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
Section: Chapter Questions
Problem 1P
Related questions
Question
- 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- μ2 = 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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff8014bf4-f640-4590-bc6d-8654f71afcc2%2Fdd69adf4-c0d7-4ca0-8e28-0725c894702e%2F0vng7rg_processed.jpeg&w=3840&q=75)
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
![](/static/compass_v2/shared-icons/check-mark.png)
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
Step by step
Solved in 4 steps
![Blurred answer](/static/compass_v2/solution-images/blurred-answer.jpg)
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