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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
appropriate confidence interval for the hypothesis test. (2)
[20]
Solution: a) From the formula table we have:
Interval for
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
Proportion
p  p0
p  p  z 2 s p
pcv  p0  z 2  p
H 0 : p  p0
z
p
H1 : p  p0
pq
p0 q0
sp 
p 
n
n
q  1 p
q  1 p
0
0
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.
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]
n

400
,
and
.
.


.
01
H 0 : p  .40, H 1 : p  .40,
z.005  2.576
p  .50
The critical value must be on either side of .40. pcv  p 0  z  p  .40  2.576.024495  .40  .0631 .
2
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
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.
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
a) p 0  .41  p 
Version 1
.50  .41 

below .3528. pval  P p  .50   P  z 
  Pz  3.65   .5  .4999  .9999
.
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.
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
,
and
.
.


.
01
H 0 : p  .41, H 1 : p  .41,
z.005  2.576
p  .50
The critical value must be on either side of .40. pcv  p 0  z  p  .41 2.576.0245917  .41 .0633 .
2
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
e2

.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
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.
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
a) p 0  .42  p 
Version 2
.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]
n

400
,
and
.
.


.
01
H 0 : p  .40, H 1 : p  .40,
z.005  2.576
p  .50
The critical value must be on either side of .42. pcv  p 0  z  p  .42  2.576.0246779  .40  .0636 .
2
The ‘reject’ zone is below .3564 and above .4836.
.50  .42 

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
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.
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
a) p 0  .43  p 
Version 3
.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]
n

400
,
and
.
.


.
01
H 0 : p  .43, H 1 : p  .43,
z.005  2.576
p  .50
The critical value must be on either side of .40. pcv  p 0  z  p  .43  2.576.0247538  .43  .0638 .
2
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 

(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
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.
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]
n

400
,
and
.
.


.
01
H 0 : p  .44, H 1 : p  .44,
z.005  2.576
p  .50
The critical value must be on either side of .44. pcv  p 0  z  p  .44  2.576.0248193  .44  .0639 .
2
The ‘reject’ zone is below .3761 and above .5039.
.50  .44 

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
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.
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]
n

400
,
and
.
.


.
01
H 0 : p  .45, H 1 : p  .45,
z.005  2.576
p  .50
The critical value must be on either side of .45. pcv  p 0  z  p  .45  2.576.0248747  .45  .0641 .
2
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 
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.
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
a) p 0  .46  p 
Version 6
.50  .46 

 Pz  1.60   .5  .4452  .9452
is below .4020. pval  P p  .50   P  z 
.
0249199 

(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]
n

400
,
and
.
.


.
01
H 0 : p  .46, H 1 : p  .46,
z.005  2.576
p  .50
The critical value must be on either side of .46. pcv  p 0  z  p  .46  2.576.0249199  .46  .0642 .
2
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 
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.
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
a) p 0  .47  p 
Version 7
.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]
n

400
,
and
.
.


.
01
H 0 : p  .47, H 1 : p  .47,
z.005  2.576
p  .50
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 
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.
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
a) p 0  .48  p 
Version 8
.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]
n

400
,
and
.
.


.
01
H 0 : p  .48, H 1 : p  .48,
z.005  2.576
p  .50
The critical value must be on either side of .48. pcv  p 0  z  p  .48  2.576.0249800  .48  .0643 .
2
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 

(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
e
2

.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 
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.
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]
n

400
,
and
.
.


.
01
H 0 : p  .49, H 1 : p  .49,
z.005  2.576
p  .50
The critical value must be on either side of .49. p  p 0  z  p  .49  2.576.0249950  .49  .0643 .
2
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.
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 
Minitab Calculation of critical values and test ratios for
all Versions of question 1 of the Take-home exam.
————— 10/17/2005 7:01:49 PM ————————————————————
Available in 252y0551s
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
7.6
8.5
6.3
7.6
8.9
5.7
7.7
9.0
5.5
7.7
9.1
4.9
8.0
9.3
7.0
8.0
9.5
7.1
8.1
9.6
7.2
8.2
9.7
7.3
8.4
9.8
7.3
8.4
10.3
7.4
8.4
a. Test that the median is below 8. State your null and alternate hypotheses clearly. (2)
b. (Extra credit) Find a 95% (or slightly more) two-sided confidence interval for the median. (2) [24]
Solution: The data sets for this problem are below. Let n be the count of the sample and x be the number
of numbers below 8.
Row
x1
x2
x3
x4
x5
x6
x7
x8
x9
x10
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
4.8
5.4
5.6
6.2
6.6
6.9
7.0
7.1
7.2
7.2
7.3
7.5
7.5
7.6
7.6
7.9
7.9
8.1
8.3
8.3
8.3
8.4
8.8
8.9
9.0
9.2
9.4
9.5
9.6
9.7
10.2
4.7
5.3
5.5
6.1
6.5
6.8
6.9
7.0
7.1
7.1
7.2
7.4
7.4
7.5
7.5
7.8
7.8
7.9
8.2
8.2
8.2
8.3
8.7
8.8
8.9
9.1
9.3
9.4
9.5
9.6
10.1
n
x
31
17
31
18
4.6
5.2
5.4
6.0
6.4
6.7
6.8
6.9
7.0
7.0
7.1
7.3
7.3
7.4
7.4
7.7
7.7
7.8
7.9
8.1
8.1
8.1
8.2
8.6
8.7
8.8
9.0
9.2
9.3
9.4
9.5
10.0
32
19
4.5
5.1
5.3
5.9
6.3
6.6
6.7
6.8
6.9
6.9
7.0
7.2
7.2
7.3
7.3
7.6
7.6
7.7
7.8
8.1
8.5
8.6
8.7
8.9
9.1
9.2
9.3
9.4
9.9
4.4
5.0
5.2
5.8
6.2
6.5
6.6
6.7
6.8
6.8
6.9
7.1
7.1
7.2
7.2
7.5
7.5
7.6
7.7
7.9
7.9
7.9
8.4
8.5
8.6
8.8
9.0
9.1
9.2
9.3
9.8
29
19
31
22
4.3
4.9
5.1
5.7
6.1
6.4
6.5
6.6
6.7
6.7
6.8
7.0
7.0
7.1
7.1
7.4
7.4
7.5
7.6
7.8
7.8
7.8
7.9
8.3
8.4
8.5
8.7
8.9
9.0
9.1
9.2
9.7
32
23
4.2
4.8
5.0
5.6
6.0
6.3
6.4
6.5
6.6
6.6
6.7
6.9
6.9
7.0
7.0
7.3
7.3
7.4
7.5
7.7
7.7
7.7
7.8
8.2
8.3
8.4
8.6
8.8
8.9
9.0
9.1
9.6
32
23
4.1
4.7
4.9
5.5
5.9
6.2
6.3
6.4
6.5
6.5
6.6
6.8
6.8
6.9
6.9
7.2
7.2
7.3
7.4
7.6
7.6
7.6
7.7
8.1
8.2
8.3
8.5
8.7
8.8
8.9
9.0
9.5
32
23
4.0
4.6
4.8
5.4
5.8
6.1
6.2
6.3
6.4
6.4
6.5
6.7
6.7
6.8
6.8
7.1
7.1
7.2
7.3
7.5
7.5
7.5
7.6
8.1
8.2
8.4
8.6
8.7
8.8
8.9
9.4
4.9
5.5
5.7
6.3
6.7
7.0
7.1
7.2
7.3
7.3
7.4
7.6
7.6
7.7
7.7
8.1
8.2
8.4
8.4
8.4
8.5
8.9
9.0
9.1
9.3
9.5
9.6
9.7
9.8
10.3
31
23
30
15
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
Hypotheses about a proportion
a median
If p is the proportion
If p is the proportion
above  0
below  0
 H 0 :   0
 H 0 : p .5
 H 0 : p .5



 H 1 : p  .5
H 1 :   0
 H 1 : p .5
It seems easiest to let p be the proportion below 8. If you defined p as the proportion above .8, the p-
 H :  8
 H : p  .5
values should be identical. Our Hypotheses are  0
and  0
. According to the outline, in
 H 1 :  8
 H 1 : p  .5
the absence of a binomial table we must use the normal approximation to the binomial distribution. If
p  p0
x
p  is our observed proportion, we use z 
. But for the sign test, p  .5 and q  1  .5  .5 .
n
p0 q0
n
So z 
x
n  .5
.25
n

x
n  .5
.5
2x  n
 x  .5n  n  x  .5n

2
.



 n  .5 
n
n
n
(For relatively small values of n , a continuity correction is advisable, so try z 
2x  1  n
, where the +
n
n
n
, and the  applies if x  . ) Values of x and n are repeated below for all ten possible
2
2
n
2x  n
2x 1  n
cases. Both z1 
and the more correct z 2 
(except when x  . ) are computed. Since
2
n
n
the alternative hypothesis is p  .5, the probabilities in the two right columns are p-values. If you used
applies if x 
z1 
p p

x
n  .5
, the values of z 1 below are correct. The exact probabilities were computed by
.25
n
Minitab in the last column.
2x  n
2x 1  n
x
n
z1 
z2 
P  z  z1 
P z  z 2 
1  F x  1
n
n
1
2
3
4
5
6
7
8
9
10
p
17
18
19
19
22
23
23
23
23
15
31
31
32
29
31
32
32
32
31
30
0.53882
0.89803
1.06066
1.67126
2.33487
2.47487
2.47487
2.47487
2.69408
0.00000
0.35921
0.71842
0.88388
1.48556
2.15526
2.29810
2.29810
2.29810
2.51447
0.00000
.5-.2054=.2946
.5-.3133=.1867
.5-.3554=.1445
.5-.4525=.0475*
.5-.4901=.0099*
.5-.4934=.0066*
.5-.4934=.0066*
.5-.4934=.0066*
.5-.4964=.0036*
.5
.5-.1404=.3596
.5-.2642=.2358
.5-.3106=.1894
.5-.4319=.0681
.5-.4846=.0154*
.5-.4893=.0107*
.5-.4893=.0107*
.5-.4893=.0107*
.5-.4940=.0060*
.5
0.360050
0.236565
0.188543
0.068023
0.014725
0.010031
0.010031
0.010031
0.005337
0.572232
Since   .05 , and the starred items are below .05, these are the cases in which we reject the hypothesis
that the median is, at least 8. It makes about as much sense to reject a hypothesis about a median
because the sample median x.50 is above or below  0 with no other information as it does to reject a
hypothesis about a mean because x is above or below  0 without more information.
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  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
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
C4
32
7.944
0.228 1.291
4.900
Sum of squares (uncorrected) of C4 = 2070.94
Q1
7.225
Median
8.000
Q3
8.975
Maximum
10.300
I do not know the values for your numbers, but the following (copied from last year’s exam) should be
useful:
 x  a    x   na, x  a2   x 2  2a x na2 . Your value of a is negative or
zero. Can you say that the population mean is below 8? Use   .05 .
If you remembered what we learned last semester, you could have done this calculation with little or
no work. The relevant formulas are:
For the mean Ex  a   Ex   a
For the variance Var x  a   Var x 
So that good old Seymour, whose student number is 976512, subtracts 0.1 from every number and gets
Mean  Ex  0.1  Ex   0.1  7.944  0.1  7.844
Variance  Var x  0.1  Var x   1.291 2
a. State your null and alternative hypotheses (1)
b. Find a critical value for the sample mean and a ‘reject’ zone. Make a diagram! (1)
c. Do you reject the null hypothesis? Use the diagram to show why. (1)
d. Find an approximate p-value for the null hypothesis. Make sure that I know where you got your results.
(1)
e. Test to see if the population standard deviation is 2. (2)
f. (Extra credit) What would the critical values be for this test of the standard deviation? (1)
g. Assume that the population standard deviation is 2, restate your critical value for the mean and create a
power curve for your test. (5)
h. Assume that the population standard deviation is 2 and find a 94% (this does not say 95%!) two sided
confidence interval for the mean. (1)
i. Using a 96% confidence level and assuming that the population standard deviation is 2, test that the mean
is below 8. (1)
j. Using a 96% confidence interval an assuming that the population standard deviation is 2, how large a
sample do you need to have an error in the mean of .005 ? (1)
[39]
Solution:
a. State your null and alternative hypotheses (1)
Since the problem asks if the mean is below   8 , and this does not contain an equality, it must
be an alternate hypothesis. Our hypotheses are
H 0 :   8 , (The average wait is at least 8 minutes.) and
H 1 :   8 (The average wait is less than 8 minutes.) This is a left-sided test.
b. Find a critical value for the sample mean and a ‘reject’ zone. Make a diagram! (1)
Given:  0  8, s  1.291, n  32, df  n  1  31, and   .05 . So
sx 
s

1.291
31
 1.696 . We need a critical value for x below 8.
 0.228 . Note that tn1  t .10
n
32
Common sense says that if the sample mean is too far below 8, we will not believe H 0 :   8 . The
formula for a critical value for the sample mean is x    t n1 s , but we want a single value
cv
0

2
x
below 8, so use xcv   0  tn1 s x  8  1.696 0.228   7.613 . Make a diagram showing an
almost Normal curve with a mean at 8 and a shaded 'reject' zone below 7.613.
I should not have to say that if you have found a critical value, the ‘reject’ zone should be either
above or below it, depending on whether it is a right or left side test.  0 , the null hypothesis
mean is never in the ‘reject’ zone. If your critical value is x cv and x is in the ‘reject’ zone, you
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
7.744
7.644
7.544
7.444
7.344
7.244
7.144
7.044
7.944
Do not
Do not
Do not
Reject
Reject
Reject
Reject
Reject
Reject
Do not
reject
reject
reject
Not below
Not below
Not below
Below cv
Below cv
Below cv
Below cv
Below cv
Below cv
Not below
reject
cv
cv
cv
cv
d. Find an approximate p-value for the null hypothesis. Make sure that I know where you got your results.
n  32 means df  31 . The closest values from the t table to the computed t are shown.
 x  0 
x 8 

The alternative hypothesis, H 1 :   8 , means that the p-value is P t 
  P t 
s
0
.228 

x


Version
Nearest t Values
Approx. p-value
x
t


31
31
1
7.844 -0.68421
p-value is between .20 and .25.
t .25  0.680 t .20  0.853
2
31
31
t .15
 1.054 t .10
 1.309
7.744 -1.12281
3
7.644 -1.56140
4
7.544 -2.00000
31
t .05
 1.309

31
t
 1.309
5
7.444 -2.43860
 2.040
6
7.344 -2.87719
7
7.244 -3.31579
8
7.144 -3.75439
9
7.044 -4.19298
10
.05
31
t .025
31
t .01
31
t .005
31
t .001
31
t .001
31
t .45
7.944 -0.24561
 2.453
 2.744
 2.453
p-value is between .01 and .025
 2.453
p-value is between .005 and .01.
 3.375
p-value is between .001 and .005.
.025
31
t .01
31
t .01
31
t .001
 3.375
p-value is between .025 and .05
p-value is between .025 and .05
p-value is below .005
 3.375
 0.127
p-value is between .10 and .15.
31
t .025
 2.040

31
t
 2.040
p-value is below .005
31
t .40
 0.256
p-value is between .40 and .45.
e. Test to see if the population standard deviation is 2. s  1.291 df  31 . , (2) The outline says
To test H 0 :    0 against H1 :    0
i. Test Ratio:  2 
ii. Critical Value:
n  1s 2
2
s cv
 02

iii. Confidence Interval:
s 2DF 
z 2  2DF 
 
or for large samples z  2  2  2DF   1
 2  02
2
n 1
or
n  1s 2
 22
12 2  02
n 1
2 
or for large samples (from table 3) s cv 
n  1s 2
12 2
 2 DF
 z  2  2 DF
.
or for large samples
s 2DF 
 z 2  2DF 
16
252y0551h 10/31/05 (Open in ‘Print Layout’ format)
If we use a test ratio,  2 
n  1s 2
 02

311.291 2
22
 12 .9168 . Since our degrees of freedom are too large
for the table, use z  2  2  2DF   1  212 .9168   231  1  25.8336  61
 5.0827  7.8102  2.7275 . If we are doing a 2-sided 5% test, we do not reject the null hypothesis if z
is between -1.960 and 1.960. It’s not, so we reject the null hypothesis.
If we use a confidence interval,
s 2DF 
z 2  2DF 
 
s 2DF 
 z 2  2DF 
. This becomes
1.2917.8740 
1.2917.8740 
or 1.034    1.718 . Since  0 is not in this interval, reject the null
 
 1.960  7.8740
1.96  7.7840
hypothesis .
f. (Extra credit) What would the critical values be for this test of the standard deviation? (1)
Table 3 says s cv 
 2 DF
 z  2  2 DF

2 231
 1.960  231

27.8740 
. The two critical values are
7.8740  1.960
27.8740 
27.8740 
15 .748
15 .748

 1.6014 and

 2.2663 . Since the standard deviation is
7.8740  1.960
9.834
7.8740  1.960
5.914
not between them, we reject the null hypothesis. pval  2Pz  2.73  2.5  .4968   .0064 .
g. Assume that the population standard deviation is 2, restate your critical value for the mean and create a
power curve for your test. (5). Our hypotheses are H 0 :   8 , (The average wait is at least 8 minutes.) and
H 1 :   8 (The average wait is less than 8 minutes.) This is a left-sided test. If   2 , n  32 ,
x 
2
 0.35355 and   .05 , our critical value is   1.645
2
 8  0.3762  7.4184 . We use the
32
32
following points: 8, 7.709, 7.4184, 7.127 and 6.836. Our results are as follows.
7.4184  8 

Px  7.4184   8  P  z 
 Pz  1.645  = .95
1  8
.35355 

Power  1  .95  .05
7.4184  7.709 

1  7.709
P x  7.4184  7.709  P  z 
  Pz  0.82  =.5 + 0.2939 =.7939
.35355


Power  1  .7939  .2061
7.4184  7.4184 

1  7.4184
P x  7.4184   7.4184  P  z 
  Pz  0  .5
.35355


Power  1  .5  .5
7.4184  7.127 

1  7.127
P x  7.4184   7.127  P  z 
  Pz  0.82  = .5-.2939 = .2061
.35355


Power  1  .2061  .7939
7.4184  6.836 

1  6.836
P x  7.4184   6.836  P  z 
  Pz  1.647  = .5 - .4500 = .0500
.35355


Power  1  .05  .95








h. Assume that the population standard deviation is 2 and find a 94% (this does not say 95%!) two sided
2
 0.35355 . To find
confidence interval for the mean. (1)   .06 . We know   x  z  x .  x 
2
32
z .03 make a Normal diagram for z showing a mean at 0 and 50% above 0, divided into 3% above z .03 and
17
252y0551h 10/31/05 (Open in ‘Print Layout’ format)
47% below z .03 . So P0  z  z.03   .4700 The closest we can come is P0  z  1.88   .4699 . So
z .03  1.88   x  z  x  x  1.88.35355  x  0.665 . Substitute your value of the sample mean.
2
i. Using a 96% confidence level and assuming that the population standard deviation is 2, test that the mean
is below 8. (1) Our hypotheses are H 0 :   8 , (The average wait is at least 8 minutes.) and H 1 :   8
(The average wait is less than 8 minutes.) This is a left-sided test. We can do this by a critical value, a test
ratio or a confidence interval.
(i)   .04 To find z .04 make a Normal diagram for z showing a mean at 0 and 50% above 0,
divided into 4% above z .04 and 46% below z .04 . So P0  z  z.04   .4600 The closest we can
come is P0  z  1.75   .4579 . So x cv   0  z  x  8  1.750.35355   7.381 If your value of
the sample mean is below 7.381, reject the null hypothesis.
x  0
x 8 
x 8

(ii) z 
. To get a p-value find P z 

 . So, for example, if
0
.
35355 
x
0.35355

7.944  8 

x  7.944 , pval  P z 
  Pz  0.16   .5  .0636  .4364 . If the p-value is below
0.35355 

x  0
x 8
.04 or if the test ratio z 
is below -1.75 , reject the null hypothesis.

x
0.35355
(iii) The one-sided confidence interval is   x  z.04 x  x  1.75.35355   x  0.6187 . If this
value is below 8, reject the null hypothesis.
j. Using a 96% confidence level and assuming that the population standard deviation is 2, how large a
sample do you need to have an error in the mean of .005 ? (1)
  .04 To find z .02 make a Normal diagram for z showing a mean at 0 and 50% above 0, divided into
2% above z .02 and 48% below z .04 . So P0  z  z.02   .4800 The closest we can come is
P0  z  2.08   .4798 From the outline n 
z 22  2
e
2

2.08 2 22
 3461 .12 . Use 3462.
.005
For Minitab Computations for Problem 3, see 252y0551s..
What follows below is excerpted from the solution to the first exam from Spring
2005.
Possible Rubric for Statistics Exams.
I have been hearing a lot about rubrics lately, and have taken a while to be assured that they are
not the materials that the third pig built his house out of. My first attempt at this came to me in a recent
assessment meeting. It is slightly expanded here.
1.
2.
3.
4.
5.
Did the student make a good effort to understand the question? This would include asking
the instructor and consulting notes and texts if he/she did not understand what was
desired.
Was the method used to solve the problem the best and most appropriate for the problem?
Was the method used correctly?
Did the student present the solution in such a way that the instructor can understand how
the student got the answers presented? This should include all formulas, equations and
tables used. Is it evident from the way the work is presented that the student understood
what he/she was doing? Is it legible?
Was the conclusion stated clearly? Was the null hypotheses rejected or not rejected? Was
a valid null and alternate hypothesis clearly stated at the beginning? What were the
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252y0551h 10/31/05 (Open in ‘Print Layout’ format)
6.
implications of the conclusion for a relevant goal, for example the decision to buy a new
product?
Did the student demonstrate knowledge of the difference between sample statistics and
population parameters? Was a statistical test using sample statistics used to evaluate a
null hypothesis containing population parameters and an equality?
In view of what was said here, it is incredible that, on every exam I give, students give me confidence
intervals and tests for means when I ask for confidence intervals and tests for medians, variances and even
proportions. Check the wording on the questions that you misunderstood. Can you identify what wording in
the question made you think it was about a mean? Can you tell me what it was? It is also remarkable that
there are any people out there who do not know that proportions, probabilities and p-values (which are
probabilities) must be between one and zero. It is also amazing to me that that so many of you cannot
express the difference between t and z . In the most practical sense a value of t  comes from the t table
and must be used with s , the sample standard deviation in confidence intervals and tests for the population
mean. There are only a few other cases where we use t  and they will be discussed later in the course. On
the other hand z  , which comes only from the bottom line of the t table, but can be calculated using the
table of the standardized Normal distribution, must be used with  , the population standard deviation, in
confidence intervals and tests for the mean. z  is also used in large sample tests for the population
proportion, population mean, population standard deviation, population median and the means of the
Poisson and Binomial distribution if the correct formulas are used, but don’t push it. The Normal
distribution should not be used if more accurate methods are available. In any case, look at Things You
Should Never Do on and Exam or Anywhere Else before you do another assignment and frequently
thereafter.
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