a)
To find p-value for this test.
a)
Answer to Problem 9.38E
The p-value = 0.0182
Explanation of Solution
Given:
Null and alternative hypothesis:
Test statistic = z = 2.36
Calculation:
Test statistic z = 2.36
Using the P-value approach:
Using excel formula, =NORMSDIST(2.36)
The p-value is p = 0.0182
b)
To check significances.
b)
Explanation of Solution
Given:
The p-value is p = 0.0182
Calculation:
The p-value is p = 0.0182
If a = 0.05 then p = 0. 0182<0.05, the result is significant.
If a = 0.01 then p = 0. 0182>0.01, the result is insignificant.
Chapter 9 Solutions
Statistics Through Applications
Additional Math Textbook Solutions
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