Chapter 10:
Hypothesis Testing
10.1
10.2
10.3
H : p  .2; H : p  .2;
0
H
H
1
0
: No change in interest rates is warranted
1
: Reduce interest rates to stimulate the economy

p
packages
H 1: p 
p
H :p
0
A
A
10.4
B
B
: There is no difference in the percentage of underfilled cereal
: Lower percentage after the change
a. European perspective:
H 0 : Genetically modified food stuffs are not safe
H
1
: They are safe
b. U.S. farmer perspective:
H 0 : Genetically modified food stuffs are safe
H
10.5
H :T
0
B
1
: They are not safe
 T G No difference in the total number of votes between Bush and
Gore
H 1 : T B  T G Bush with more votes
10.6 A random sample is obtained from a population with a variance of 625 and the
sample mean is computed. Test the null hypothesis H 0 :
 100 versus the
alternative
H :
1

 100 . Compute the critical value xc and state the decision
rule
a. n = 25. Reject
H
0
if x  xc  0  z 
n = 100 +1.645(25)/
108.225
b. n = 16. Reject
H
0
if x  xc  0  z 
n = 100 +1.645(25)/ 16 =
110.28125
c. n = 44. Reject
H
0
if x  xc  0  z 
n = 100 +1.645(25)/ 44 =
106.1998
25 =
198
Solutions Manual for Statistics for Business & Economics, 6 th Edition
a. n = 32 Reject
H
if x  xc  0  z 
0
n = 100 +1.645(25)/ 32 =
107.26994
10.7 A random sample of n = 25 is obtained from a population with a variance  2 and
the sample mean is computed. Test the null hypothesis H 0 :
 100 versus the
alternative
H :
1

 100 with alpha = .05. Compute the critical value xc and
state the decision rule
a.  2 = 225. Reject H 0 if x  xc  0  z 
n = 100 +1.645(15)/ 25 =
104.935
b.  2 = 900. Reject
H
0
if x  xc  0  z 
n = 100 +1.645(30)/ 25 =
109.87
c.  2 = 400. Reject
H
0
if x  xc  0  z 
n = 100 +1.645(20)/ 25 =
106.58
d.  2 = 600. Reject
H
0
if x  xc  0  z 
n = 100 +1.645(24.4949)/ 25
= 108.0588
10.8
Using the results from the above two exercises, indicate how the critical value xc
is influenced by sample size. Next indicate how the critical value is influenced by
the population variance.
The critical value xc is farther away from the hypothesized value the smaller the
sample size n. This is due to the increase in the standard error with a smaller
sample size.
The critical value xc is farther away from the hypothesized value the larger the
population variance. This is due to the increased standard error with a larger
population variance.
10.9
A random sample is obtained from a population with variance = 400 and the
sample mean is computed to be 70. Consider the null hypothesis H 0 :
 80
versus the alternative
H :
x  0
 n
x  0
b. n = 16. z 
 n
x  0
c. n = 44. z 
 n
x  0
d. n = 32. z 
 n
a. n = 25. z 
1

 80 . Compute the p-value
70  80
20 25
70  80
=
20 16
70  80
=
20 44
70  80
=
20 32
=
= -2.50. p  value  P( z p  2.50) = .0062
= -2.00. p  value  P( z p  2.00) = .0228
= -3.32. p  value  P( z p  3.32) = .0004
= -2.83. p  value  P( z p  2.83) = .0023
Chapter 10: Hypothesis Testing
10.10 A random sample of n = 25, variance =  2 and the sample mean is = 70.
Consider the null hypothesis H 0 :
 80 versus the alternative H 1 :

Compute the p-value
x  0
a.  2 = 225. z 
 n
x  0
b.  2 = 900. z 
 n
x  0
c.  2 = 400. z 
 n
x  0
d.  2 = 600. z 
 n
.0207
10.11
H :
H :
 50 ;
0
1
70  80
= -3.33. p  value  P( z p  3.33) = .0004
15 25
70  80
=
= -1.67. p  value  P( z p  1.67) = .0475
30 25
70  80
=
= -2.50. p  value  P( z p  2.50) = .0062
20 25
70  80
=
= -2.04. p  value  P( z p  2.04) =
24.4949 25
 16 ; reject
H
H :
1
 50 ; reject
48.2  50
= -1.8, therefore, Reject
3 9
Z
10.13 a.
H :
H :
0
 3;
 80 .
=
15.84  16
= -1.6, therefore, Reject
.4 16
Z
10.12
 16 ;
0

199
H :
1
 3 ; reject
if Z.10 < -1.28.
0
H
H
H
at the 10% level.
if Z.10 < -1.28
0
H
0
0
0
at the 10% level.
if Z.05 > 1.645
3.07  3
= 1.4, therefore, Do Not Reject H 0 at the 5% level.
.4 64
b. p-value = 1 – FZ(1.4) = 1 - .9192 = .0808
c. the p-value would be higher – the graph should show that the p-value now
corresponds to the area in both of the tails of the distribution whereas before it
was the area in one of the tails.
d. A one-sided alternative is more appropriate since we are not interested in
detecting possible low levels of impurity, only high levels of impurity.
Z
10.14 Test
a.
H :
0
 100 ;
H :
1
 100 , using n = 25 and alpha = .05
x  0
106  100
= 2.00. Since 2.00 is
 tn 1, 2 ,
s n
15 25
greater than the critical value of 1.711, there is sufficient evidence to reject the
null hypothesis.
x  106, s  15 . Reject if
200
Solutions Manual for Statistics for Business & Economics, 6 th Edition
x  0
104  100
= 2.00. Since 2.00 is
 tn 1, 2 ,
s n
10 25
greater than the critical value of 1.711, there is sufficient evidence to reject the
null hypothesis.
c. Assuming a one-tailed lower tailed test, x  95, s  10 . Reject if
x  0
95  100
= -2.50. Since -2.50 is less than the critical value of
 tn 1, 2 ,
s n
10 25
-1.711, there is sufficient evidence to reject the null hypothesis.
x  0
d. Assuming a one-tailed lower test, x  92, s  18 . Reject if
 tn 1, 2 ,
s n
92  100
= -2.22. Since -2.22 is less than the critical value of -1.711, there is
18 25
sufficient evidence to reject the null hypothesis.
b. x  104, s  10 . Reject if
10.15 Test
H :
0
 100 ;
H :
1
 100 , using n = 36 and alpha = .05
x  0
106  100
= 2.40. Since 2.40 is
 tn 1, 2 ,
s n
15 36
greater than -1.697, there is insufficient evidence to reject the null hypothesis.
x  0
104  100
b. x  104, s  10 . Reject if
= 2.40. Since 2.40 is
 tn 1, 2 ,
s n
10 36
greater than -1.697, there is insufficient evidence to reject the null hypothesis.
x  0
95  100
c. x  95, s  10 . Reject if
= -3.00. Since -3.00 is less
 tn 1, 2 ,
s n
10 36
than the critical value of -1.697, there is sufficient evidence to reject the null
hypothesis.
x  0
92  100
d. x  92, s  18 . Reject if
= -2.67. Since -2.67 is less
 tn 1, 2 ,
s n
18 36
than the critical value of -1.697, there is sufficient evidence to reject the null
hypothesis.
a. x  106, s  15 . Reject if
10.16
H :
0
 3; H 1 :   3;
2.4  3
= -3.33, p-value is < .005. Reject
1.8 100
alpha
t
H
0
at any common level of
Chapter 10: Hypothesis Testing
10.17
H :
0
 4; H 1 :   4; reject
H
0
if –2.576 > t1561,.005 > -2.576
4.27  4
= 8.08, p-value is < .010. Reject
1.32 1562
alpha.
t
10.18
H :
0
.078  0
= 3.38, p-value is < .010. Reject
.201 76
alpha
H :
0
t
10.20
 3; H 1 :   3; reject
H
0
0
H
H
0
at any common level of
0
at any common level of alpha.
 0; H 1 :   0;
2.91  0
= -3.35, p-value is < .005. Reject
11.33 170
alpha.
H
H :
if |t15, .05/2 | > 2.131
t
10.21
at any common level of
0
if t171,.01 > 2.326
3.31  3
= 5.81, p-value is < .005. Reject
.7 172
H :
H
 0; H 1 :   0;
t
10.19
0
 125.32; H 1 :   125.32; reject
H
0
0
at any common level of
131.78  125.32
= 1.017, p-value is > .200. Do not reject
25.4 16
level.
t
10.22
10.23
H
0
at the .05
a. No, the 95% confidence level provides for 2.5% of the area in either tail. This
does not correspond to a one-tailed hypothesis test with an alpha of 5% which
has 5% of the area in one of the tails.
b. Yes.
H :
0
 10; H 1 :   10;
8.82  10
= -1.554, p-value is between .100 and .050. Do not reject
2.4013 10
at common levels of alpha.
t
10.24
201
H :
0
t
 20; H 1 :   20; reject
H
0
if |t8, .05/2 | > 2.306
20.3556  20
= 1.741, therefore, do not reject
.6126 9
H
0
at the 5% level
H
0
202
Solutions Manual for Statistics for Business & Economics, 6 th Edition
H :
10.25
0
t
 78.5; H 1 :   78.5; reject
H
0
if |t7, .10/2 | > 1.895
74.5  78.5
= -1.815, therefore, do not reject
6.2335 8
H
0
at the 10% level
10.26 The population values must be assumed to be normally distributed.
H 0 :  50; H 1 :  50; reject H 0 if t19, .05 < -1.729

t
10.27 a.

41.3  50
= -3.189, therefore, reject
12.2 20
H :
0
H
0
at the 5% level
 400; H 1 :   400;
381.35  400
= -1.486, p-value = .0797, therefore, reject H 0 at alpha
48.60 15
levels greater than 7.97%
b. Yes, with a larger sample size, the standard error would be smaller and hence,
the calculated value of t would be larger. This would yield a smaller p-value
and hence the company’s claim could be rejected at a lower significance level
than part.
t
10.28 A random sample is obtained to test the null hypothesis of the proportion of
women who said yes to a new shoe model. H 0 : p  .25; H 1 : p  .25; . What
value of the sample proportion is required to reject the null hypothesis with alpha =
.03?
a. n = 400. Reject H 0 if pˆ  pˆ c  p0  z p0 (1  p0 ) / n = .25 +1.88
(.25)(1  .25) / 400 = .2907
b. n = 225. Reject
H
0
if pˆ  pˆ c  p0  z p0 (1  p0 ) / n = .25 +1.88
(.25)(1  .25) / 225 = .30427
c. n = 625. Reject
H
0
if pˆ  pˆ c  p0  z p0 (1  p0 ) / n = .25 +1.88
(.25)(1  .25) / 625 = .28256
d. n = 900. Reject
H
0
if pˆ  pˆ c  p0  z p0 (1  p0 ) / n = .25 +1.88
(.25)(1  .25) / 900 = .2771
10.29 A random sample is obtained to test the null hypothesis of the proportion of
women who would purchase an existing shoe model.
H 0 : p  .25; H 1 : p  .25; . What value of the sample proportion is required to
reject the null hypothesis with alpha = .05?
Chapter 10: Hypothesis Testing
H
a. n = 400. Reject
0
203
if pˆ  pˆ c  p0  z p0 (1  p0 ) / n = .25 – 1.645
(.25)(1  .25) / 400 = .2144
H
b. n = 225. Reject
0
if pˆ  pˆ c  p0  z p0 (1  p0 ) / n = .25 – 1.645
(.25)(1  .25) / 225 = .2025
H
c. n = 625. Reject
0
if pˆ  pˆ c  p0  z p0 (1  p0 ) / n = .25 – 1.645
(.25)(1  .25) / 625 = .2215
H
d. n = 900. Reject
0
if pˆ  pˆ c  p0  z p0 (1  p0 ) / n = .25 – 1.645
(.25)(1  .25) / 900 = .22626
10.30
H : p  .25; H : p  .25;
0
1
.2908  .25
= 1.79, p-value = 1 – FZ(1.79) = 1 - .9633 = .0367
(.25)(.75) / 361
Therefore, reject H 0 at alpha greater than 3.67%
z
10.31
H : p  .25; H : p  .25; reject H
0
1
0
if z.05 < -1.645
.173  .25
= -5.62, p-value = 1 – FZ(5.62) = 1 – 1.0000 = .0000
(.25)(.75) / 998
Therefore, reject H 0 at the 5% alpha level
z
10.32
H : p  .5; H : p  .5;
0
1
.45  .5
= -1.26, p-value = 2[1 – FZ(1.26)] = 2[1 – .8962] = .2076
z
(.5)(.5) /160
The probability of finding a random sample with a sample proportion this far or
further from .5 if the null hypothesis is really true is .2076
10.33
H : p  .5; H : p  .5; reject H
0
1
0
if |z.10/2 | > 1.645
.5226  .5
= .64, p-value = 2[1 – FZ(.64)] = 2[1 – .7389] = .5222
(.5)(.5) /199
Therefore, do not reject H 0 at the 10% alpha level. The p-value shows the
z
probability of finding a random sample with a sample proportion this far or farther
from .5 if the null hypothesis is really true is .5222
10.34
H : p  .5; H : p  .5;
0
1
.56  .5
= .85, p-value = 1 – FZ(.85) = 1 – .8023 = .1977
z
(.5)(.5) / 50
Therefore, reject H 0 at alpha levels in excess of 19.77%
204
10.35
Solutions Manual for Statistics for Business & Economics, 6 th Edition
H : p  .75; H : p  .75;
0
1
.686  .75
= -1.94, p-value = 1 – FZ(1.94) = 1 – .9738 = .0262
(.25)(.75) /172
Therefore, reject H 0 at alpha levels in excess of 2.62%
z
10.36
H : p  .75; H : p  .75;
0
1
.6931  .75
= -1.87, p-value = 1 – FZ(1.87) = 1 – .9693 = .0307
(.75)(.25) / 202
Therefore, reject H 0 at alpha levels in excess of 3.07%
z
10.37 Compute the probability of Type II error and the power for the following
a.   5.10 .   P( x  xc |    * = P( x  5.041|  *  5.10) =

5.041  5.10 
P z 

.1 16 

= P(z ≤ -2.36) = .0091. Power = 1 – .0091 = .9909
b.   5.03 .   P( x  xc |    * = P( x  5.041|  *  5.03) =

5.041  5.03 
P z 

.1 16 

= P(z ≤ .44) = .6700. Power = 1 – .6700 = .3300
c.   5.15 .   P( x  xc |    * = P( x  5.041|  *  5.15) =

5.041  5.15 
P z 

.1 16 

= P(z ≤ -4.36) = .0000. Power = 1 – .0000 = 1.0000
b.   5.07 .   P( x  xc |    * = P( x  5.041|  *  5.07) =

5.041  5.07 
P z 

.1 16 

= P(z ≤ -1.16) = .3770. Power = 1 – .3770 = .6230
10.38 What is the probability of Type II error if the actual proportion is


.46  p*
z
a. P  .52 .   P(.46  pˆ  .54 | p  p* ) = P 
 p* (1  p* )

n



.54  p 
p* (1  p* ) 

n

*
Chapter 10: Hypothesis Testing

 .46  .52
.54  .52
= P
z
.52(1  .52)
 .52(1  .52)

600
600
.8349
205


 = P(-2.94 ≤ z ≤ .98) = .4984 + .3365 =




b. P  .58 .   P(.46  pˆ  .54 | p  p* ) = P 


= P(-5.96 ≤ z ≤ -1.99) = .5000 – .4767 = .0233


c. P  .53 .   P(.46  pˆ  .54 | p  p* ) = P 


= P(-3.44 ≤ z ≤ .49) = .4997 + .1879 = .6876


d. P  .48 .   P(.46  pˆ  .54 | p  p* ) = P 


= P(-.98 ≤ z ≤ 2.94) = .3365 + .4984 = .8349


e. P  .43 .   P(.46  pˆ  .54 | p  p* ) = P 


= P(1.48 ≤ z ≤5.44) = .5000 – .4306 = .0694

.46  .58
.54  .58 

z
.58(1  .58)
.58(1  .58) 

600
600

.46  .53
.54  .53 

z
.53(1  .53)
.53(1  .53) 

600
600

.46  .48
.54  .48 

z
.48(1  .48)
.48(1  .48) 

600
600

.46  .43
.54  .43 

z
.43(1  .43)
.43(1  .43) 

600
600
X  50
< -1.28 or when X < 48.72. Given an X = 48.2
3 9
hours, the decision is to reject the null hypothesis.
48.72  49
b. The power of the test = 1 -  = 1 – P(Z >
) = 1 – P(Z > -.28) = .3897
3 9
10.39 a.
H
0
is rejected when
X 3
> 1.645 or when X > 3.082. Since the sample
.4 64
mean is 3.07% which is less than the critical value, the decision is do not reject the
null hypothesis.
3.082  3.1
b. The  = P(Z <
) = 1 – FZ(.36) = .3594. Power of the test = 1 -  =
.4 64
.6406
10.40 a.
H
0
is rejected when
206
Solutions Manual for Statistics for Business & Economics, 6 th Edition
X 4
< 2.275 or when 3.914 < X <
1.32 1562
4.086. Since the sample mean was 4.27, which is greater than the upper critical
value, the decision is to reject the null hypothesis.
3.914  3.95
4.086  3.95
b.  = P(
<Z<
) = P(-1.08 > Z > 4.07) = .8599
1.32 1562
1.32 1562
H
10.41 a.
0
is rejected when –2.275 <
p  .5
< -1.28 or when p < .477
.25 / 802
.477  .45
The power of the test = 1 -  = 1 – P(Z >
) = 1-P(Z > 1.54) =
(.45)(.55) / 802
.9382
H
10.42
0
is rejected when
p  .25
< -1.645 or when p < .2275. Since
(.25)(.75) / 998
the sample proportion is .173 which is less than the critical value, the decision is
to reject the null hypothesis.
.2275  .2
b. The power of the test = 1 -  = 1 – P(Z >
) = 1-P(Z > 2.17) =
(.2)(.8) / 998
.9850
10.43
a.
H
0
is rejected when
p  .5
> 1.645 or when .442 > p > .558.
.25 /199
Since the sample proportion is .5226 which is within the critical values. The
decision is that there is insufficient evidence to reject the null hypothesis.
.442  .6
.558  .6
b.  = P(
< Z<
) = 1-P(-4.55 < Z < -1.21) = .1131
(.6)(.4) /199
(.6)(.4) /199
10.44 a.
H
0
is rejected when –1.645 >
30.8  32
) = P(Z < -2.4) = 0.0082
3 36
30.8  32
b.   P(Z 
) = P(Z < -1.2) =0.1151
3 9
30.8  31
c.   P( Z 
) = P(Z > -.4) =0.6554
3 36
10.45 a.   P(Z 
10.46 a.   P( Z 
.14  .10
) = P(Z > 1.33) = .0918
(.1)(.9) /100
Chapter 10: Hypothesis Testing
207
.14  .10
) = P(Z > 2.67) = .0038. The smaller probability of a
(.1)(.9) / 400
Type I error is due to the larger sample size which lowers the standard error of the
mean.
.14  .20
c.   P( Z 
) = P(Z < -1.5) = .0668
(.2)(.8) /100
d. i) lower, ii) higher
b.   P( Z 
10.47
a. The null hypothesis is the statement that is assumed to be true unless there is
sufficient evidence to suggest that the null hypothesis can be rejected. The
alternative hypothesis is that the statement that will be accepted if there is
sufficient evidence to reject the null hypothesis
b. A simple hypothesis assumes a specific value for the population parameter
that is being tested. A composite hypothesis assumes a range of values for the
population parameter.
c. One sided alternatives can be either a one-tailed upper (> greater than) or a
one-tailed lower (< less than) statement about the population parameter. Two
sided alternatives are made up of both greater than or less than statements and
are written as ( not equal to).
d. A Type I error is falsely rejecting the null hypothesis. To make a Type I error,
the truth must be that the null hypothesis is really true and yet you conclude to
reject the null and accept the alternative. A Type II error is falsely not
rejecting the null hypothesis when in fact the null hypothesis is false. To
make a Type II error, the null hypothesis must be false (the alternative is true)
and yet you conclude to not reject the null hypothesis.
e. Significance level is the chosen level of significance that established the
probability of a making a Type I error. This is represented by alpha. The
power of the test is the ability of the hypothesis test to identify correctly a false
null hypothesis and reject it.
10.48 The p-value indicates the likelihood of getting the sample result at least as far
away from the hypothesized value as the one that was found, assuming that the
distribution is really centered on the null hypothesis. The smaller the p-value, the
stronger the evidence against the null hypothesis.
10.49 a. X  45, s  10.5409
b.
t
10.50
H :
0
 40; H 1 :   40; reject
H
0
if t(9,.05) > 1.833
45  40
= 1.50, therefore, do not reject
10.5409 10
H
0
at the 5% level
a. False. The significance level is the probability of making a Type I error –
falsely rejecting the null hypothesis when in fact the null is true.
b. True
208
Solutions Manual for Statistics for Business & Economics, 6 th Edition
c. True
d. False. The power of the test is the ability of the test to correctly reject a false
null hypothesis.
e. False. The rejection region is farther away from the hypothesized value at the
1% level than it is at the 5% level. Therefore, it is still possible to reject at the
5% level but not at the 1% level.
f. True
g. False. The p-value tells the strength of the evidence against the null
hypothesis.
10.51 a. X  333/ 9  37; sx  312 8 = 6.245
H :
0
 40; H 1 :   40; reject
H
0
if t8,.05 < -1.86
37  40
= -1.44, therefore, do not reject
6.245 9
t
H
0
at the 5% level
776  800
) = P(Z < -2) = .0228
120 100)
776  740
b.   P( Z 
) = P(Z > 3) = .0014
120 100
c. i) smaller ii) smaller
d. i) smaller ii) larger
10.52 a.   P( Z 
10.53 a.
H : p  .25; H : p  .25; reject H
0
1
0
if z.05 < -1.645
.215  .25
= -1.90, therefore, reject H 0 at the 5% level
(.25)(.75) / 545
p  .25
b. H 0 is rejected when
< -1.645 or when p < .2195
(.25)(.75) / 545
.2195  .2
i)
power = 1 – P(Z >
) = 1 – P(Z > 1.14) = .8729
(.2)(.8) / 545
.2195  .25
ii)
power = 1 – P(Z >
) = 1 – P(Z > -1.64) = .0505
(.25)(.75) / 545
.2195  .3
iii)
power = 1 – P(Z >
) = 1 – P(Z > -4.1) = .0000
(.3)(.7) / 545
z
10.54
H : p  .5; H : p  .5;
0
1
.4808  .5
= -.39, p-value = 2[1-FZ(.39)] = 2[1-.6517] = .6966
z
(.5)(.5) /104
Therefore, reject H 0 at levels in excess of 69.66%
Chapter 10: Hypothesis Testing
10.55
209
H : p  .5; H : p  .5;
0
1
.576  .5
= 1.51, p-value = 1-FZ(1.51) = 1-.9345 = .0655
(.5)(.5) / 99
Therefore, reject H 0 at levels in excess of 6.55%
z
10.56
H : p  .25; H : p  .25;
0
reject
H
0
if z.05 > 1.645
.3333  .25
= 2.356, therefore, reject
(.25)(.75) /150
z
10.57
1
H
0
at the 5% level
H : p  .2; H : p  .2;
0
1
.2746  .2
= 2.22, p-value = 1-FZ(2.22) = 1-.9868 = .0132
(.2)(.8) /142
Therefore, reject H 0 at levels in excess of 1.32%
z
10.58 Cost Model where W = Total Cost: W = 1,000 + 5X
W  1,000  5(400)  3,000
125
 2W  (5)2 (625)  15, 625,  W  125,  W 
 25
25
H 0 : W  3000; H1: W  3000;
Using the test statistic criteria: (3050 – 3000)/25 = 2.00 which yields a p-value of
.0228, therefore, reject H 0 at the .05 level.
Using the sample statistic criteria: X crit  3, 000  (25)(1.645)  3041.1 ,
X calc  3, 050 , since X calc  3, 050 > X crit  3041.1 , therefore, reject
H
0
at the
.05 level.
10.59
H
0
:   39, H 1 :   39
40  39
 1.19 . Probability of X = 40 given that  is 39 is .1170.
50
71
Therefore, the Vice President’s claim is not very strong.
t40 
10.60 Assume that the population of matched differences are normally distributed
H o :  x   y  0; H1 :  x   y  0;
1.3667  0
= 1.961.
2.414 12
Reject H 0 at the 10% level since 1.961 > 1.796 = t(11, .05)
t
210
Solutions Manual for Statistics for Business & Economics, 6 th Edition
10.61 The hypothesis test is:
H :
0
 100; H 1 :   100; reject
H
0
if Z.05 > 1.645
Note: two zero values can be removed since a loaf of bread cannot weigh zero
grams:
Using Minitab:
One-Sample T: Dbread
Test of mu = 100 vs mu < 100
Variable
N
Mean
Dbread
37
101.19
Variable
Dbread
95.0% Upper Bound
110.29
At the .05 level of significance, do not reject
10.62
H
0
StDev
32.79
T
0.22
H
SE Mean
5.39
P
0.587
0
:   40, H 1 :   40; X  49.73  42.86 reject
H
0
One-Sample T: Salmon Weight
Test of mu = 40 vs mu > 40
Variable
Salmon Weigh
Variable
Salmon Weigh
N
39
Mean
49.73
95.0% Lower Bound
46.86
StDev
10.60
T
5.73
SE Mean
1.70
P
0.000
At the .05 level of significance we have strong enough evidence to reject Ho that
the true mean weight of salmon is no different than 40 in favor of Ha that the true
mean weight is significantly greater than 40.
X crit  Ho  tcrit ( S x ) : 40 + 1.686(1.70) = 42.8662
Population mean for  = .50 (power=.50): tcrit = 0.0: 42.8662 + 0.0(1.70) =
42.8662
Population mean for  = .25 (power=.75): tcrit = .681: 42.8662 + .681(1.70) =
44.0239
Population mean for  = .10 (power=.90): tcrit = 1.28: 42.8662 + 1.28(1.70) =
45.0422
Population mean for  = .05 (power=.95): tcrit = 1.645: 42.8662 + 1.645(1.70)
= 45.6627
Chapter 10: Hypothesis Testing
Power curve
For beta = .50 .25 .10 and .05
1.0
0.9
0.8
Power
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
40
41
42
43
PopMean
44
45
46
211