Solution Found!
In a test of performed using the computer, SPSS reports a
Chapter 7, Problem 26E(choose chapter or problem)
In a test of \(H_{0}: \mu=75\) performed using the computer, SPSS reports a two-tailed p-value of .1032. Make the appropriate conclusion for each of the following situations:
a. \(H_{\mathrm{a}}: \mu<75, z=-1.63, \alpha=.05\)
b. \(H_{\mathrm{a}}: \mu<75, z=1.63, \alpha=.10\)
c. \(H_{\mathrm{a}}: \mu>75, z=1.63, \alpha=.10\)
d. \(H_{\mathrm{a}}: \mu \neq 75, z=-1.63, \alpha=.01\)
Text Transcription:
H_0: mu = 75
H_a: mu < 75, z = -1.63, alpha = .05
H_a: mu < 75, z = 1.63, alpha = .10
H_a: mu > 75, z =1.63, alpha = .10
H_a: mu neq 75, z = -1.63, alpha = .01
Questions & Answers
QUESTION:
In a test of \(H_{0}: \mu=75\) performed using the computer, SPSS reports a two-tailed p-value of .1032. Make the appropriate conclusion for each of the following situations:
a. \(H_{\mathrm{a}}: \mu<75, z=-1.63, \alpha=.05\)
b. \(H_{\mathrm{a}}: \mu<75, z=1.63, \alpha=.10\)
c. \(H_{\mathrm{a}}: \mu>75, z=1.63, \alpha=.10\)
d. \(H_{\mathrm{a}}: \mu \neq 75, z=-1.63, \alpha=.01\)
Text Transcription:
H_0: mu = 75
H_a: mu < 75, z = -1.63, alpha = .05
H_a: mu < 75, z = 1.63, alpha = .10
H_a: mu > 75, z =1.63, alpha = .10
H_a: mu neq 75, z = -1.63, alpha = .01
ANSWER:Solution:
Step 1 of 4:
We have, :
= 75 and p-value = 0.1032
-
Let,
:
= 75 against
:
= 0.05.
let,
P(z-1.63) + P(z
1.63) = 0.1032
Since it is one-tailed test
p = P(z-1.63)
=
= 0.0516
Since, p-value is greater than . there is no sufficient evidence to indicate
< 75 at
= 0.05.