Refer image to questions -  What are the degrees of freedom? (choose from below) 46 45 43 92     - What is the correct decision for this test?(choose from below) Since p-value > 0.05, do not reject H0 Since p-value < 0.05, reject H0 Since p-value > 0.05, reject H0 Since p-value < 0.05, do not reject H0   - An appropriate conclusion for this test is (choose from below) -The first pulse rates of students who did not run are significantly higher than the first pulse rates of students who did run -The first pulse rates of students who did not run are the same as the second pulse rates of students who did not run -The first pulse rates of students who did not run could be the same as the second pulse rates of students who did not run -The first pulse rates of students who did not run are significantly lower than the second pulse rates of students who did not run

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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Refer image to questions

-  What are the degrees of freedom? (choose from below)

46 45 43 92

   

- What is the correct decision for this test?(choose from below)

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

 

- An appropriate conclusion for this test is (choose from below)

-The first pulse rates of students who did not run are significantly higher than the first pulse rates of students who did run
-The first pulse rates of students who did not run are the same as the second pulse rates of students who did not run
-The first pulse rates of students who did not run could be the same as the second pulse rates of students who did not run
-The first pulse rates of students who did not run are significantly lower than the second pulse rates of students who did not run
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

From given Image, we get

Hypothesis : 

Null hypothesis (H0) : True difference in means is equal to 0 

Alternative hypothesis (H1) : True difference in means is not equal to 0 

Test statistic : t = 0.16638

Degrees of freedom (df) = 46

P-value = 0.8686

 

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