- What is the appropriate test to answer this question?   -The null and alternative hypotheses are (choose from below) H0: π = 0.5, H1: π ≠ 0.5 H0: β = 0, H1: β ≠ 0 H0: μ = 0, H1: μ ≠ 0 H0: μ = 70, H1: μ ≠ 70 H0: μ1- μ2 = 0, H1: μ1- μ2 ≠ 0

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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- What is the appropriate test to answer this question?  

-The null and alternative hypotheses are (choose from below)

H0: π = 0.5, H1: π ≠ 0.5
H0: β = 0, H1: β ≠ 0
H0: μ = 0, H1: μ ≠ 0
H0: μ = 70, H1: μ ≠ 70
H0: μ1- μ= 0, H1: μ1- μ2 ≠ 0

-  How can you justify the assumption for this test? (choose from below)

-Since nπ and n(1-π) are both ≥5, the Central Limit Theorem applies and we can assume that p arises from a normal distribution
-The shape of the histogram indicates that the scores may be from a normal distribution and/or the sample size is reasonably large so the sample mean will be from a normal population
-The boxplots indicate that the samples are fairly symmetric and the spreads are similar, so we can assume both sets of data come from normal distributions with the same spread
-The points are randomly scattered around a straight line, so the assumptions of linearity and constant spread are satisfied.  The shape of the histogram indicates that the residuals come from a normal distribution
- None of these

 

- What is the value of the test statistic? (choose from below)

0.0292 0.53406 0.88127 0.7407 5.333
Activity.table<-table(pulseB_f$Activity)
Activity.table
A lot Moderate
slight
4
20
7
prop. test (20,27,0. 5, correct=TRUE)
Activity Levels of Female Students
1-sample proportions test with continuity correction
data: 20 out of 27, null probability 0. 5
x-squared
alternative hypothesis: true p is not equal to 0.5
95 percent confidence interval:
0. 5340628 0.8812706
sample estimates:
5. 3333, df = 1, p-value = 0.02092
%3D
p.
0.7407407
A Lot Moderate Slight
Transcribed Image Text:Activity.table<-table(pulseB_f$Activity) Activity.table A lot Moderate slight 4 20 7 prop. test (20,27,0. 5, correct=TRUE) Activity Levels of Female Students 1-sample proportions test with continuity correction data: 20 out of 27, null probability 0. 5 x-squared alternative hypothesis: true p is not equal to 0.5 95 percent confidence interval: 0. 5340628 0.8812706 sample estimates: 5. 3333, df = 1, p-value = 0.02092 %3D p. 0.7407407 A Lot Moderate Slight
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