252y0313a 10/02/03 Don’t waste paper by copying more of this document than you need.
IV. Solution to first problem on TAKE HOME SECTION
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.
1. An airline president is tracking late arrivals and believes that the proportion is at most p 0 . Suppose
that a sample of 200 flights is selected and 82 were late. Do the following:
a) To find p 0 , take the third digit of your Social Security Number divide by 100 and add it to .30.
For example, my Social Security Number is 265398248 and the third digit is 5, so the value of p 0
that you would use is 30  5% or .35. (no point credit for this section.)
b) Using the value of p 0 that you found in a) prepare to conduct a test to determine whether there
is evidence that the proportion is at most p 0 by stating your null and alternative hypotheses. (1)
c) Find a critical value for the sample proportion for the hypotheses in b), using a significance
level of 1% specify what your ‘reject’ region is and use it to test the null hypothesis. (2)
d) (Extra credit) Assume that the actual population proportion is p 0  .03 . (Since my p 0 was .35,
I would assume that p1 was .38) , find the power of the test in c). If I had used a lower significance
level, explain whether the power would be higher, lower, or the same. (3.5)
e) Compute a test ratio for the hypotheses in c) and test the hypotheses using a significance level of
1% (2)
f) Use your test ratio in e) to get a p-value for the hypothesis in c) and explain whether and why
you would reject the null hypothesis if the significance level was 3%. (2)
g) Test the hypotheses in c) using an appropriate confidence interval and a significance level of 1%
(2)
h) If you were doing a 2-sided 99% confidence interval for the proportion of flights that were late
and wanted the proportion to be known within .01 , how large a sample would you use if you expected
the proportion to be about 20%? What if you thought the proportion was about 4%? (2)
i) (Extra credit) do a power curve for the test in c), using a few carefully chosen values of p1 that
are above your p 0 . (4.5)
Solution: From the formula table we have:
Interval for
Confidence
Hypotheses
Interval
Proportion
p  p  z 2 s p
H 0 : p  p0
pq
n
q  1 p
sp 
H1 : p  p0
Test Ratio
z
p  p0
p
Critical Value
pcv  p0  z 2  p
p0 q0
n
q0  1  p0
p 
Note: Rather than compute the standard error of p more than twice in each solution, I have inserted a table
p1 q1
at the end of this document. To use the solutions find your value of p 0 . A 2-page
n
solution for each version of the problem will follow.
of  p 
1
252y0313a 10/02/03
a) p0  .30 The problem says that n  200 , x  82 and p 
sp 
pq

n
x
82

 .41 . q  1  p  .59. so, for later
n 200
.41.59 
 .0012095  .0347779
200
p0 q0
.30 .70 

 .00105  .0324037
n
200
c) Since the alternate hypothesis is H1 : p  p0 , the critical value must be above p0 , and we must use a
b) H 0 : p  .30, H1 : p  .30 and, for later,  p 
right-tail test.   .01 and z  z.01  2.327 . Use
pcv  p0  z p  .30  2.327.0324037  .30  .0754  .3754. The ‘reject, region must be above .3754.
Make a diagram showing a Normal curve with a mean at p0  .30 and a shaded reject region above .3754.
Since p  .41 is in this ‘reject’ region, reject H 0 .
p1 q1
.33.67 

 .0011055  .0332491
n
200
not reject H 0 : p  .30 if p  pcv  .3754 . So   P p  .3754 p  .33
d) p1  p0  .03  .30  .03  .33 . From this  p 
We do
.3754  .33 

 P z 
  Pz  1.37   .5  .4147  .9147 . Power  1    .0853.
.0332491 

Since pcv  p0  z p  .30  2.327.0324037  .30  .0754  .3754, if I had used a lower significance
level, I would use a larger value of z , which means a larger value of pcv , which would make my test
ratio larger, which would make  larger and the power smaller.
e) The test ratio formula is z 
p  p0
p
. p  .41 . We already know that  p  .0324037 and that
z  z.01  2.327 . Our ‘reject’ region is above 2.327. We make a diagram showing a Normal curve with a
mean of zero and shade the area above 2.327. Since z 
p  p0
p

.41  .30
 3.39 is in the ‘reject region,
.0324037
we reject H 0 .
f) Since this is a right-tail test, p  value  Pz  3.39   .5  .4997  .0003 . Since this p-value is below
3%, reject H 0 .
g) Since the alternate hypotheses is H1 : p  .30 , the confidence interval is p  p  z s p
 .41  2.327 .0347779
 or
p  .3291. Since p0  .30 is not on this interval, reject H 0 .
h) From the outline: The usually suggested formula is n 
forgets that we covered.
p  .20, e  .01,   .01, z 2  z.005  2.576 and n 
p  .04, e  .01,   .01, z 2  z.005  2.576 and
pqz 2
e2
, ………… This is the formula everyone
.20 .80 2.576 2
.012
.04 .96 2.576 2
n
.012
 10617 .2  10618 .
 2548 .1  2549 .
2
252y0313a 10/02/03
i) We have  for the following values of 1 : .30, .33, .3754, since for 1   0 , power =  , and for
1  xcv , power = 0 . Let’s try two points above the critical value, say .41 and .45.
Our table will read as below, with supporting computations following.
1

Power 1  
.30
.99
.01
.33
.9147 .0853
.3754 .5
.5
.41
.1736 .8264
.45
.0197 .9803
Graph off your y-axis with intervals of 0.1 for power between 0 and 1.0 and set up your x-axis with
divisions of .2 for 1 between .30 and .45 and graph the ‘Power’ points.
  P p  .3754 p  .41  P z 


.3754  .41 
  Pz  0.94   .5  .3264  .1736 .
.0347779 
We can write
  P p  .3754 p  .45  P z 


.3754  .45 
  Pz  2.06   .5  .4803  .0197 .
.0351781 
3
252y0313a 10/02/03
a) p0  .31 The problem says that n  200 , x  82 and p 
later s p 
pq

n
x
82

 .41 . q  1  p  .59. so, for
n 200
.41.59 
 .0012095  .0347779
200
p0 q0
.31.69 

 .00106950  .0327032
n
200
c) Since the alternate hypothesis is H1 : p  p0 , the critical value must be above p0 , and we must use a
b) H 0 : p  .31, H1 : p  .31 and, for later,  p 
right-tail test.   .01 and z  z.01  2.327 . Use
pcv  p0  z  p  .31 2.327.0327032  .31 .0761  .3861. The ‘reject, region must be above .3861.
Make a diagram showing a Normal curve with a mean at p0  .31 and a shaded reject region above .3861.
Since p  .41 is in this ‘reject’ region, reject H 0 .
d) p1  p0  .03  .31  .03  .34 . From this  p 
.34 .66 
 .001122  .0334963
200
p1q1

n

We do

not reject H 0 : p  .31 if p  p cv  .3861 . So   P p  .3861 p  .34
.3861  .34 

 P z 
  Pz  1.38   .5  .4162  .9162 . Power  1    .0838.
.0334963 

Since pcv  p0  z  p  .31 2.327.0327032  .31 .0761  .3861, if I had used a lower significance
level, I would use a larger value of z , which means a larger value of pcv , which would make my test
ratio larger, which would make  larger and the power smaller.
e) The test ratio formula is z 
p  p0
p
. p  .41 . We already know that  p  .0327032 and that
z  z.01  2.327 . Our ‘reject’ region is above 2.327. We make a diagram showing a Normal curve with a
mean of zero and shade the area above 2.327. Since z 
p  p0
p

.41  .31
 3.06 is in the ‘reject region,
.0327032
we reject H 0 .
f) Since this is a right-tail test, p  value  Pz  3.06   .5  .4989  .0011 . Since this p-value is below
3%, reject H 0 .
g) Since the alternate hypotheses is H1 : p  .31 , the confidence interval is p  p  z s p
 .41  2.327 .0347779
 or
p  .3291. Since p0  .31 is not on this interval, reject H 0 .
h) From the outline: The usually suggested formula is n 
forgets that we covered.
p  .20, e  .01,   .01, z 2  z.005  2.576 and n 
p  .04, e  .01,   .01, z 2  z.005  2.576 and
pqz 2
e2
, ………… This is the formula everyone
.20 .80 2.576 2
.012
.04 .96 2.576 2
n
.012
 10617 .2  10618 .
 2548 .1  2549 .
4
252y0313a 10/02/03
i) We have  for the following values of 1 : .31, .34, .3861, since for 1   0 , power =  , and for
1  xcv , power = 0 . Let’s try two points above the critical value, say .42 and .46.
Our table will read as below, with supporting computations following.
1

Power 1  
.31
.99
.01
.34
.9162 .0838
.3861 .5
.5
.42
.1660 .8340
.46
.0179 .9821
Graph off your y-axis with intervals of 0.1 for power between 0 and 1.0 and set up your x-axis with
divisions of .2 for 1 between .31 and .50 and graph the ‘Power’ points.
  P p  .3861 p  .42  P z 


.3861  .42 
  Pz  0.97   .5  .3340  .1660 .
.0348999 
We can write
  P p  .3861 p  .46  P z 


.3861  .46 
  Pz  2.10   .5  .4821  .0179 .
.0352420 
5
252y0313a 10/02/03
a) p0  .32 The problem says that n  200 , x  82 and p 
later s p 
pq

n
x
82

 .41 . q  1  p  .59. so, for
n 200
.41.59 
 .0012095  .0347779
200
p0 q0
.32 .68 

 .001088  .0329848
n
200
c) Since the alternate hypothesis is H1 : p  p0 , the critical value must be above p0 , and we must use a
b) H 0 : p  .32, H1 : p  .32 and, for later,  p 
right-tail test.   .01 and z  z.01  2.327 . Use
pcv  p0  z  p  .32  2.327.0329848  .30  .0768  .3968. The ‘reject, region must be above .3968.
Make a diagram showing a Normal curve with a mean at p0  .32 and a shaded reject region above .3968.
Since p  .41 is in this ‘reject’ region, reject H 0 .
d) p1  p0  .03  .32  .03  .35 . From this  p 
p1q1
.35 .65 

 .0011375  .0337268
n
200

We do

not reject H 0 : p  .32 if p  p cv  .3968 . So   P p  .3968 p  .35
.3968  .35 

 P z 
  Pz  1.38   .5  .4162  .9162 . Power  1    .0838.
.0337268 

Since pcv  p0  z  p  .32  2.327.0329848  .30  .0768  .3968. if I had used a lower significance
level, I would use a larger value of z , which means a larger value of pcv , which would make my test
ratio larger, which would make  larger and the power smaller.
e) The test ratio formula is z 
p  p0
p
. p  .41 . We already know that  p  .0329848 and that
z  z.01  2.327 . Our ‘reject’ region is above 2.327. We make a diagram showing a Normal curve with a
p  p0
.41  .32

 2.73 is in the ‘reject region,
mean of zero and shade the area above 2.327. Since z 
p
.0329848
we reject H 0 .
f) Since this is a right-tail test, p  value  Pz  2.73   .5  .4968  .0032 . Since this p-value is below 3%,
reject H 0 .
g) Since the alternate hypotheses is H 1 : p  .32 , the confidence interval is p  p  z s p
 .41  2.327 .0347779
 or
p  .3291. Since p0  .32 is not on this interval, reject H 0 .
h) From the outline: The usually suggested formula is n 
forgets that we covered.
p  .20, e  .01,   .01, z 2  z.005  2.576 and n 
p  .04, e  .01,   .01, z 2  z.005  2.576 and
pqz 2
e2
, ………… This is the formula everyone
.20 .80 2.576 2
.012
.04 .96 2.576 2
n
.012
 10617 .2  10618 .
 2548 .1  2549 .
6
252y0313a 10/02/03
i) We have  for the following values of 1 : .32, .35, .3968, since for 1   0 , power =  , and for
1  xcv , power = 0 . Let’s try two points above the critical value, say .43 and .47.
Our table will read as below, with supporting computations following.
1

Power 1  
.32
.99
.01
.35
.9162 .0838
.3968 .5
.5
.43
.1771 .8289
.47
.0192 .9808
Graph off your y-axis with intervals of 0.1 for power between 0 and 1.0 and set up your x-axis with
divisions of .2 for 1 between .32 and .50 and graph the ‘Power’ points.
  P p  .3968 p  .43  P z 


.3968  .43 
  Pz  0.95   .5  .3289  .1771 .
.0350071 
We can write
  P p  .3968 p  .47  P z 


.3968  .47 
  Pz  2.07   .5  .4808  .0192 .
.0352916 
7
252y0313a 10/02/03
a) p0  .33 The problem says that n  200 , x  82 and p 
later s p 
pq

n
x
82

 .41 . q  1  p  .59. so, for
n 200
.41.59 
 .0012095  .0347779
200
p0 q0
.33.67 

 .0011055  .0332491
n
200
c) Since the alternate hypothesis is H1 : p  p0 , the critical value must be above p0 , and we must use a
b) H 0 : p  .33, H1 : p  .33 and, for later,  p 
right-tail test.   .01 and z  z.01  2.327 . Use
pcv  p0  z  p  .33  2.327.0332491  .33  .0774  .4074. The ‘reject, region must be above .4074.
Make a diagram showing a Normal curve with a mean at p0  .33 and a shaded reject region above .4074.
Since p  .41 is in this ‘reject’ region, reject H 0 .
d) p1  p0  .03  .33  .03  .36 . From this  p 
.36 .64 
 .001152  .0339411
200
p1q1

n

We do

not reject H 0 : p  .33 if p  pcv  .4074 . So   P p  .4074 p  .36
.4074  .36 

 P z 
  Pz  1.39   .5  .4177  .9177 . Power  1    .0823.
.033941 

Since pcv  p0  z  p  .33  2.327.0332491  .33  .0774  .4074. if I had used a lower significance
level, I would use a larger value of z , which means a larger value of pcv , which would make my test
ratio larger, which would make  larger and the power smaller.
e) The test ratio formula is z 
p  p0
p
. p  .41 . We already know that  p  .0324491 and that
z  z.01  2.327 . Our ‘reject’ region is above 2.327. We make a diagram showing a Normal curve with a
mean of zero and shade the area above 2.327. Since z 
p  p0
p

.41  .33
 2.41 is in the ‘reject region,
.0332491
we reject H 0 .
f) Since this is a right-tail test, p  value  Pz  2.41  .5  .4920  .0080 . Since this p-value is below
3%, reject H 0 .
g) Since the alternate hypotheses is H 1 : p  .33 , the confidence interval is p  p  z s p
 .41  2.327 .0347779
 or
p  .3291. Since p0  .33 is on this interval, do not reject H 0 . This is an
unusual result, which I knew was possible, but had never seen before. If we use  p  .0324491 in place of
s p  .0347779, we get a rejection.
h) From the outline: The usually suggested formula is n 
forgets that we covered.
p  .20, e  .01,   .01, z 2  z.005  2.576 and n 
p  .04, e  .01,   .01, z 2  z.005  2.576 and
pqz 2
e2
, ………… This is the formula everyone
.20 .80 2.576 2
.012
.04 .96 2.576 2
n
.012
 10617 .2  10618 .
 2548 .1  2549 .
8
252y0313a 10/02/03
i) We have  for the following values of 1 : .33, .36, .4074, since for 1   0 , power =  , and for
1  xcv , power = 0 . Let’s try two points above the critical value, say .44 and .48.
Our table will read as below, with supporting computations following.
1

Power 1  
.33
.99
.01
.36
.9177 .0823
.4074 .5
.5
.44
.1762 .8238
.48
.0197 .9803
Graph off your y-axis with intervals of 0.1 for power between 0 and 1.0 and set up your x-axis with
divisions of .2 for 1 between .33 and .50 and graph the ‘Power’ points.
  P p  .4074 p  .44  P z 


.4074  .44 
  Pz  0.93   .5  .3238  .1762 .
.0350999 
We can write
  P p  .4074 p  .48  P z 


.4074  .48 
  Pz  2.06   .5  .4803  .0197 .
.0353270 
9
252y0313a 10/02/03
a) p0  .34 The problem says that n  200 , x  82 and p 
later s p 
pq

n
x
82

 .41 . q  1  p  .59. so, for
n 200
.41.59 
 .0012095  .0347779
200
p0 q0
.34 .66 

 .001122  .0334963
n
200
c) Since the alternate hypothesis is H1 : p  p0 , the critical value must be above p0 , and we must use a
b) H 0 : p  .34, H1 : p  .34 and, for later,  p 
right-tail test.   .01 and z  z.01  2.327 . Use
pcv  p0  z  p  .34  2.327.0334963  .34  .0779  .4179. The ‘reject, region must be above .4179.
Make a diagram showing a Normal curve with a mean at p0  .34 and a shaded reject region above .4179.
Since p  .41 is not in this ‘reject’ region, do not reject H 0 .
d) p1  p0  .03  .34  .03  .37 . From this  p 
.37 .63 
 .0011655  .0341394
200
p1q1

n

We do

not reject H 0 : p  .34 if p  pcv  .4179 . So   P p  .4179 p  .37
.4179  .37 

 P z 
  Pz  1.40   .5  .4192  .9192 . Power  1    .0808.
.0341394 

Since pcv  p0  z  p  .34  2.327.0334963  .34  .0779  .4179. if I had used a lower significance
level, I would use a larger value of z , which means a larger value of pcv , which would make my test
ratio larger, which would make  larger and the power smaller.
e) The test ratio formula is z 
p  p0
p
. p  .41 . We already know that  p  .0334963 and that
z  z.01  2.327 . Our ‘reject’ region is above 2.327. We make a diagram showing a Normal curve with a
mean of zero and shade the area above 2.327. Since z 
p  p0
p

.41  .34
 2.09 is not in the ‘reject
.0334963
region, we do not reject H 0 .
f) Since this is a right-tail test, p  value  Pz  2.09   .5  .4817  .0183 . Since this p-value is below
3%, reject H 0 .
g) Since the alternate hypotheses is H1 : p  .30 , the confidence interval is p  p  z s p
 .41  2.327 .0347779
 or
p  .3291. Since p0  .34 is on this interval, do not reject H 0 .
h) From the outline: The usually suggested formula is n 
forgets that we covered.
p  .20, e  .01,   .01, z 2  z.005  2.576 and n 
p  .04, e  .01,   .01, z 2  z.005  2.576 and
pqz 2
e2
, ………… This is the formula everyone
.20 .80 2.576 2
.012
.04 .96 2.576 2
n
.012
 10617 .2  10618 .
 2548 .1  2549 .
10
252y0313a 10/02/03
i) We have  for the following values of 1 : .34, .37, .4179, since for 1   0 , power =  , and for
1  xcv , power = 0 . Let’s try two points above the critical value, say .45 and .49.
Our table will read as below, with supporting computations following.
1

Power 1  
.34
.99
.01
.37
.9177 .0823
.4179 .5
.5
.45
.1762 .8238
.49
.0207 .9793
Graph off your y-axis with intervals of 0.1 for power between 0 and 1.0 and set up your x-axis with
divisions of .2 for 1 between .34 and .50 and graph the ‘Power’ points.
  P p  .4179 p  .45  P z 


.4174  .45 
  Pz  0.93   .5  .3238  .1762 .
.0351781 
We can write
  P p  .4179 p  .49  P z 


.4179  .49 
  Pz  2.04   .5  .4793  .0207 .
.0353483 
11
252y0313a 10/02/03
a) p0  .35 The problem says that n  200 , x  82 and p 
later s p 
pq

n
x
82

 .41 . q  1  p  .59. so, for
n 200
.41.59 
 .0012095  .0347779
200
p0 q0
.35.65 

 .0011375  .0337268
n
200
c) Since the alternate hypothesis is H1 : p  p0 , the critical value must be above p0 , and we must use a
right-tail test.   .01 and z  z.01  2.327 . Use
pcv  p0  z  p  .35  2.327.0337268  .35  .0785  .4285. The ‘reject, region must be above .4285.
b) H 0 : p  .35, H 1 : p  .35 and, for later,  p 
Make a diagram showing a Normal curve with a mean at p0  .35 and a shaded reject region above .4285.
Since p  .41 is not in this ‘reject’ region, do not reject H 0 .
p1 q1
.38.62 

 .0011780  .0343220
n
200
 ..4285 . So   P p  .4285 p  .38
d) p1  p 0  .03  .35  .03  .38. From this  p 
not reject H 0 : p  .35 if p  p cv
We do
.4285  .38 

 P z 
  Pz  1.41  .5  .4207  .9207 . Power  1    .0793.
.034322 

Since pcv  p0  z  p  .35  2.327.0337268  .35  .0785  .4285, if I had used a lower significance
level, I would use a larger value of z , which means a larger value of pcv , which would make my test
ratio larger, which would make  larger and the power smaller.
e) The test ratio formula is z 
p  p0
p
. p  .41 . We already know that  p  .0337268 and that
z  z.01  2.327 . Our ‘reject’ region is above 2.327. We make a diagram showing a Normal curve with a
p  p0
.41  .35

 1.78 is not in the ‘reject
mean of zero and shade the area above 2.327. Since z 
p
.0337268
region, we do not reject H 0 .
f) Since this is a right-tail test, p  value  Pz  1.78   .5  .4625  .0375 . Since this p-value is above 3%,
do not reject H 0 .
g) Since the alternate hypotheses is H 1 : p  .35 , the confidence interval is p  p  z s p
 .41  2.327 .0347779
 or
p  .3291. Since p 0  .35 is on this interval, do not reject H 0 .
h) From the outline: The usually suggested formula is n 
forgets that we covered.
p  .20, e  .01,   .01, z 2  z.005  2.576 and n 
p  .04, e  .01,   .01, z 2  z.005  2.576 and
pqz 2
e2
, ………… This is the formula everyone
.20 .80 2.576 2
.012
.04 .96 2.576 2
n
.012
 10617 .2  10618 .
 2548 .1  2549 .
12
252y0313a 10/02/03
i) We have  for the following values of 1 : .35, .38, .4285, since for 1   0 , power =  , and for
1  xcv , power = 0 . Let’s try two points above the critical value, say .46 and .50.
Our table will read as below, with supporting computations following.
1

Power 1  
.35
.99
.01
.38
.9207 .0793
.4285 .5
.5
.46
.1867 .8133
.50
.0217 .9783
Graph off your y-axis with intervals of 0.1 for power between 0 and 1.0 and set up your x-axis with
divisions of .2 for 1 between .35 and .55 and graph the ‘Power’ points.
  P p  .4285 p  .46  P z 


.4285  .46 
  Pz  0.89   .5  .3133  .1867 .
.0352420 
We can write
  P p  .4285 p  .50  P z 


.4285  .50 
  Pz  2.02   .5  .4783  .0217 .
.0353553 
13
252y0313a 10/02/03
a) p0  .36 The problem says that n  200 , x  82 and p 
later s p 
pq

n
x
82

 .41 . q  1  p  .59. so, for
n 200
.41.59 
 .0012095  .0347779
200
p0 q0
.36.64 

 .0011520  .0339411
n
200
c) Since the alternate hypothesis is H1 : p  p0 , the critical value must be above p0 , and we must use a
right-tail test.   .01 and z  z.01  2.327 . Use
pcv  p0  z  p  .36  2.327.0339411  .36  .0790  .4390. The ‘reject, region must be above .4390.
b) H 0 : p  .36, H 1 : p  .36 and, for later,  p 
Make a diagram showing a Normal curve with a mean at p0  .30 and a shaded reject region above .4390.
Since p  .41 is not in this ‘reject’ region, do not reject H 0 .
p1 q1
.39.61

 .0011895  .0347779
n
200
 .4390 . So   P p  .4390 p  .39
d) p1  p 0  .03  .36  .03  .39 . From this  p 
not reject H 0 : p  .36 if p  p cv
We do
.4390  .39 

 P z 
  Pz  1.41  .5  .4207  .9207 . Power  1    .0793.
.0347779 

Since pcv  p0  z  p  .36  2.327.0339411  .36  .0790  .4390. if I had used a lower significance
level, I would use a larger value of z , which means a larger value of pcv , which would make my test
ratio larger, which would make  larger and the power smaller.
e) The test ratio formula is z 
p  p0
p
. p  .41 . We already know that  p  .0324037 and that
z  z.01  2.327 . Our ‘reject’ region is above 2.327. We make a diagram showing a Normal curve with a
p  p0
.41  .36

 1.47 is not in the ‘reject
mean of zero and shade the area above 2.327. Since z 
p
.0339411
region, we do not reject H 0 .
f) Since this is a right-tail test, p  value  Pz  1.47   .5  .4292  .0708 . Since this p-value is above 3%,
do not reject H 0 .
g) Since the alternate hypotheses is H1 : p  .30 , the confidence interval is p  p  z s p
 .41  2.327 .0347779
 or
p  .3291. Since p 0  .36 is on this interval, do not reject H 0 .
h) From the outline: The usually suggested formula is n 
forgets that we covered.
p  .20, e  .01,   .01, z 2  z.005  2.576 and n 
p  .04, e  .01,   .01, z 2  z.005  2.576 and
pqz 2
e2
, ………… This is the formula everyone
.20 .80 2.576 2
.012
.04 .96 2.576 2
n
.012
 10617 .2  10618 .
 2548 .1  2549 .
14
252y0313a 10/02/03
i) We have  for the following values of 1 : .36, .39, .4390, since for 1   0 , power =  , and for
1  xcv , power = 0 . Let’s try two points above the critical value, say .47 and .51.
Our table will read as below, with supporting computations following.
1

Power 1  
.36
.99
.01
.39
.9207 .0793
.4390 .5
.5
.47
.1894 .8106
.51
.0222 .9778
Graph off your y-axis with intervals of 0.1 for power between 0 and 1.0 and set up your x-axis with
divisions of .2 for 1 between .36 and .55 and graph the ‘Power’ points.
  P p  .4390 p  .47  P z 


.4390  .47 
  Pz  0.88   .5  .3106  .1894 .
.0352916 
We can write
  P p  .4390 p  .51  P z 


.4390  .51 
  Pz  2.01  .5  .4778  .0222 .
.0353483 
15
252y0313a 10/02/03
a) p0  .37
later s p 
The problem says that n  200 , x  82 and p 
pq

n
x
82

 .41 . q  1  p  .59. so, for
n 200
.41.59 
 .0012095  .0347779
200
p0 q0
.37 .63 

 .0011655  .0341394
n
200
c) Since the alternate hypothesis is H1 : p  p0 , the critical value must be above p0 , and we must use a
right-tail test.   .01 and z  z.01  2.327 . Use
pcv  p0  z  p  .37  2.327.0341394  .37  .0794  .4494. The ‘reject, region must be above .4494.
b) H 0 : p  .37, H 1 : p  .37 and, for later,  p 
Make a diagram showing a Normal curve with a mean at p 0  .37 and a shaded reject region above .4494.
Since p  .41 is not in this ‘reject’ region, do not reject H 0 .
p1 q1
.40 .60 

 .0012000  .0346410
n
200
 .4494 . So   P p  .4494 p  .40
d) p1  p 0  .03  .37  .03  .40. From this  p 
not reject H 0 : p  .37 if p  p cv
We do
.4494  .40 

 P z 
  Pz  1.43   .5  .4236  .9236 . Power  1    .0764.
.0346410 

Since pcv  p0  z  p  .37  2.327.0341394  .37  .0794  .4494, if I had used a lower significance
level, I would use a larger value of z , which means a larger value of pcv , which would make my test
ratio larger, which would make  larger and the power smaller.
e) The test ratio formula is z 
p  p0
p
. p  .41 . We already know that  p  .0341394 and that
z  z.01  2.327 . Our ‘reject’ region is above 2.327. We make a diagram showing a Normal curve with a
p  p0
.41  .37

 1.17 is not in the ‘reject
mean of zero and shade the area above 2.327. Since z 
p
.0341394
region, we do not reject H 0 .
f) Since this is a right-tail test, p  value  Pz  1.17   .5  .3790  .1210 . Since this p-value is above 3%,
do not reject H 0 .
g) Since the alternate hypotheses is H1 : p  .30 , the confidence interval is p  p  z s p
 .41  2.327 .0347779
 or
p  .3291. Since p 0  .37 is on this interval, do not reject H 0 .
h) From the outline: The usually suggested formula is n 
forgets that we covered.
p  .20, e  .01,   .01, z 2  z.005  2.576 and n 
p  .04, e  .01,   .01, z 2  z.005  2.576 and
pqz 2
e2
, ………… This is the formula everyone
.20 .80 2.576 2
.012
.04 .96 2.576 2
n
.012
 10617 .2  10618 .
 2548 .1  2549 .
16
252y0313a 10/02/03
i) We have  for the following values of 1 : .37, .40, .4494, since for 1   0 , power =  , and for
1  xcv , power = 0 . Let’s try two points above the critical value, say .48 and .52.
Our table will read as below, with supporting computations following.
1

Power 1  
.37
.99
.01
.40
.9236 .0764
.4494 .5
.5
.48
.1922 .8078
.52
.0228 .9772
Graph off your y-axis with intervals of 0.1 for power between 0 and 1.0 and set up your x-axis with
divisions of .2 for 1 between .37 and .55 and graph the ‘Power’ points.
  P p  .4494 p  .48  P z 


.4494  .48 
  Pz  0.87   .5  .3078  .1922 .
.0353270 
We can write
  P p  .4494 p  .52  P z 


.4494  .52 
  Pz  2.00   .5  .4772  .0228 .
.0353270 
17
252y0313a 10/02/03
a) p0  .38 The problem says that n  200 , x  82 and p 
later s p 
pq

n
x
82

 .41 . q  1  p  .59. so, for
n 200
.41.59 
 .0012095  .0347779
200
p0 q0
.38.72 

 .0011780  .0343220
n
200
c) Since the alternate hypothesis is H1 : p  p0 , the critical value must be above p0 , and we must use a
right-tail test.   .01 and z  z.01  2.327 . Use
pcv  p0  z  p  .38  2.327.0343220  .38  .0799  .4599. The ‘reject, region must be above .4599.
b) H 0 : p  .38, H 1 : p  .38 and, for later,  p 
Make a diagram showing a Normal curve with a mean at p 0  .38 and a shaded reject region above .4599
Since p  .41 is not in this ‘reject’ region, do not reject H 0 .
Since pcv  p0  z  p  .38  2.327.0343220  .38  .0799  .4599, if I had used a lower significance
level, I would use a larger value of z , which means a larger value of pcv , which would make my test
ratio larger, which would make  larger and the power smaller.
p1 q1
.41.69 

 .0012095  .0347779
n
200
 .4599 . So   P p  .4599 p  .41
d) p1  p 0  .03  .38  .03  .41. From this  p 
not reject H 0 : p  .38 if p  p cv
We do
.4599  .41 

 P z 
  Pz  1.43   .5  .4236  .9236 . Power  1    .0764.
.0347779 

p  p0
e) The test ratio formula is z 
. p  .41 . We already know that  p  .0343220 and that
p
z  z.01  2.327 . Our ‘reject’ region is above 2.327. We make a diagram showing a Normal curve with a
p  p0
.41  .38

 0.87 is not in the ‘reject
mean of zero and shade the area above 2.327. Since z 
p
.0343220
region, we do not reject H 0 .
f) Since this is a right-tail test, p  value  Pz  0.87   .5  .3078  .1922 . Since this p-value is above 3%,
do not reject H 0 .
g) Since the alternate hypotheses is H1 : p  .30 , the confidence interval is p  p  z s p
 .41  2.327 .0347779
 or
p  .3291. Since p 0  .38 is on this interval, do not reject H 0 .
h) From the outline: The usually suggested formula is n 
forgets that we covered.
p  .20, e  .01,   .01, z 2  z.005  2.576 and n 
p  .04, e  .01,   .01, z 2  z.005  2.576 and
pqz 2
e2
, ………… This is the formula everyone
.20 .80 2.576 2
.012
.04 .96 2.576 2
n
.012
 10617 .2  10618 .
 2548 .1  2549 .
18
252y0313a 10/02/03
i) We have  for the following values of 1 : .38, .41, .4598, since for 1   0 , power =  , and for
1  xcv , power = 0 . Let’s try two points above the critical value, say .49 and .53.
Our table will read as below, with supporting computations following.
1

Power 1  
.38
.99
.01
.41
.9236 .0764
.4598 .5
.5
.49
.1977 .8023
.53
.0233 .9767
Graph off your y-axis with intervals of 0.1 for power between 0 and 1.0 and set up your x-axis with
divisions of .2 for 1 between .38 and .55 and graph the ‘Power’ points.
  P p  .4598 p  .49  P z 


.4598  .49 
  Pz  0.85   .5  .3023  .1977 .
.0353483 
We can write
  P p  .4598 p  .53  P z 


.4598  .53 
  Pz  1.99   .5  .4767  .0233 .
.0352916 
19
252y0313a 10/02/03
a) p0  .39 The problem says that n  200 , x  82 and p 
later s p 
pq

n
x
82

 .41 . q  1  p  .59. so, for
n 200
.41.59 
 .0012095  .0347779
200
p0 q0
.39 .61

 .0011895  .0347779
n
200
c) Since the alternate hypothesis is H1 : p  p0 , the critical value must be above p0 , and we must use a
b) H 0 : p  .39, H1 : p  .39 and, for later,  p 
right-tail test.   .01 and z  z.01  2.327 . Use
pcv  p0  z p  .39  2.327.0347779  .39  .0809  .4709. The ‘reject, region must be above .4709.
Make a diagram showing a Normal curve with a mean at p0  .39 and a shaded reject region above .4709.
Since p  .41 is not in this ‘reject’ region, do not reject H 0 .
d) p1  p0  .03  .39  .03  .42 . From this  p 
.42 .58 
 .001218  .0348999
200
p1q1

n

We do

not reject H 0 : p  .39 if p  pcv  .4709 . So   P p  .4709 p  .42
.4709  .42 

 P z 
  Pz  1.46   .5  .4279  .9279 . Power  1    .0721.
.0348999 

Since pcv  p0  z p  .39  2.327.0347779  .39  .0809  .4709. if I had used a lower significance
level, I would use a larger value of z , which means a larger value of pcv , which would make my test
ratio larger, which would make  larger and the power smaller.
e) The test ratio formula is z 
p  p0
p
. p  .41 . We already know that  p  .0347779 and that
z  z.01  2.327 . Our ‘reject’ region is above 2.327. We make a diagram showing a Normal curve with a
mean of zero and shade the area above 2.327. Since z 
p  p0
p

.41  .39
 0.575 is not in the ‘reject
.0347779
region, we do not reject H 0 .
f) Since this is a right-tail test, p  value  Pz  0.575   .5  .2174  .2826 . Since this p-value is above
3%, do not reject H 0 .
g) Since the alternate hypotheses is H1 : p  .30 , the confidence interval is p  p  z s p
 .41  2.327 .0347779
 or
p  .3291. Since p0  .39 is on this interval, do not reject H 0 .
h) From the outline: The usually suggested formula is n 
forgets that we covered.
p  .20, e  .01,   .01, z 2  z.005  2.576 and n 
p  .04, e  .01,   .01, z 2  z.005  2.576 and
pqz 2
e2
, ………… This is the formula everyone
.20 .80 2.576 2
.012
.04 .96 2.576 2
n
.012
 10617 .2  10618 .
 2548 .1  2549 .
20
252y0313a 10/02/03
i) We have  for the following values of 1 : .39, .42, .4709. Let’s try two points above the critical value,
say .51 and .55.
i) We have  for the following values of 1 : .39, .42, .4709, since for 1   0 , power =  , and for
1  xcv , power = 0 . Let’s try two points above the critical value, say .50 and .54.
Our table will read as below, with supporting computations following.
1

Power 1  
.39
.99
.01
.42
.9279 .0721
.4709 .5
.5
.50
.2061 .7939
.54
.0250 .9750
Graph off your y-axis with intervals of 0.1 for power between 0 and 1.0 and set up your x-axis with
divisions of .2 for 1 between .39 and .55 and graph the ‘Power’ points.
  P p  .4709 p  .50  P z 


.4709  .50 
  Pz  0.82   .5  .2939  .2061 .
.0353553 
We can write
  P p  .4709 p  .54  P z 


.4709  .54 
  Pz  1.96   .5  .4750  .0250 .
.0352420 
21
252y0313a 10/02/03
Table of  p 
n  200
p1 q1
n
p1
q1
.30
.31
.32
.33
.34
.35
.36
.37
.38
.39
.40
.41
.42
.43
.44
.45
.46
.47
.48
.49
.50
.51
.52
.53
.54
.55
.70
.69
.68
.67
.66
.65
.64
.63
.62
.41
.60
.59
.58
.57
.56
.55
.54
.53
.52
.51
.50
.49
.48
.47
.46
.45
p1 q1
n
0.00105000
0.00106950
0.00108800
0.00110550
0.00112200
0.00113750
0.00115200
0.00116550
0.00117800
0.00118950
0.00120000
0.00120950
0.00121800
0.00122550
0.00123200
0.00123750
0.00124200
0.00124550
0.00124800
0.00124950
0.00125000
0.00124950
0.00124800
0.00124550
0.00124200
0.00123750
p 
p1 q1
n
0.0324037
0.0327032
0.0329848
0.0332491
0.0334963
0.0337268
0.0339411
0.0341394
0.0343220
0.0347779
0.0346410
0.0347779
0.0348999
0.0350071
0.0350999
0.0351781
0.0352420
0.0352916
0.0353270
0.0353483
0.0353553
0.0353483
0.0353270
0.0352916
0.0352420
0.0351781
22