Refer questions to image - 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

MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
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Refer questions to image

- 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- 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)

-4.041 0.5267 0.63779 73.60 7.7928
t. test (pulsec_NR$Pulsel-pulsec_NR$Gender , var. equal=TRUE)
Two sample t-test
data: pulsec_NR$Pulsel by pulsec_NR$Gender
t = 0.63779, df = 47, p-value = 0. 5267
alternative hypothesis: true difference in means is not equal to
95 percent confidence interval:
-4.041071 7.792795
sample estimates:
mean in group F mean in group M
73. 60000
First pulse rates of Male and Female Non Runners
71.72414
60
70
80
90
100
Transcribed Image Text:t. test (pulsec_NR$Pulsel-pulsec_NR$Gender , var. equal=TRUE) Two sample t-test data: pulsec_NR$Pulsel by pulsec_NR$Gender t = 0.63779, df = 47, p-value = 0. 5267 alternative hypothesis: true difference in means is not equal to 95 percent confidence interval: -4.041071 7.792795 sample estimates: mean in group F mean in group M 73. 60000 First pulse rates of Male and Female Non Runners 71.72414 60 70 80 90 100
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