u- 36 H-20 u+ 20 4+30 -- - - -- -3 Z score 15. Based on a 5% level of significance (see standard normal distribution above) circle whether the standardized test statistic z indicates that you should REJECT or FAIL TO REJECT the null hypothesis. a. z = 1.668 REJECT or FAIL TO REJECT b. z = 1.433 REJECT or FAIL TO REJECT C. z = -1.499 REJECT or TO REJ CT d. z = 3 REJECT or FAIL TO REJECT

A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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Please show me how to understand and solve. Also, I’m using a TI-84 Plus calculator, I need to know the step by step process. I don’t need this just written out on paper, but I also need to be shown the every move made into the calculator (if it’s needed with this problem). Like, ‘STAT’ then ‘TESTS’ then ‘2: T-Test’.
5%
μ-30
34
+20 +30
-3-2
-2 -1
」。
2 3
Z score
15. Based on a 5% level of significance (see standard normal distribution above) circle whether the standardized test
statistic z indicates that you should REJECT or FAIL TO REJECT the null hypothesis.
a.
z = 1.668
REJECT)or FAIL TO REJECT
b. z 1.433
REJECT or FAIL TO REJECT
C. z = -1.499
REJECT or FAIL TO REJECT
d. z = 3
REJECT or FAIL TO REJECT
Transcribed Image Text:5% μ-30 34 +20 +30 -3-2 -2 -1 」。 2 3 Z score 15. Based on a 5% level of significance (see standard normal distribution above) circle whether the standardized test statistic z indicates that you should REJECT or FAIL TO REJECT the null hypothesis. a. z = 1.668 REJECT)or FAIL TO REJECT b. z 1.433 REJECT or FAIL TO REJECT C. z = -1.499 REJECT or FAIL TO REJECT d. z = 3 REJECT or FAIL TO REJECT
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