must be above .40. p cv
 p
0
 z



 p
.
01

.
.
45 z

.
01

2 .
327
2 .
327
.

.
0248747

= .5079 . The ‘reject’ zone is above .5029. pval

P
 p

.
50


P

 z

.
50

.
45
.
0248747



P
 z

2 .
01


.
5

.
4778

.
0452
(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.
H
0
: p

.
45 , H
1
: p

.
45 , n

400 , p

.
50 and
The critical value must be on either side of .45.
 p cv

.
01

. p
0
 z z

.
005

2
The ‘reject’ zone is below .3859 and above .5141. pval

2 P
 p

.
50


2 P

 z

.
50

.
45
.
0248747



2 P
 z

2 .
01
 
.
0452


.
0904 p

2 .
576

.
45

.
2 .
576

.
0248747

[12]

.
45

.
0641 . b) (i) H
1
: p

.
45 ,
 
P
 p

.
5079 p
1

.
50

 p
 pq n

.
5
400

.
5
20

.
025 .
 
P

 z
(ii) H
1

:
.
5079
.
025

.
5

 p

.
45 ,

P
 z

0 .
32


.
5

.
1255

.
6255 power

1

.
6255

.
3745
 
P

.
3859
 p

.
5141 p
1

.
50


P


.
3859
.
025

.
5
 z

.
5141

.
025
.
5



P


4 .
56
 z
 
0 .
56

 c)
.
5
 d) p
0 s p
.
2123


.
45
 p q n n
.
2877


2
Power pqz
2 e
.
5
 

400
.
45

.
5

20
.
1
55


.
2877
2 .
576
.
005
2
.
025 .


2 p
.
7123


65694 p
 z

.
2 s p
This is above 400, so the sample size is inadequate.

.
5

2 .
327
 

.
4418 . If the null hypothesis is
H
0
: p

.
45 , p could be between .4418 and .45 so we must not reject the null hypothesis.
7

252y0551h 10/31/05 (Open in ‘Print Layout’ format)
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.
Version 6 a) p
0

.
46
 p
 p
0 q
0 n

.
46

.
0006210

.
0249199
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
The critical value must be below .46. p cv
 p
and
0
 z



 p
.
01

.
.
46 z

.
01

2 .
327
2 .
327
.

.
0249199

= .4020. The ‘reject’ zone is below .4020. pval

P
 p

.
50


P

 z

.
50

.
46
.
0249199



P
 z

1 .
60


.
5

.
4452

.
9452
(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
The critical value must be above .46. p cv
 p
0
 z



 p
.
01

.
.
46 z

.
01

2 .
327
2 .
327
.

.
0249199

= .5180 . The ‘reject’ zone is above .5180. pval

P
 p

.
50


P

 z

.
50

.
46
.
0249199



P
 z

1 .
60


.
5

.
4452

.
0548
(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.
H
0
: p

.
46 , H
1
: p

.
46 , n

400 , p

.
50 and
The critical value must be on either side of .46.
 p cv

.
01
 p
0
.
 z z

.
005

2
The ‘reject’ zone is below .3958 and above .5242. pval

2 P
 p

.
50


2 P

 z

.
50

.
46
.
0249199



2 P
 z

1 .
60
 
.
0548


.
1096 p

2 .
576

.
46

.
2 .
576

.
0249199

[12]

.
46

.
0642 . b) (i) H
1
: p

.
46 ,
 
P
 p

.
5180 p
1

.
50

 p
 pq n

.
5
400

.
5
20

.
025 .
 
P

 z
(ii) H
1

:
.
5180
.
025

.
5

 p

.
46 ,

P
 z

0 .
72


.
5

.
2642

.
7642 Power

1

.
7642

.
2358
 
P

.
3958
 p

.
5242 p
1

.
50


P


.
3958
.
025

.
5
 c)
.
5

.
3340 d) p
0 s p


.
46
 n p q n
.
8340


Power pqz
2
2 e
.
5
 


400
.
46
.
5
20

.
1
54

.
8340
2 .
576


2
.
005

2
.
025 .
p
.
1660


65933 p
 z

 z

.
5242
.
025

.
5


.
5


P


.
1 This is above 400, so the sample size is inadequate. s p

2 .
327
 

4 .
17
 z

0 .
97

.
4418 If the null hypothesis is
H
0
: p

.
46 , p could be between .4418 and .46, so we must not reject the null hypothesis.
8

252y0551h 10/31/05 (Open in ‘Print Layout’ format)
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.
Version 7 a) p
0

.
47
 p
 p
0 n q
0 
.
47
 
400

.
0006228

.
0249550
(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
The critical value must be below .47. p cv
 p
and
0
 z



 p
.
01

.
.
47 z

.
01

2 .
327
2 .
327
.

.
0249550

= .4119. The ‘reject’ zone is below .4119. pval

P
 p

.
50


P

 z

.
50

.
47
.
0249550



P
 z

1 .
20


.
5

.
3849

.
8849
(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
The critical value must be above .40. p cv
 p
0
 z



 p
.
01

.
.
47 z
.
01


2 .
327
2 .
327
.

.
0249550

= .5281 . The ‘reject’ zone is above .5281. pval

P
 p

.
50


P

 z

.
50

.
46
.
0249550



P
 z

1 .
20


.
5

.
3849 = .1151
(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.
H
0
: p

.
47 , H
1
: p

.
47 , n

400 , p

.
50 and

The critical value must be on either side of .47.

.
47 ,

P
 p

.
5281 p
1

.
50




.
01 p
0
.
 z z

.
005
2
 p


2 .
576
.
47

.
2 .
576

.
0249550

[12]

.
40

.
0643 .
The ‘reject’ zone is below .4057 and above .5343. pval

2 P
 p

.
50


2 P

 z

.
.
50

.
47
024495



2 P
 z

1 .
20
 
.
1151


.
2302 b) (i) H
1
: p
 p cv p
 pq n

.
5
 
400

.
5
20

.
025 .
 
P

 z
(ii) H
1

:
.
5281
.

025
.
5

 p

.
47 ,

P
 z

1 .
12


.
5

.
3708

.
8708 Power

1

.
8708

.
1292


.
4999 c) d)
 p
0 s p
P

.
4057


.
47


.
4147 n p q n
 p


.
5343

.
9146 pqz
2
2 e
.
5
 


400 p
1

.
50


P


.
4057
.
025

.
5

Power
.
47 .
53

1

.
9146
2 .
576

2

2

.
005
.
0854
66118 .
9 z

.
5343
.
025

.
5



P


3 .
77
 z

1 .
37

This is above 400, so the sample size is inadequate.
.
5
20

.
025 .
p
 p
 z
 s p

.
5

2 .
327
 

.
4418 If the null hypothesis is
H
0
: p

.
47 , p could be between .4418 and .47, so we must not reject the null hypothesis.
9

252y0551h 10/31/05 (Open in ‘Print Layout’ format)
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.
Version 8 a) p
0

.
48
 p
 p
0 q
0 n

.
48

.
0006240

.
0249800
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
The critical value must be below .40. p cv
 p
and
0
 z



 p
.
01

.
.
48 z

.
01

2 .
327
2 .
327
.

.
0249800

= .4219. The ‘reject’ zone is below .4219. pval

P
 p

.
50


P

 z

.
50

.
48
.
0249800



P
 z

0 .
80


.
5

.
2881

.
7881
(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
The critical value must be above .40. p cv
 p
0
 z



 p
.
01

.
.
48 z

.
01

2 .
327
2 .
327
.

.
0249800

= .5381 . The ‘reject’ zone is above .5381. pval

P
 p

.
50


P

 z

.
50

.
48
.
0249800



P
 z

0 .
80


.
5

.
2119

.
2881
(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.
H
0
: p

.
48 , H
1
: p

.
48 , n

400 , p

.
50 and
The critical value must be on either side of .48.
 p cv

.
01

. p
0
 z z

.
005

2
The ‘reject’ zone is below .4157 and above .5443. pval

2 P
 p

.
50


2 P

 z

.
50

.
48
.
0249800



2 P
 z

0 .
80


.
5762 p

2 .
576

.
48

.
2 .
576

.
0249800

[12]

.
48

.
0643 . b) (i) H
1
: p

.
48 ,
 
P
 p

.
5381 p
1

.
50

 p
 pq n

.
5
 
400

.
5
20

.
025 .
 
P

 z

.
5381
.

025
.
5



P
 z
(ii)
 
P

H
1
:
.
4156 p

 p
.
48 ,

.
5443 p
1

1 .
53


.
5

.
4370

.
50


.
9370 Power

1

.
9370

.
6300

 c) d)
P
P



.
4156
.
025

6 .
52

 p
0 s p


.
48 n p q n
.
5 z

 z

.
5443
.
025

.
5



1 .
48


.
5


.
4306
P



3 .
37
.
0694

 pqz
2
2 e
.
5
 
400


.
48
 
2 .
576

2
.
5
20
.
005
2

.
025 .
p


 z

1 .
77
Power


.
4996

1

.
9612
.
5


.
4616

.
0388
66251 .
6 This is above 400, so the sample size is inadequate. p
 z
 s p

2 .
327
 

.
9612

.
4418 . If the null hypothesis is
H
0
: p

.
48 , p could be between .4418 and .48, so we must not reject the null hypothesis.
10

252y0551h 10/31/05 (Open in ‘Print Layout’ format)
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.
Version 9 a) p
0

.
49
 p
 p
0 n q
0 
.
49
 
400

.
0006248

.
0249950
(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
The critical value must be below .49. p cv
 p
and
0
 z



 p
.
01

.
.
49 z

.
01

2 .
327
2 .
327
.

.
0249950

= .4318. The ‘reject’ zone is below .4318. pval

P
 p

.
50


P

 z

.
50

.
49
.
0249950



P
 z

0 .
40


.
5

.
1554

.
6554
(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
The critical value must be above .49. p cv
 p
0
 z



 p
.
01

.
.
49 z

.
01

2 .
327
2 .
327
.

.
0249950

= .5481 . The ‘reject’ zone is above .5481. pval

P
 p

.
50


P

 z

.
50

.
49
.
0249950



P
 z

0 .
40


.
5

.
1554

.
3446
(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.
H
0
: p

.
49 , H
1
: p

.
49 , n

400 , p

.
50 and

The critical value must be on either side of .49. p


.
01 p
0

. z

2 z
.
005
 p
The ‘reject’ zone is below .4256 and above .5544. pval

2 P
 p

.
50


2 P

 z

.
50

.
49
.
0249950



2 P
 z

0 .
40
 
.
3446


.
6892


2 .
576
.
49

.
2 .
576

.
0249950

[12]

.
49

.
0643 . b) (i) H
1
: p

.
49 ,
 
P
 p

.
5481 p
1

.
50

 p
 pq n

.
5
 
400

.
5
20

.
025 .
 
P

 z
(ii) H
1

:
..
5481
.
025

.
5

 p

.
40 ,

P
 z

1 .
93


.
5

.
4732

.
9732 Power

1

.
9732

.
0268
 c)


.
4986 p
0
P

.
4256

.
49


.
4854 n
 p

.
5544

.
9840 pqz
2 e
2
 p
1

.
50


P


.
4256
.
025

.
5

Power
.
49 .
 

1

.
9840
2 .
576

2

2

.
005
.
0160
66331 .
2 z

.
5544
.
025

.
5



P


2 .
98
 z

2 .
18

. This is above 400, so the sample size is inadequate. d) s p
 p q

.
5
 

.
5

.
025 .
p
 p
 z
 s p

.
5

2 .
327
 

.
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.
————— 10/17/2005 7:01:49 PM ————————————————————
Available in 252y0551s
11

252y0551h 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 6.3 5.7 5.5 4.9 7.0 7.1 7.2 7.3 7.3 7.4
7.6 7.6 7.7 7.7 8.0 8.0 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 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 x
1 x
2 x
3 x
4 x
5 x
6 x
7 x
8 x
9 x
10
1 4.8 4.7 4.6 4.5 4.4 4.3 4.2 4.1 4.0 4.9
2 5.4 5.3 5.2 5.1 5.0 4.9 4.8 4.7 4.6 5.5
3 5.6 5.5 5.4 5.3 5.2 5.1 5.0 4.9 4.8 5.7
4 6.2 6.1 6.0 5.9 5.8 5.7 5.6 5.5 5.4 6.3
5 6.6 6.5 6.4 6.3 6.2 6.1 6.0 5.9 5.8 6.7
6 6.9 6.8 6.7 6.6 6.5 6.4 6.3 6.2 6.1 7.0
7 7.0 6.9 6.8 6.7 6.6 6.5 6.4 6.3 6.2 7.1
8 7.1 7.0 6.9 6.8 6.7 6.6 6.5 6.4 6.3 7.2
9 7.2 7.1 7.0 6.9 6.8 6.7 6.6 6.5 6.4 7.3
10 7.2 7.1 7.0 6.9 6.8 6.7 6.6 6.5 6.4 7.3
11 7.3 7.2 7.1 7.0 6.9 6.8 6.7 6.6 6.5 7.4
12 7.5 7.4 7.3 7.2 7.1 7.0 6.9 6.8 6.7 7.6
13 7.5 7.4 7.3 7.2 7.1 7.0 6.9 6.8 6.7 7.6
14 7.6 7.5 7.4 7.3 7.2 7.1 7.0 6.9 6.8 7.7
15 7.6 7.5 7.4 7.3 7.2 7.1 7.0 6.9 6.8 7.7
16 7.9 7.8 7.7 7.6 7.5 7.4 7.3 7.2 7.1 8.1
17 7.9 7.8 7.7 7.6 7.5 7.4 7.3 7.2 7.1 8.2
18 8.1 7.9 7.8 7.7 7.6 7.5 7.4 7.3 7.2 8.4
19 8.3 8.2 7.9 7.8 7.7 7.6 7.5 7.4 7.3 8.4
20 8.3 8.2 8.1 8.1 7.9 7.8 7.7 7.6 7.5 8.4
21 8.3 8.2 8.1 8.5 7.9 7.8 7.7 7.6 7.5 8.5
22 8.4 8.3 8.1 8.6 7.9 7.8 7.7 7.6 7.5 8.9
23 8.8 8.7 8.2 8.7 8.4 7.9 7.8 7.7 7.6 9.0
24 8.9 8.8 8.6 8.9 8.5 8.3 8.2 8.1 8.1 9.1
25 9.0 8.9 8.7 9.1 8.6 8.4 8.3 8.2 8.2 9.3
26 9.2 9.1 8.8 9.2 8.8 8.5 8.4 8.3 8.4 9.5
27 9.4 9.3 9.0 9.3 9.0 8.7 8.6 8.5 8.6 9.6
28 9.5 9.4 9.2 9.4 9.1 8.9 8.8 8.7 8.7 9.7
29 9.6 9.5 9.3 9.9 9.2 9.0 8.9 8.8 8.8 9.8
30 9.7 9.6 9.4 9.3 9.1 9.0 8.9 8.9 10.3
31 10.2 10.1 9.5 9.8 9.2 9.1 9.0 9.4
32 10.0 9.7 9.6 9.5 n 31 31 32 29 31 32 32 32 31 30 x 17 18 19 19 22 23 23 23 23 15
12

252y0551h 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 a median



H
H
1
0
:

:

 
0
 
0
If p is the proportion If p is the proportion above

0 below

0



H
H
1
0
:
: p p
Hypotheses about a proportion

.
5

.
5



H
H
1
0
:
: p p

.
5

.
5
It seems easiest to let p be the proportion below 8. If you defined p as the proportion above .8, the pvalues should be identical. Our Hypotheses are

H

 H
1
0
:

:



8
and
8



H
H
1
0
:
: p p


.
5
.
5
. According to the outline, in the absence of a binomial table we must use the normal approximation to the binomial distribution. If p
 x n
is our observed proportion, we use z
 p p

0 p q
0
0
. But for the sign test, p

.
5 and q

1

.
5

.
5 . n
So z
 x n

.
5

.
25 x n

.
5
.
5
 
 n
.
5 n



.
5 n
 x

.
5 n
2 n

2 x
 n
.
n n n
(For relatively small values of n , a continuity correction is advisable, so try z

2 x

1
 n
, where the + n applies if x
 n
2
, and the

applies if cases. Both z
1

2 x
 n n x
 n
2
. ) Values of x and n are repeated below for all ten possible
and the more correct z
2

2 x

1 n
 n
(except when x
 n
2
.
) are computed. Since the alternative hypothesis is p

.
5 , the probabilities in the two right columns are p-values. If you used z
1
 p

 p p
 x n

.
5
.
25
, the values of z
1
below are correct. The exact probabilities were computed by n
Minitab in the last column. x n z
1

2 x
 n z
2

2 x

1
 n
P
 z
 z
1

P
 z
 z
2

1

F
 x

1
 n n
1 17 31 0.53882 0.35921 .5-.2054=.2946 .5-.1404=.3596 0.360050
2 18 31 0.89803 0.71842 .5-.3133=.1867 .5-.2642=.2358 0.236565
3 19 32 1.06066 0.88388 .5-.3554=.1445 .5-.3106=.1894 0.188543
4 19 29 1.67126 1.48556 .5-.4525=.0475* .5-.4319=.0681 0.068023
5 22 31 2.33487 2.15526 .5-.4901=.0099* .5-.4846=.0154* 0.014725
6 23 32 2.47487 2.29810 .5-.4934=.0066* .5-.4893=.0107* 0.010031
7 23 32 2.47487 2.29810 .5-.4934=.0066* .5-.4893=.0107* 0.010031
8 23 32 2.47487 2.29810 .5-.4934=.0066* .5-.4893=.0107* 0.010031
9 23 31 2.69408 2.51447 .5-.4964=.0036* .5-.4940=.0060* 0.005337
10 15 30 0.00000 0.00000 .5 .5 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.
13

252y0551h 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

2 z

.
2 n
. In this formula z

3
 z
.
025
We use this formula with the following results.
Row n sqrtn k k * rounded
1 31 5.56776 10.5436 10
2 31 5.56776 10.5436 10
3 32 5.65685 10.9563 10
4 29 5.38516 9.7225 9
5 31 5.56776 10.5436 10
6 32 5.65685 10.9563 10
7 32 5.65685 10.9563 10
8 32 5.65685 10.9563 10
9 31 5.56776 10.5436 10
10 30 5.47723 10.1323 10 down n
 k *

1 interval
22
22
23
23
22
23
23
23
22
21
7 .
2
7 .
1
7 .
0
6 .
9
6 .
8
6 .
7
6 .
6
6 .
5
6 .
4
7 .
3
 
 
 
 
 
 
 
 
 
 










8 .
4
8 .
3
8
8
8
7
7
7
7
8 .
.
.
.
.
.
.
.
2
6
4
9
8
7
5
5

1 .
96 .
14

252y0551h 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 Q1 Median Q3 Maximum
C4 32 7.944 0.228 1.291 4.900 7.225 8.000 8.975 10.300
Sum of squares (uncorrected) of C4 = 2070.94
I do not know the values for your numbers, but the following (copied from last year’s exam) should be useful:
  x
 a


 
 na ,
  x
 a

2 
  x
2
   zero. Can you say that the population mean is below 8? Use
 

.
05 na
.
2
. Your value of a is negative or
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
For the variance
 x

Var a

   x


E a
 x

 a
Var
 
So that good old Seymour, whose student number is 976512, subtracts 0.1 from every number and gets
Mean

E
 x

0 .
1
  

0 .
1

7 .
944

0 .
1

7 .
844
Variance

Var
 x

0 .
1


Var
  
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 s x
 s n

1 .
291
32

0 .
228 .
Note that t
 

 t
 
.
10

1 .
696
Common sense says that if the sample mean is too far below 8, we will not believe formula for a critical value for the sample mean is below 8, so use x cv
 
0
 t
 
 n

1 s x

8


1 .
696
 x cv
0 .
228



.

7
We need a critical value for
0
.
 t

 
2

1
613 s x
H
0
: x

below 8.

8 . The
, but we want a single value
.
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
15

252y0551h 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 Do not reject
7.744 Do not reject
7.644 Do not reject
7.544 Reject
7.444 Reject
7.344 Reject
7.244 Reject
7.144 Reject
7.044 Reject
7.944 Do not reject
Not below cv
Not below cv
Not below cv
Below cv
Below cv
Below cv
Below cv
Below cv
Below cv
Not below 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.
The alternative hypothesis, H
1
:
 
8 , means that the p-value is P 

 t
Version x t
1 7.844 -0.68421
Nearest t
 
.
25

0 .
680 t Values t
.
20

0 .
853

2 7.744 -1.12281 t
.
15

1 .
054 t
.
10

1 .
309 x
 
0 s x




P

 t
 x

8
0 .
228


Approx. p-value p-value is between .20 and .25.
p-value is between .10 and .15.
3 7.644 -1.56140
4 7.544 -2.00000
5 7.444 -2.43860
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 t t
.
05 t
.
05 t
.
025
.
01

1 .
309 t
.
025

2 .
040

1 .
309 t
.
025

2 .
040

2 .
040 t
.
01

2 .
453 t
.
01

2 .
453

2 .
453 t t
.
005

2 .
744 t
.
001

3 .
375 t
.
001

3 .
375 t
.
001

3 .
375
.
45

0 .
127 t
.
40

0 .
256 p-value is between .025 and .05
p-value is between .025 and .05
p-value is between .01 and .025
p-value is between .005 and .01.
p-value is between .001 and .005.
p-value is below .005
p-value is below .005
p-value is between .40 and .45.
e. Test to see if the population standard deviation is 2. s

1.291 df
To test H
0
:
  
0
against i. Test Ratio:
 2 
 n

1
 s

H
1
2
0
2
:
  
0
or for large samples z

2
 2  ii. Critical Value: s
2 cv

 2
 n
2


1
0
2 or

1
2
 
2 n


1
0
2

31 .
, (2) The outline says
2
 

1 or for large samples (from table 3) s cv


 z

2

2 DF
2 DF iii. Confidence Interval: z

2 s

2
 
2
 
  
 z
 s
2
 n

1
 s


2

2
2
2

 
2
 
 2 
 n

1
 s

1
2
 
2
2
or for large samples
.
16

252y0551h 10/31/05 (Open in ‘Print Layout’ format)
If we use a test ratio,
 2 
 n

1
 s
2
 2
0

31

1 .
291
2
2

2

12 .
9168 .
Since our degrees of freedom are too large for the table, use

5 .
0827

7 .
8102 z

2
 2
 
2 .
7275

2
 

1

2

12 .
9168


2
 

1

25 .
8336

61
. 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,

1 .
291
1 .
960

7 .
8740


7 .
8740
  
1 .
291
2

7 .
8740
1 .
96
 z
 s
7 .
7840


2
 
2
or
 
1 .
034
 
 


 z
 s
2
1 .
718
2

 
2
 
. Since
. This becomes

0
is not in this interval, reject the null hypothesis . f. (Extra credit) What would the critical values be for this test of the standard deviation? (1)
Table 3 says
2

7 .
8740
7 .
8740

 s
1 .
960 cv



15
 z

2

.
748
9 .
834
2 DF

2 DF

1 .
6014

and
2
1 .
960
2

2
 

2
 
7 .
8740
7 .
8740
 not between them, we reject the null hypothesis.


1 .
960 pval
7 .
8740

2

7 .
8740
15

.
748

2
5 .
914
P
 z


1 .
960

. The two critical values are
2 .
2663

2 .
73
. Since the standard deviation is
.
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
32 32 following points: 8, 7.709, 7.4184, 7.127 and 6.836. Our results are as follows.

7 .
4184 . We use the

1

8

1

7.709

1

7.4184

1

7.127

1

6.836
P
 x

7 .
4184
Power
 
8



1

.
95

.
05
P

 z
P
 x

7 .
4184
 
7 .
127



P

 z
Power
P
 x

1

.
2061

7 .
4184
 

.
7939
6 .
836


P

 z
Power

1

.
05

.
95
7 .
.
4184

35355
8




7 .
4184

.
35355
P
 z

P
 x

7 .
4184
 
7 .
709


Power
P
 x

1

.
7939

7 .
4184
 

.
2061
7 .
4184


P

 z
P

 z

Power

1

.
5

.
5

7 .
4184

7 .
709
.
35355
7 .
4184
.
35355




7 .
4184 

P
 z

 
0 .
82

=.5 + 0.2939 =.7939
P
 z

0


.
5

7 .
127
7 .
4184
.

6 .
836
35355





1 .
645

= .95


P
 z
P
 z

0 .
82

= .5-.2939 = .2061

1 .
647

= .5 - .4500 = .0500 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)
 
.
06 . We know
  x
 z

2
 x
.
 x

2

0 .
35355 . To find
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
17

252y0551h 10/31/05 (Open in ‘Print Layout’ format)
47% below z
.
03

1 .
88 z
.
03
. So
  x
 z
P


0
2

 z
 z
.
03 x
 x

1

.

88
.
4700 The closest we can come is
.
35355

 x

P

0
 z

1 .
88


.
4699 . So
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 come is P

0
 z

1 .
75
 z
.
04
and 46% below

.
4579 . So x cv
 
0 z
.
04
. So
 z

 x
P


0
8
 z
 z

1 .
75

.
04


.
4600
0 .
35355


The closest we can
7 .
381 If your value of the sample mean is below 7.381, reject the null hypothesis.
(ii) z
 x

 x

0  x

8
0 .
35355
. To get a p-value find P

 z
 x

8
0 .
35355


. So, for example, if x

7 .
944 , pval

P
 z

7 .
944

8
0 .
35355

P
 z
.04 or if the test ratio z
 x

 x

0  x

8
0 .
35355
(iii) The one-sided confidence interval is

 
0 .
16
is below -1.75 , reject the null hypothesis.
 x


 z
.
04

.
5

.
0636 x
 x

.
4364

1 .
75
. If the p-value is below

.
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

2
2
 2 e
2

2 .
08
2
 
2
.
005

3461 .
12 .
Use 3462.
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.
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.
2.
Was the method used to solve the problem the best and most appropriate for the problem?
3.
Was the method used correctly?
4.
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?
5.
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
18

252y0551h 10/31/05 (Open in ‘Print Layout’ format) implications of the conclusion for a relevant goal, for example the decision to buy a new product?
6.
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.
19