What is the appropriate test to answer this question? -The null and alternative hypotheses are (choose from below) H0: π = 0.7, H1: π ≠ 0.7 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 - 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) 0.90938 1.833 18.818 0.04833 3.415
What is the appropriate test to answer this question? -The null and alternative hypotheses are (choose from below) H0: π = 0.7, H1: π ≠ 0.7 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 - 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) 0.90938 1.833 18.818 0.04833 3.415
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
Refer image to questions
-What is the appropriate test to answer this question?
-The null and alternative hypotheses are (choose from below)
H0: π = 0.7, H1: π ≠ 0.7
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
- 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)
0.90938 | 1.833 | 18.818 | 0.04833 | 3.415 |
![Notrun. Im<-1m(pulseB_NRSPulse2-pulseB_NR$Pulsel)
summary (Notrun. Im)
Before and After Pulse rates of Students who did not Run
Residuals; Students who Did not Run
-10
-5
10
60
70
80
90
resid(Notrun. Im)
Before Pulse Rate
call:
1m(formula
pulseB_NR$Pulse2
pulseB_NR$Pulsel)
coefficients:
Estimate
(Intercept)
pulseB_NR$Pulsel 0.90938
std. Error t value Pr(>|t|)
1. 833
0. 04833 18. 818
6. 52063
3.55765
0.0734
<2e-16
Residual standard error: 3.415 on 45 degrees of freedom
Multiple R-squared: 0.8872, Adjusted R-squared:
F-statistic: 354.1 on 1 and 45 DF, p-value: < 2.2e-16
0.8847
After Pulse Rate
06 08 02 09 09](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff8014bf4-f640-4590-bc6d-8654f71afcc2%2Fd2c1c1ce-c6f6-42b9-81bd-78a2b02d6e6d%2Fct7fa2i_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Notrun. Im<-1m(pulseB_NRSPulse2-pulseB_NR$Pulsel)
summary (Notrun. Im)
Before and After Pulse rates of Students who did not Run
Residuals; Students who Did not Run
-10
-5
10
60
70
80
90
resid(Notrun. Im)
Before Pulse Rate
call:
1m(formula
pulseB_NR$Pulse2
pulseB_NR$Pulsel)
coefficients:
Estimate
(Intercept)
pulseB_NR$Pulsel 0.90938
std. Error t value Pr(>|t|)
1. 833
0. 04833 18. 818
6. 52063
3.55765
0.0734
<2e-16
Residual standard error: 3.415 on 45 degrees of freedom
Multiple R-squared: 0.8872, Adjusted R-squared:
F-statistic: 354.1 on 1 and 45 DF, p-value: < 2.2e-16
0.8847
After Pulse Rate
06 08 02 09 09
Expert Solution
![](/static/compass_v2/shared-icons/check-mark.png)
Step 1
We have to answer the above questions by using the given output as reference.
There are two dependent samples which are before and after pulse rates.
The independent variable is before pulse rate.
The dependent variable is after pulse rate.
This is simple linear regression model.
Step by step
Solved in 2 steps with 1 images
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