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Class 2:
10.2
H
H
Hypothesis Testing
0
: No change in interest rates is warranted
1
: Reduce interest rates to stimulate the economy
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
a. n = 32 Reject
107.26994
≥ 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 =
H
0
if x > xc = μ0 + zα σ
25 =
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 =
104.935
b. σ 2 = 900. Reject
109.87
c. σ 2 = 400. Reject
106.58
d. σ 2 = 600. Reject
= 108.0588
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
198
th
Solutions Manual for Statistics for Business & Economics, 6 Edition
10.9A 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
H :μ
alternative
1
≤ 80 . Compute the p-value
The probability value of p-value is the smallest significance level at which the null
hypothesis can be rejected.
x - μ0
, given that H 0 is true)
σ/ n
x - μ0
= P(Z >
| μ = μ0 )
σ/ n
p - value = P(Z >
x − μ0
70 − 80
=
20 25
σ n
x − μ0
70 − 80
b. n = 16. z =
=
σ n
20 16
x − μ0
70 − 80
c. n = 44. z =
=
σ n
20 44
x − μ0
70 − 80
d. n = 32. z =
=
σ n
20 32
a. n = 25. z =
10.15 Test
H :μ
0
= 100 ;
H :μ
1
= -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
< 100 , using n = 36 and alpha = .05
x − μ0
106 − 100
= 2.40. Since 2.40 is
s n
15 36
greater than -1.697, there is insufficient evidence to reject the null hypothesis.
x − μ0
104 − 100
= 2.40. Since 2.40 is
b. x = 104, s = 10 . Reject if
< tn −1,α 2 ,
s n
10 36
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 ,
s n
10 36
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 ,
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
< tn −1,α 2 ,
Chapter 10: Hypothesis Testing
199
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 β = P(Z <
) = 1 – FZ(.36) = .3594. Power of the test = 1 - β =
.4 64
.6406
10.40a.
H
0
is rejected when
⎛
x −μ*⎞
⎟
β = P(x < x c | μ = μ*) = P⎜⎜ z < c
σ / n ⎟⎠
⎝