From the image you see,   Q1) What is the value of the test statistic? Choose one from below -16.451 -0.741 -2.361 -3.288 -2.319      Q2) What are the degrees of freedom? Choose one from below -24 -66 -25 -26   -65         Q3) What is the correct decision for this test? Choose one from below -Since p-value < 0.05, do not reject H0 -Since p-value > 0.05, do not reject H0 -Since p-value > 0.05, reject H0 -Since p-value < 0.05, reject H0     Q3a) An appropriate conclusion for this test is: Choose one from below -The true mean height of all  basketball players in Australia is lower than 170cm -The true mean height of all  basketball players in Australia is higher than 170cm -The true mean height of all  basketball players in Australia could be 170cm -None of the these

MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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From the image you see,

 

Q1) What is the value of the test statistic? Choose one from below

-16.451
-0.741
-2.361
-3.288
-2.319

  

 

Q2) What are the degrees of freedom? Choose one from below

-24

-66

-25

-26
 
-65

   

 

 

Q3) What is the correct decision for this test? Choose one from below

-Since p-value < 0.05, do not reject H0
-Since p-value > 0.05, do not reject H0
-Since p-value > 0.05, reject H0
-Since p-value < 0.05, reject H0

 

 

Q3a) An appropriate conclusion for this test is: Choose one from below

-The true mean height of all  basketball players in Australia is lower than 170cm
-The true mean height of all  basketball players in Australia is higher than 170cm
-The true mean height of all  basketball players in Australia could be 170cm
-None of the these

 

 

 

 

Histogram of height
4.
166
168
170
172
174
176
178
height
Kouanbaiy
Transcribed Image Text:Histogram of height 4. 166 168 170 172 174 176 178 height Kouanbaiy
data: height
t = ????, df =
alternative hypothesis: true mean is not equal to 170
95 percent confidence interval:
170.5682 172.4830
??, p-value
= 0.003096
sample estimates:
mean of x
171.5256
> sd(height)
[1] 2.319408
Transcribed Image Text:data: height t = ????, df = alternative hypothesis: true mean is not equal to 170 95 percent confidence interval: 170.5682 172.4830 ??, p-value = 0.003096 sample estimates: mean of x 171.5256 > sd(height) [1] 2.319408
Expert Solution
Step 1

Given:

Ho: mu = 170

H1: mu is not equal to 170

 

Level of significance = 0.05

95% CI of mu = [ 170.5682 , 172.4830 ]

 

Mean of x = x bar = 171.5256

std dev=s = 2.319408 

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