- What is the appropriate test to answer this question? (refer to the 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: 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) 3.0615 0.005067 77.037 72.31 81.76
- What is the appropriate test to answer this question? (refer to the 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: 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) 3.0615 0.005067 77.037 72.31 81.76
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 to the 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: 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)
3.0615 | 0.005067 | 77.037 | 72.31 | 81.76 |
![pulseB_f<-subset (pulseB, Gender=="F")
> t. test (pulseB_f$Pulsel, mu=70)
One sample t-test
data: pulseB_f$Pulsel
t = 3.0615, df = 26, p-value = 0.005067
alternative hypothesis: true mean is not equal to 70
95 percent confidence interval:
72. 31226 81.76181
sample estimates:
First pule rates; Female Students
mean of x
77.03704
60
70
80
90
100
pulseB_ISPulse1
Frequency
0 2 4 6](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff8014bf4-f640-4590-bc6d-8654f71afcc2%2F2509a963-5312-48ce-a790-874d6f762572%2Ft0zi22p_processed.jpeg&w=3840&q=75)
Transcribed Image Text:pulseB_f<-subset (pulseB, Gender=="F")
> t. test (pulseB_f$Pulsel, mu=70)
One sample t-test
data: pulseB_f$Pulsel
t = 3.0615, df = 26, p-value = 0.005067
alternative hypothesis: true mean is not equal to 70
95 percent confidence interval:
72. 31226 81.76181
sample estimates:
First pule rates; Female Students
mean of x
77.03704
60
70
80
90
100
pulseB_ISPulse1
Frequency
0 2 4 6
Expert Solution
![](/static/compass_v2/shared-icons/check-mark.png)
One sample mean t test:
One sample t test is suitable method to compare the sample mean with the population mean. And t - test assumes that
- samples observations should be independent to each other
- Samples observations should be approximate to normality
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