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
H :p
≥
p
B
packages H 1: p <
p
B
0
A
A
10.4
: 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
H
1
: They are safe
b. U.S. farmer perspective: H 0 : Genetically modified food stuffs are safe
10.5
H :T 0
B
: 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 : μ = 100 versus the alternative
H :μ 1
rule a. n = 25. Reject 108.225 b. n = 16. Reject 110.28125 c. n = 44. Reject 106.1998
≥ 100 . Compute the critical value xc and state the decision
H
0
if x > xc = μ0 + zα σ
n = 100 +1.645(25)/
H
0
if x > xc = μ0 + zα σ
n = 100 +1.645(25)/ 16 =
H
0
if x > xc = μ0 + zα σ
n = 100 +1.645(25)/ 44 =
25 =
198
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Solutions Manual for Statistics for Business & Economics, 6 Edition
a. n = 32 Reject 107.26994
H
if x > xc = μ0 + zα σ
0
n = 100 +1.645(25)/ 32 =
10.7 A random sample of n = 25 is obtained from a population with a variance σ 2 and the sample mean is computed. Test the null hypothesis H 0 : μ = 100 versus the alternative
H :μ 1
≥ 100 with alpha = .05. Compute the critical value xc and
state the decision rule a. σ 2 = 225. Reject H 0 if x > xc = μ0 + zα σ
n = 100 +1.645(15)/ 25 =
H
0
if x > xc = μ0 + zα σ
n = 100 +1.645(30)/ 25 =
H
0
if x > xc = μ0 + zα σ
n = 100 +1.645(20)/ 25 =
H
0
if x > xc = μ0 + zα σ
n = 100 +1.645(24.4949)/ 25
104.935 b. σ 2 = 900. Reject 109.87 c. σ 2 = 400. Reject 106.58 d. σ 2 = 600. Reject = 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 : μ = 80 versus the alternative x − μ0
H :μ 1
≤ 80 . Compute the p-value
70 − 80 σ n 20 25 x − μ0 70 − 80 = b. n = 16. z = 20 16 σ n x − μ0 70 − 80 = c. n = 44. z = 20 44 σ n x − μ0 70 − 80 = d. n = 32. z = 20 32 σ n
a. n = 25. z =
=
= -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
199
10.10 A random sample of n = 25, variance = σ 2 and the sample mean is = 70. Consider the null hypothesis H 0 : μ = 80 versus the alternative H 1 : μ ≤ 80 . Compute the p-value x − μ0 a. σ 2 = 225. z = σ n x − μ0 b. σ 2 = 900. z = σ n x − μ0 c. σ 2 = 400. z = σ n x − μ0 d. σ 2 = 600. z = σ n .0207 10.11
H :μ
H :μ
≥ 50 ;
0
1
< 16 ; reject
H
H :μ 1
< 50 ; reject
H :μ 0
= 3;
H :μ 1
> 3 ; reject
if Z.10 < -1.28.
0
H
H
48.2 − 50 = -1.8, therefore, Reject 3 9
Z=
10.13 a.
H :μ
15.84 − 16 = -1.6, therefore, Reject .4 16
Z=
10.12
≥ 16 ;
0
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
=
H
if Z.10 < -1.28
0
H
at the 10% level.
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 :μ 0
≤ 100 ;
H :μ 1
> 100 , using n = 25 and alpha = .05 x − μ0
106 − 100 = 2.00. Since 2.00 is 15 25 s n greater than the critical value of 1.711, there is sufficient evidence to reject the null hypothesis.
x = 106, s = 15 . Reject if
> tn −1,α 2 ,
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Solutions Manual for Statistics for Business & Economics, 6 Edition
x − μ0 104 − 100 = 2.00. Since 2.00 is > tn −1,α 2 , 10 25 s n 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 − μ0 95 − 100 = -2.50. Since -2.50 is less than the critical value of > tn −1,α 2 , 10 25 s n -1.711, there is sufficient evidence to reject the null hypothesis. x − μ0 d. Assuming a one-tailed lower test, x = 92, s = 18 . Reject if > tn −1,α 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 :μ 0
= 100 ;
H :μ 1
< 100 , using n = 36 and alpha = .05
x − μ0
106 − 100 = 2.40. Since 2.40 is 15 36 s n greater than -1.697, there is insufficient evidence to reject the null hypothesis. x − μ0 104 − 100 b. x = 104, s = 10 . Reject if = 2.40. Since 2.40 is < tn −1,α 2 , 10 36 s n greater than -1.697, there is insufficient evidence to reject the null hypothesis. x − μ0 95 − 100 c. x = 95, s = 10 . Reject if = -3.00. Since -3.00 is less < tn −1,α 2 , 10 36 s n than the critical value of -1.697, there is sufficient evidence to reject the null hypothesis. x − μ0 92 − 100 d. x = 92, s = 18 . Reject if = -2.67. Since -2.67 is less < tn −1,α 2 , 18 36 s n 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 :μ 0
< tn −1,α 2 ,
≥ 3; H 1 : μ < 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 :μ 0
= 4; H 1 : μ ≠ 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 :μ 0
.078 − 0 = 3.38, p-value is < .010. Reject .201 76 alpha
H :μ 0
t=
H
at any common level of
0
= 0; H 1 : μ ≠ 0;
t=
10.19
201
≤ 3; H 1 : μ > 3; reject
H
0
H
0
at any common level of
if t171,.01 > 2.326
3.31 − 3 = 5.81, p-value is < .005. Reject .7 172
H
0
at any common level of
alpha. 10.20
H :μ 0
= 0; H 1 : μ < 0;
−2.91 − 0 = -3.35, p-value is < .005. Reject 11.33 170 alpha. t=
10.21
H :μ 0
= 125.32; H 1 : μ ≠ 125.32; reject
H
0
H
0
at any common level of
if |t15, .05/2 | > 2.131
131.78 − 125.32 = 1.017, p-value is > .200. Do not reject 25.4 16 level. t=
H
0
at the .05
10.22 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. 10.23
H :μ 0
= 10; H 1 : μ < 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
H :μ 0
t=
= 20; H 1 : μ ≠ 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
th
Solutions Manual for Statistics for Business & Economics, 6 Edition
10.25
H :μ 0
t=
= 78.5; H 1 : μ ≠ 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 : μ ≥ 50; H 1 : μ < 50; reject H 0 if t19, .05 < -1.729 t=
10.27 a.
41.3 − 50 = -3.189, therefore, reject 12.2 20
H :μ 0
H
0
at the 5% level
≥ 400; H 1 : μ < 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 + zα p0 (1 − p0 ) / n = .25 +1.88 (.25)(1 − .25) / 400 = .2907
b. n = 225. Reject
H
0
if pˆ > pˆ c = p0 + zα p0 (1 − p0 ) / n = .25 +1.88
(.25)(1 − .25) / 225 = .30427
c. n = 625. Reject
H
0
if pˆ > pˆ c = p0 + zα p0 (1 − p0 ) / n = .25 +1.88
(.25)(1 − .25) / 625 = .28256
d. n = 900. Reject
H
0
if pˆ > pˆ c = p0 + zα 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 − zα p0 (1 − p0 ) / n = .25 – 1.645
(.25)(1 − .25) / 400 = .2144
H
b. n = 225. Reject
0
if pˆ < pˆ c = p0 − zα p0 (1 − p0 ) / n = .25 – 1.645
(.25)(1 − .25) / 225 = .2025
H
c. n = 625. Reject
0
if pˆ < pˆ c = p0 − zα p0 (1 − p0 ) / n = .25 – 1.645
(.25)(1 − .25) / 625 = .2215
H
d. n = 900. Reject
0
if pˆ < pˆ c = p0 − zα 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 (.5)(.5) / 50 Therefore, reject H 0 at alpha levels in excess of 19.77%
z=
204 10.35
th
Solutions Manual for Statistics for Business & Economics, 6 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 z= (.75)(.25) / 202 Therefore, reject H 0 at alpha levels in excess of 3.07%
10.37 Compute the probability of Type II error and the power for the following a. μ = 5.10 . β = P( x ≤ xc | μ = μ * = P ( x ≤ 5.041| μ * = 5.10) = ⎛ 5.041 − 5.10 ⎞ P⎜ z ≤ ⎟ .1 16 ⎠ ⎝ = P(z ≤ -2.36) = .0091. Power = 1 – .0091 = .9909 b. μ = 5.03 . β = P( x ≤ xc | μ = μ * = P ( x ≤ 5.041| μ * = 5.03) =
⎛ 5.041 − 5.03 ⎞ P⎜ z ≤ ⎟ .1 16 ⎠ ⎝ = P(z ≤ .44) = .6700. Power = 1 – .6700 = .3300 c. μ = 5.15 . β = P( x ≤ xc | μ = μ * = P ( x ≤ 5.041| μ * = 5.15) = ⎛ 5.041 − 5.15 ⎞ P⎜ z ≤ ⎟ .1 16 ⎠ ⎝ = P(z ≤ -4.36) = .0000. Power = 1 – .0000 = 1.0000 b. μ = 5.07 . β = P( x ≤ xc | μ = μ * = P ( x ≤ 5.041| μ * = 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 . β = P(.46 ≤ pˆ ≤ .54 | p = p ) = P ⎢ ⎢ p* (1 − p* ) ⎢ n ⎣
⎤ ⎥ .54 − p* ⎥ p* (1 − p* ) ⎥ ⎥ n ⎦
Chapter 10: Hypothesis Testing
⎡ ⎢ .46 − .52 = P⎢ ≤z≤ ⎢ .52(1 − .52) ⎢⎣ 600 .8349
205
⎤ .54 − .52 ⎥ ⎥ = P(-2.94 ≤ z ≤ .98) = .4984 + .3365 = .52(1 − .52) ⎥ ⎥⎦ 600
⎡ ⎢ b. P = .58 . β = P(.46 ≤ pˆ ≤ .54 | p = p* ) = P ⎢ ⎢ ⎢⎣ = P(-5.96 ≤ z ≤ -1.99) = .5000 – .4767 = .0233 ⎡ ⎢ c. P = .53 . β = P(.46 ≤ pˆ ≤ .54 | p = p* ) = P ⎢ ⎢ ⎢⎣ = P(-3.44 ≤ z ≤ .49) = .4997 + .1879 = .6876 ⎡ ⎢ d. P = .48 . β = P(.46 ≤ pˆ ≤ .54 | p = p* ) = P ⎢ ⎢ ⎢⎣ = P(-.98 ≤ z ≤ 2.94) = .3365 + .4984 = .8349 ⎡ ⎢ e. P = .43 . β = P(.46 ≤ pˆ ≤ .54 | p = p* ) = P ⎢ ⎢ ⎢⎣ = P(1.48 ≤ z ≤5.44) = .5000 – .4306 = .0694
.46 − .58 ≤z≤ .58(1 − .58) 600
⎤ .54 − .58 ⎥ ⎥ .58(1 − .58) ⎥ ⎥⎦ 600
.46 − .53 ≤z≤ .53(1 − .53) 600
⎤ .54 − .53 ⎥ ⎥ .53(1 − .53) ⎥ ⎥⎦ 600
.46 − .48 ≤z≤ .48(1 − .48) 600
⎤ .54 − .48 ⎥ ⎥ .48(1 − .48) ⎥ ⎥⎦ 600
.46 − .43 ≤z≤ .43(1 − .43) 600
⎤ .54 − .43 ⎥ ⎥ .43(1 − .43) ⎥ ⎥⎦ 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 ) = 1 – P(Z > -.28) = .3897 b. The power of the test = 1 - β = 1 – P(Z > 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 ) = 1 – FZ(.36) = .3594. Power of the test = 1 - β = b. The β = P(Z < .4 64 .6406
10.40 a.
H
0
is rejected when
206
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Solutions Manual for Statistics for Business & Economics, 6 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 Z > 4.07) = .8599 b. β = P( 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 ) = 1-P(Z > 1.54) = The power of the test = 1 - β = 1 – P(Z > (.45)(.55) / 802 .9382
H
10.42
10.43
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 ) = 1-P(Z > 2.17) = b. The power of the test = 1 - β = 1 – P(Z > (.2)(.8) / 998 .9850
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 < Z< ) = 1-P(-4.55 < Z < -1.21) = .1131 b. β = P( (.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 ) = P(Z < -1.2) =0.1151 b. α = P( Z < 3 9 30.8 − 31 c. β = P( Z > ) = P(Z > -.4) =0.6554 3 36
10.45 a. α = P( Z <
10.46 a. α = 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 ) = P(Z < -1.5) = .0668 c. β = P( Z < (.2)(.8) /100 d. i) lower, ii) higher
b. α = 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 :μ 0
= 40; H 1 : μ > 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
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Solutions Manual for Statistics for Business & Economics, 6 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 :μ 0
≥ 40; H 1 : μ < 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 ) = P(Z > 3) = .0014 b. β = P( Z > 120 100 c. i) smaller ii) smaller d. i) smaller ii) larger
10.52 a. α = 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 power = 1 – P(Z > ) = 1 – P(Z > 1.14) = .8729 i) (.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 z= (.5)(.5) / 99 Therefore, reject H 0 at levels in excess of 6.55%
10.56
H : p ≤ .25; H : p > .25; reject H 0
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 μW = 1, 000 + 5(400) = 3, 000 125 σ 2W = (5)2 (625) = 15, 625, σ W = 125, σ 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 .05 level. 10.59
H
0
H
0
at the
: μ ≤ 39, H 1 : μ > 39
40 − 39 = 1.19 . Probability of X = 40 given that μ 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 : μ x − μ y = 0; H1 : μ x − μ 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=
10.61 The hypothesis test is:
H :μ 0
≥ 100; H 1 : μ < 100; reject
H
0
if Z.05 > 1.645
210
th
Solutions Manual for Statistics for Business & Economics, 6 Edition
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
: μ ≤ 40, H 1 : μ > 40; X = 49.73 > 42.86 reject One-Sample T: Salmon Weight
H
0
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 β = .50 (power=.50): tcrit = 0.0: 42.8662 + 0.0(1.70) = 42.8662 Population mean for β = .25 (power=.75): tcrit = .681: 42.8662 + .681(1.70) = 44.0239 Population mean for β = .10 (power=.90): tcrit = 1.28: 42.8662 + 1.28(1.70) = 45.0422 Population mean for β = .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
