What is the appropriate test to answer this question?(refer image)

MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
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- What is the appropriate test to answer this question?(refer image)

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

H0: π = 0.7, H1: π ≠ 0.7
H0: β = 0, H1: β ≠ 0
H0: μd = 0, H1: μd ≠ 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)

0.8686 0.0851 0.16638 0.9445 1.1147

  

pulseB_NR<-subset (pulseB, Ran =="No")
t.test (pulseB_NR$Pulsel, pulseB_NR$Pulse2, paired=T)
Paired t-test
data: pulseB_NR$Pulsel and pulseB_NR$Pulse2
t = 0.16638, df = 46, p-value = 0.8686
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0. 9444994 1.1147122
sample estimates:
mean of the differences
Difference in Pulse Rates of NonRunners
0.08510638
-10
-5
diff_Notran
Frequency
4 8 12
Transcribed Image Text:pulseB_NR<-subset (pulseB, Ran =="No") t.test (pulseB_NR$Pulsel, pulseB_NR$Pulse2, paired=T) Paired t-test data: pulseB_NR$Pulsel and pulseB_NR$Pulse2 t = 0.16638, df = 46, p-value = 0.8686 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -0. 9444994 1.1147122 sample estimates: mean of the differences Difference in Pulse Rates of NonRunners 0.08510638 -10 -5 diff_Notran Frequency 4 8 12
Expert Solution
Step 1

Our aim is to find the following from the given output : 

a) The null and alternative hypothesis

b) Assumption

c) The value of test test statistic

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