252y0313b 10/02/03
IV. Solution to second 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.
1scratchh
2. Robert N Carver presents us with a data set representing departure delays for flights between JFK
and LAX airports. The sample of 52 flights gives us a sample mean delay of 2.21 minutes. Assume a
population standard deviation with a value of (10 + the third digit of your social security number).
(Example: My Social Security Number is 265398248 and the third digit is 5, so the population
standard deviation will be   10  5  15 ). Test the assertion that the mean delay is less than 2.5
minutes. (Use a significance level of 10% in this problem.)
a) State your null and alternative hypotheses (1)
b) Find a critical value of the sample mean and specify your ‘reject’ region. Make a diagram. (1)
c) Do you reject the null hypothesis? Use your diagram to show why. (1)
d) Create a power curve for this test. (Note that negative values of the mean are not impossible and
would indicate an early departure.) (6)
e) Assume that you want a 90% 2-sided confidence interval for the mean delay, with an error of
1.0 . How large a sample would you need? (2)
f) Find a p-value for the null hypothesis.(2)
From the formula table we have:
Interval for
Confidence
Interval
Mean (
  x  z 2  x
known)
Hypotheses
Test Ratio
H0 :   0
z
H1 :    0
x  0
x
Critical Value
xcv    z  2  x
A one-page solution of each version of the problem follows.
Don’t waste paper. Only copy this page and one other page!
1
252y0313b 10/02/03
Solution:   10.00
a) H1 :   2.5 , so the null hypothesis is H 0 :   2.5 . The problem said n  52, x  2.21,   .10 , so we
can use z  z.10  1.282 or z  z.05  1.645 .
2
b) This is a one-sided test, so we are only worried about values of the sample mean below 2.5 . We use
x 


10 .00

10 2
 1.923  1.3868 . Because of the alternate hypothesis, we want a critical value
52
n
52
below 2.5, so we use xcv    z  x  2.5  1.282 1.3868   2.5  1.778  0.722 . The ‘reject’ region is
below 0.722. Make a diagram of a Normal curve centered at 2.5 with a shaded area below 0.722.
c) Since x  2.21 is above .722 and thus not in the ‘reject’ region, do not reject H 0.
d) Since the alternate hypothesis is H1 :   2.5 , all values of 1 must be below 2.5. If we use
.51.778   0.9 as a rough metric, let 1 be 2.5, 1.6, .722, -0.2, -1.1.
0.722  1 

 . So if
We do not reject H 0 if x  x cv =.722, so   Px  0.722   1   P z 
1.3868 . 


1  2.5
  P z 
1  1.6
  P z 
1  0.722
  P z 
1  0.2
  P z 
1  1.1
  P z 








0.722  2.5 
  Pz  1.282   .90
1.3868 . 
1      .10
0.722  1.6 
  Pz  0.63   .5  .2357  .7357
1.3868 . 
1    .2643
0.722  0.722 
  Pz  0   .5
1.3868 . 
1    .5
0.722   0.2 
  Pz  0.66   .5  .2454  .2546
1.3868 .

1    .7454
0.722   1.1 
  Pz  1.32   .5  .4066  .0934
1.3868 . 
1    .9066

Note that it is not necessary to do the first and third calculation. It is always true that if 1   0 , power =
 and that if 1  x cv , power = .5.
You now have the power for 5 points less than or equal to 2.5. Make a diagram. Put zero through one on
the y-axis and -1.5 to 2.5 on the x-axis.
e) The formula is n 
z 2 2
e2
. z  z.05  1.645 ,   10 , and e  1. n 
2
1.645 2 10 2
1
 270 .6 . So use
271.
f) H1 :   2.5 , so the null hypothesis is H 0 :   2.5 . The problem said n  52, x  2.21, and we know
z
x  0
x
and  x  1.3868 .
2.21  2.5 

So pvalue  Px  2.21  P z 
  Pz  0.21  .5  .0832  .4168 .
1.3868 

2
252y0313b 10/02/03
Solution:   11.00
a) H1 :   2.5 , so the null hypothesis is H 0 :   2.5 . The problem said n  52, x  2.21,   .10 , so we
can use z  z.10  1.282 or z  z.05  1.645 .
2
b) This is a one-sided test, so we are only worried about values of the sample mean below 2.5 . We use
x 


11 .00

11 2
 2.327  1.5254 . Because of the alternate hypothesis, we want a critical
52
n
52
value below 2.5, so we use xcv    z  x  2.5  1.282 1.5254   2.5  1.956  0.544 . The ‘reject’ region
is below 0.544. Make a diagram of a Normal curve centered at 2.5 with a shaded area below 0.544.
c) Since x  2.21 is above .544 and thus not in the ‘reject’ region, do not reject H 0.
d) Since the alternate hypothesis is H1 :   2.5 , all values of 1 must be below 2.5. If we use
.51.956   1.0 as a rough metric, let 1 be 2.5, 1.5, 0.544, -0.5, -1.5.
0.544  1 

 . So if
We do not reject H 0 if x  x cv =.544, so   Px  0.544   1   P z 
1.5254 . 


1  2.5
  P z 
1  1.5
  P z 
1  0.544
  P z 
1  0.5
  P z 
1  1.5
  P z 








0.544  2.5 
  Pz  1.282   .90
1.5254 
1      .10
0.544  1.5 
  Pz  0.63  .5  .2357  .7357
1.5254 . 
1    .2643
0.544  0.544 
  Pz  0  .5
1.5254 . 
1    .5
0.544   0.5 
  Pz  0.68   .5  .2517  .2483
1.5254 .

1    .7517
0.544   1.5 
  Pz  1.34   .5  .4099  .0901
1.5254 .

1    .9099

Note that it is not necessary to do the first and third calculation. It is always true that if 1   0 , power =
 and that if 1  x cv , power = .5.
You now have the power for 5 points less than or equal to 2.5. Make a diagram. Put zero through one on
the y-axis and -1.5 to 2.5 on the x-axis.
1.645 2 112  327 .4 . So use
z 2 2
e) The formula is n 
.
z
  z .05  1.645 ,   11, and e  1. n 
2
1
e2
328.
f) H1 :   2.5 , so the null hypothesis is H 0 :   2.5 . The problem said n  52, x  2.21, and we know
x  0
z
and  x  1.5254 .
x
2.21  2.5 

So pvalue  Px  2.21  P z 
  Pz  0.19   .5  .0753  .4247 .
1.5254 

3
252y0313b 10/02/03
Solution:   12.00
a) H1 :   2.5 , so the null hypothesis is H 0 :   2.5 . The problem said n  52, x  2.21,   .10 , so we
can use z  z.10  1.282 or z  z.05  1.645 .
2
b) This is a one-sided test, so we are only worried about values of the sample mean below 2.5 . We use
x 


12 .00

12 2
 2.769  1.6641 . Because of the alternate hypothesis, we want a critical value
52
n
52
below 2.5, so we use xcv    z  x  2.5  1.282 1.6641   2.5  2.133  0.367 . The ‘reject’ region is
below 0.367. Make a diagram of a Normal curve centered at 2.5 with a shaded area below 0.367.
c) Since x  2.21 is above .367 and thus not in the ‘reject’ region, do not reject H 0.
d) Since the alternate hypothesis is H1 :   2.5 , all values of 1 must be below 2.5. If we use
.51.778   1.1 as a rough metric, let 1 be 2.5, 1.4, 0.367, -0.8, -1.9.
0.367  1 

 . So if
We do not reject H 0 if x  x cv =.367, so   Px  0.367   1   P z 
1.6641 


1  2.5
  P z 
1  1.4
  P z 
1  0.367
  P z 
1  0.8
  P z 
1  1.9
  P z 








0.367  2.5 
  Pz  1.282   .90
1.6641 
1      .10
0.367  1.4 
  Pz  0.62   .5  .2324  .7324
1.6641 
1    .2676
0.367  0.367 
  Pz  0  .5
1.6641

1    .5
0.367   0.8 
 Pz  0.70   .5  .2580  .2420
1.6641

1    .7580
0.367   1.9  
  Pz  1.36   .5  .4131  .0869
1.6641

1    .9131

Note that it is not necessary to do the first and third calculation. It is always true that if 1   0 , power =
 and that if 1  x cv , power = .5.
You now have the power for 5 points less than or equal to 2.5. Make a diagram. Put zero through one on
the y-axis and -2.0 to 2.5 on the x-axis.
1.645 2 12 2  389 .7 . So use
z 2 2


10
,
e


1
.
e) The formula is n 
.
,
and
z

z

1
.
645
n


.05
2
1
e2
390.
f) H1 :   2.5 , so the null hypothesis is H 0 :   2.5 . The problem said n  52, x  2.21, and we know
x  0
z
and  x  1.6641 .
x
2.21  2.5 

So pvalue  Px  2.21  P z 
  Pz  0.17   .5  .0675  .4325 .
1.6641 

4
252y0313b 10/02/03
Solution:   13.00
a) H1 :   2.5 , so the null hypothesis is H 0 :   2.5 . The problem said n  52, x  2.21,   .10 , so we
can use z  z.10  1.282 or z  z.05  1.645 .
2
b) This is a one-sided test, so we are only worried about values of the sample mean below 2.5 . We use
x 


13 .00

13 2
 3.250  1.8028 . Because of the alternate hypothesis, we want a critical
52
n
52
value below 2.5, so we use xcv    z  x  2.5  1.282 1.8028   2.5  2.311  0.189 . The ‘reject’ region
is below 0.189. Make a diagram of a Normal curve centered at 2.5 with a shaded area below 0.189.
c) Since x  2.21 is above 0.189 and thus not in the ‘reject’ region, do not reject H 0.
d) Since the alternate hypothesis is H1 :   2.5 , all values of 1 must be below 2.5. If we use
.52.311   1.2 as a rough metric, let 1 be 2.5, 1.3, 0.189, -1.1, -2.3.
0.189  1 

 . So if
We do not reject H 0 if x  x cv =.722, so   Px  0.189   1   P z 
1.8028 . 


1  2.5
  P z 
1  1.3
  P z 
1  0.189
  P z 
1  1.1
  P z 
1  2.3
  P z 








0.189  2.5 
  Pz  1.282   .90
1.8028 . 
1      .10
0.189  1.3 
  Pz  0.61  .5  .2291  .7291
1.8028 . 
1    .2709
0.189  0.189 
  Pz  0  .5
1.8028 . 
1    .5
0.189   1.1 
  Pz  0.71  .5  .1772  .3228
1.8028 . 
1    .6772
0.189   2.3 
  Pz  1.31  .5  .4049  .0951
1.8028 .

1    .9049

Note that it is not necessary to do the first and third calculation. It is always true that if 1   0 , power =
 and that if 1  x cv , power = .5.
You now have the power for 5 points less than or equal to 2.5. Make a diagram. Put zero through one on
the y-axis and -2.5 to 2.5 on the x-axis.
z 2 2
1.645 2 132  457 .3 . So use
e) The formula is n 
.
z
  z .05  1.645 ,   10 , and e  1. n 
2
1
e2
458.
f) H1 :   2.5 , so the null hypothesis is H 0 :   2.5 . The problem said n  52, x  2.21, and we know
x  0
z
and  x  1.8028 .
x
2.21  2.5 

So pvalue  Px  2.21  P z 
  Pz  0.16   .5  .0636  .4364 .
1.8028 

5
252y0313b 10/02/03
Solution:   14.00
a) H1 :   2.5 , so the null hypothesis is H 0 :   2.5 . The problem said n  52, x  2.21,   .10 , so we
can use z  z.10  1.282 or z  z.05  1.645 .
2
b) This is a one-sided test, so we are only worried about values of the sample mean below 2.5 . We use
x 


14 .00

14 2
 3.769  1.9415 . Because of the alternate hypothesis, we want a critical
52
n
52
value below 2.5, so we use xcv    z  x  2.5  1.282 1.9415   2.5  2.489  0.011 The ‘reject’ region
is below 0.011. Make a diagram of a Normal curve centered at 2.5 with a shaded area below 0.011.
c) Since x  2.21 is above 0.011 and thus not in the ‘reject’ region, do not reject H 0.
d) Since the alternate hypothesis is H1 :   2.5 , all values of 1 must be below 2.5. If we use
.52.489   1.2 as a rough metric, let 1 be 2.5, 1.3, 0.011, -1.1, -2.3.
0.011  1 

 . So if
We do not reject H 0 if x  x cv =.722, so   Px  0.011   1   P z 
1.9415 


1  2.5
  P z 
1  1.3
  P z 
1  0.011
  P z 
1  1.1
  P z 
1  2.3
  P z 








0.011  2.5 
  Pz  1.282   .90
1.9415 
1      .10
0.011  1.3 
   Pz  0.67   .5  .2486  .7486
1.9415 
1    .2514
0.011  .011 
  Pz  0  .5
1.9415 
1    .5
0.011   1.1 
  Pz  0.57   .5  .2157  .2843
1.9415

1    .7157
0.011   2.3 
 Pz  1.19   .5  .3830  .1170
1.9415

1    .8830

Note that it is not necessary to do the first and third calculation. It is always true that if 1   0 , power =
 and that if 1  x cv , power = .5.
You now have the power for 5 points less than or equal to 2.5. Make a diagram. Put zero through one on
the y-axis and -2.5 to 2.5 on the x-axis.
e) The formula is n 
z 2 2
e2
. z  z.05  1.645 ,   14 , and e  1. n 
2
1.645 2 14 2
1
 530 .4 . So use
531.
f) H1 :   2.5 , so the null hypothesis is H 0 :   2.5 . The problem said n  52, x  2.21, and we know
z
x  0
x
and  x  1.9415 .
2.21  2.5 

So pvalue  Px  2.21  P z 
  Pz  0.15   .5  .0596  .4404 .
1.9415 

6
252y0313b 10/02/03
Solution:   15.00
a) H1 :   2.5 , so the null hypothesis is H 0 :   2.5 . The problem said n  52, x  2.21,   .10 , so we
can use z  z.10  1.282 or z  z.05  1.645 .
2
b) This is a one-sided test, so we are only worried about values of the sample mean below 2.5 . We use
x 


15 .00

15 2
 4.326  2.0801 . Because of the alternate hypothesis, we want a critical
52
n
52
value below 2.5, so we use xcv    z  x  2.5  1.282 2.0801   2.5  2.667  0.167 . The ‘reject’
region is below -0.167. Make a diagram of a Normal curve centered at 2.5 with a shaded area below
-0.167..
c) Since x  2.21 is above -0.167 and thus not in the ‘reject’ region, do not reject H 0.
d) Since the alternate hypothesis is H1 :   2.5 , all values of 1 must be below 2.5. If we use
.52.667   1.3 as a rough metric, let 1 be 2.5, 1.2, -0.167, -1.4, -2.7.
 0.167  1 

 . So if
We do not reject H 0 if x  x cv = -0.167, so   Px  0.  0.167   1   P z 
2.0801 


1  2.5
  P z 
1  1.2
  P z 
1  0.167
  P z 
1  1.4
  P z 








  P z 
 0.167  2.5 
  Pz  1.282   .90
2.0801 
1      .10
 0.167  1.2 
  Pz  0.66   .5  .2454  .7454
2.0801 
1    .7546
 0.167   0.167  
  Pz  0  .5
2.0801

1    .5
 0.167   1.4 
  Pz  0.59   .5  .2224  .2776
2.0801

1    .7224
 0.167   2.7  
  Pz  1.22   .5  .3888  .1112
2.0801

1    .8888

Note that it is not necessary to do the first and third calculation. It is always true that if 1   0 , power =
 and that if 1  x cv , power = .5.
You now have the power for 5 points less than or equal to 2.5. Make a diagram. Put zero through one on
the y-axis and -3.0 to 2.5 on the x-axis.
1  2.7
e) The formula is n 
z 2 2
e
2
. z  z.05  1.645 ,   15 , and e  1. n 
2
1.645 2 15 2
1
 608 .9 . So use
609.
f) H1 :   2.5 , so the null hypothesis is H 0 :   2.5 . The problem said n  52, x  2.21, and we know
z
x  0
x
and  x  2.0801 .
2.21  2.5 

So pvalue  Px  2.21  P z 
  Pz  0.14   .5  .0557  .4443 .
2.0801 

7
252y0313b 10/02/03
Solution:   16.00
a) H1 :   2.5 , so the null hypothesis is H 0 :   2.5 . The problem said n  52, x  2.21,   .10 , so we
can use z  z.10  1.282 or z  z.05  1.645 .
2
b) This is a one-sided test, so we are only worried about values of the sample mean below 2.5 . We use
x 


10 .00
n
52

16 2
 4.923  2.2188 . Because of the alternate hypothesis, we want a critical
52
value below 2.5, so we use x cv    z  x  2.5  1.2822.2188   2.5  2.845  0.345. The ‘reject’
region is below -0.345. Make a diagram of a Normal curve centered at 2.5 with a shaded area below
0.345 .
c) Since x  2.21 is above -0.345 and thus not in the ‘reject’ region, do not reject H 0.
d) Since the alternate hypothesis is H1 :   2.5 , all values of 1 must be below 2.5. If we use
.52.845   1.4 as a rough metric, let 1 be 2.5, 1.1, -0.345, -1.7, -3.1.
 0.345  1 

 . So if
We do not reject H 0 if x  x cv =.722, so   Px  0.345   1   P z 
2.2188 . 


1  2.5
  P z 
1  1.1
  P z 
1  0.345
  P z 
1  1.7
  P z 
1  3.1
  P z 








 0.345  2.5 
  Pz  1.282   .90
2.2188 . 
1      .10
 0.345  1.1 
  Pz  0.65   .5  .2422  .7422
2.2188 . 
1    .2578
 0.345   0.345  
  Pz  0  .5
2.2188 .

1    .5
 0.345   1.7  
  Pz  0.61  .5  .2291  .2709
2.2188 .

1    .7291
 0.345   3.1 
  Pz  1.24   .5  .3925  .1075
2.2188 .

1    .8925

Note that it is not necessary to do the first and third calculation. It is always true that if 1   0 , power =
 and that if 1  x cv , power = .5.
You now have the power for 5 points less than or equal to 2.5. Make a diagram. Put zero through one on
the y-axis and -3.5 to 2.5 on the x-axis.
e) The formula is n 
z 2 2
e2
. z  z.05  1.645 ,   16 , and e  1. n 
2
1.645 2 16 2
1
 692 .7 . So use
693.
f) H1 :   2.5 , so the null hypothesis is H 0 :   2.5 . The problem said n  52, x  2.21, and we know
z
x  0
x
and  x  2.2188 .
2.21  2.5 

So pvalue  Px  2.21  P z 
  Pz  0.13   .5  .0517  .4483 .
2.2188 

8
252y0313b 10/02/03
Solution:   17.00
a) H1 :   2.5 , so the null hypothesis is H 0 :   2.5 . The problem said n  52, x  2.21,   .10 , so we
can use z  z.10  1.282 or z  z.05  1.645 .
2
b) This is a one-sided test, so we are only worried about values of the sample mean below 2.5 . We use
x 


17 .00

17 2
 5.558  2.3575 . Because of the alternate hypothesis, we want a critical
52
n
52
value below 2.5, so we use xcv    z  x  2.5  1.282 2.3575   2.5  3.022  0.522 . The ‘reject’
region is below -0.522. Make a diagram of a Normal curve centered at 2.5 with a shaded area below
-0.522.
c) Since x  2.21 is above -0.522 and thus not in the ‘reject’ region, do not reject H 0.
d) Since the alternate hypothesis is H1 :   2.5 , all values of 1 must be below 2.5. If we use
.53.022   1.5 as a rough metric, let 1 be 2.5, 1.0, -0.522, -2.0, -3.5.
 0.522  1 

 . So if
We do not reject H 0 if x  x cv =-0.522, so   Px  0.522   1   P z 
2.3575 


1  2.5
  P z 
1  1.0
  P z 
1  0.522
  P z 
1  2.0
  P z 








  P z 
 0.522  2.5 
  Pz  1.282   .90
2.3575

1      .10
 0.522  1.0 
  Pz  0.64   .5  .2389  .7389
2.3575 
1    .2611
 0.522   0.522  
  Pz  0  .5
2.3575

1    .5
 0.522   2.0 
 Pz  0.63   .5  .2357  .2643
2.3575

1    .7357
 0.522   3.5 
  Pz  1.26   .5  .3962  .1038
2.3575

1    .8962

Note that it is not necessary to do the first and third calculation. It is always true that if 1   0 , power =
 and that if 1  x cv , power = .5.
You now have the power for 5 points less than or equal to 2.5. Make a diagram. Put zero through one on
the y-axis and -4.0 to 2.5 on the x-axis.
1  3.5
e) The formula is n 
z 2 2
e
2
. z  z.05  1.645 ,   10 , and e  1. n 
2
1.645 2 17 2
1
 782 .04 . So use
783.
f) H1 :   2.5 , so the null hypothesis is H 0 :   2.5 . The problem said n  52, x  2.21, and we know
z
x  0
x
and  x  2.3575 .
2.21  2.5 

So pvalue  Px  2.21   z 
  Pz  0.12   .5  .0478  .4522 .
2.3575 

9
252y0313b 10/02/03
Solution:   18.00
a) H1 :   2.5 , so the null hypothesis is H 0 :   2.5 . The problem said n  52, x  2.21,   .10 , so we
can use z  z.10  1.282 or z  z.05  1.645 .
2
b) This is a one-sided test, so we are only worried about values of the sample mean below 2.5 . We use
x 


18 .00

18 2
 6.231  2.4961 . Because of the alternate hypothesis, we want a critical
52
n
52
value below 2.5, so we use xcv    z  x  2.5  1.282 2.4961   2.5  3.200  0.700 . The ‘reject’
region is below -0.700. Make a diagram of a Normal curve centered at 2.5 with a shaded area below
- 0.700.
c) Since x  2.21 is above -0.700 and thus not in the ‘reject’ region, do not reject H 0.
d) Since the alternate hypothesis is H1 :   2.5 , all values of 1 must be below 2.5. If we use
.53.200   1.6 as a rough metric, let 1 be 2.5, 0.9, -0.700, -2.3, -3.9.
 0.700  1 

 . So if
We do not reject H 0 if x  x cv =-0.700, so   Px  0.700   1   P z 
2.4961 

 0.700  1 

  P z 

2.4961 


1  2.5
  P z 
1  0.9
  P z 
1  0.700
  P z 
1  2.3
  P z 








  P z 
 0.700  2.5 
  Pz  1.282   .90
2.4961 
1      .10
 0.700  0.9 
  Pz  0.6  .5  .2389  .7389
2.4961 
1    .2611
 0.700   0.700  
  Pz  0  .5
2.4961

1    .5
 0.700   2.3 
  Pz  0.64   .5  .2389  .2611
2.4961

1    .7389
 0.700   3.9 
  Pz  1.28   .5  .3997  .1003
2.4961

1    .8997

Note that it is not necessary to do the first and third calculation. It is always true that if 1   0 , power =
 and that if 1  x cv , power = .5.
You now have the power for 5 points less than or equal to 2.5. Make a diagram. Put zero through one on
the y-axis and -4.5 to 2.5 on the x-axis.
1  3.9
e) The formula is n 
z 2 2
e2
. z  z.05  1.645 ,   10 , and e  1. n 
2
1.645 2 18 2
1
 876 .8 . So use
877.
f) H1 :   2.5 , so the null hypothesis is H 0 :   2.5 . The problem said n  52, x  2.21, and we know
z
x  0
x
and  x  2.4961 .
2.21  2.5 

So pvalue  Px  2.21  P z 
  Pz  0.11  .5  .0438  .4562 .
2.4961 

10
252y0313b 10/02/03
Solution:   19.00
a) H1 :   2.5 , so the null hypothesis is H 0 :   2.5 . The problem said n  52, x  2.21,   .10 , so we
can use z  z.10  1.282 or z  z.05  1.645 .
2
b) This is a one-sided test, so we are only worried about values of the sample mean below 2.5 . We use
x 


19 .00

19 2
 4.942  2.6348 . Because of the alternate hypothesis, we want a critical
52
n
52
value below 2.5, so we use xcv    z  x  2.5  1.282 2.6348   2.5  3.378  0.878 . The ‘reject’
region is below -0.878. Make a diagram of a Normal curve centered at 2.5 with a shaded area below
-0.878.
c) Since x  2.21 is above -0.878 and thus not in the ‘reject’ region, do not reject H 0.
d) Since the alternate hypothesis is H1 :   2.5 , all values of 1 must be below 2.5. If we use
.53.378   1.7 as a rough metric, let 1 be 2.5, 0.8, -0.878, -2.6, -4.3.
 0.878  1 

 . So if
We do not reject H 0 if x  x cv =.-0.878, so   Px  0.878   1   P z 
2.6348 

 0.878  1 

  P z 

2.6348 


1  2.5
  P z 
1  0.8
  P z 
1  0.878
  P z 
1  2.6
  P z 








  P z 
 0.878  2.5 
  Pz  1.282   .90
2.6348 
1      .10
 0.878  0.8 
  Pz  0.64   .5  .2389  .7389
2.6348 
1    .2611
 0.878   0.878  
  Pz  0  .5
2.6348

1    .5
 0.878   2.6  
  Pz  0.65   .5  .2422  .2578
2.6348

1    .7422
 0.878   4.3 
  Pz  1.29   .5  .4015  .0985
2.6348

1    .9015

Note that it is not necessary to do the first and third calculation. It is always true that if 1   0 , power =
 and that if 1  x cv , power = .5.
You now have the power for 5 points less than or equal to 2.5. Make a diagram. Put zero through one on
the y-axis and -4.5 to 2.5 on the x-axis.
1  4.3
e) The formula is n 
z 2 2
e2
. z  z.05  1.645 ,   19 , and e  1. n 
2
1.645 2 19 2
1
 976 .9 . So use
977.
f) H1 :   2.5 , so the null hypothesis is H 0 :   2.5 . The problem said n  52, x  2.21, and we know
z
x  0
x
and  x  2.6348 .
2.21  2.5 

So pvalue  Px  2.21  P z 
  Pz  0.11  .5  .0438  .4562 .
2.6348 

11