Hypotheses HO: list the null hypothesis here: μ ≤.25 Ha: list the alternative hypothesis here: μ> .25 Assumptions List assumptions here and show the tests for the assumptions. 1) Random: Sampling is assumed to be random 2) Normal: We have to check the large counts condition. There are 56 people who do aerobics and 108 people do don't do aerobics. Since 0.25*164 and (1-0.25)*164 are both larger than 10, then the assumption for normal distribution is guaranteed. Code: X Record. Number Age Sex Aer Smk Caf Pr Rct Starting_Sal HRs.per.week Major 11 1 20 1 1 2 1 72 125 FIN 22 2 18 1 2 2 80 150 FIN 33 3 21 1 2 2 60 100 FIN 44 4 18 1 2 2 60 160 55 18 1 1 2 2 53 150 66 18 1 1 2 2 81 140 == "1")) 5 6 > length(which(Student_data_BUS$Aer [1] 56 P 2 0.3414634 38723 40407 39988 40166 39580 39982 > length(Student_data_BUS$Aer) [1] 164 > prop.test(56,164, p.25, conf.level .95, alternative = "greater", correct = FALSE) 1-sample proportions test without continuity correction data: 56 out of 164, null probability 0.25 X-squared= 7.3171, df = 1, p-value = 0.003415 alternative hypothesis: true p is greater than 0.25 95 percent confidence interval: 0.2835709 1.0000000 sample estimates: 58 62 55 61 ACCT 58 ACCT 57 ACCT Conclusion & Interpretation Our p-value of 0.003415 is less than 0.05, so we reject the null hypothesis. There is sufficient evidence to show that the true proportion of people who do aerobics is greater than 25%.
Hypotheses HO: list the null hypothesis here: μ ≤.25 Ha: list the alternative hypothesis here: μ> .25 Assumptions List assumptions here and show the tests for the assumptions. 1) Random: Sampling is assumed to be random 2) Normal: We have to check the large counts condition. There are 56 people who do aerobics and 108 people do don't do aerobics. Since 0.25*164 and (1-0.25)*164 are both larger than 10, then the assumption for normal distribution is guaranteed. Code: X Record. Number Age Sex Aer Smk Caf Pr Rct Starting_Sal HRs.per.week Major 11 1 20 1 1 2 1 72 125 FIN 22 2 18 1 2 2 80 150 FIN 33 3 21 1 2 2 60 100 FIN 44 4 18 1 2 2 60 160 55 18 1 1 2 2 53 150 66 18 1 1 2 2 81 140 == "1")) 5 6 > length(which(Student_data_BUS$Aer [1] 56 P 2 0.3414634 38723 40407 39988 40166 39580 39982 > length(Student_data_BUS$Aer) [1] 164 > prop.test(56,164, p.25, conf.level .95, alternative = "greater", correct = FALSE) 1-sample proportions test without continuity correction data: 56 out of 164, null probability 0.25 X-squared= 7.3171, df = 1, p-value = 0.003415 alternative hypothesis: true p is greater than 0.25 95 percent confidence interval: 0.2835709 1.0000000 sample estimates: 58 62 55 61 ACCT 58 ACCT 57 ACCT Conclusion & Interpretation Our p-value of 0.003415 is less than 0.05, so we reject the null hypothesis. There is sufficient evidence to show that the true proportion of people who do aerobics is greater than 25%.
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
Can someone check to make sure this is all correct? (Hypothesis, Assumption and Conclusion/Interpretation0
![Test the hypothesis that more than 25% of college students do aerobics at the a = 0.05 level.
Hypotheses
HO: list the null hypothesis here: μ ≤.25
Ha: list the alternative hypothesis here: u >.25
Assumptions
List assumptions here and show the tests for the assumptions.
1) Random: Sampling is assumed to be random
2)
A 22
Normal: We have to check the large counts condition. There are 56 people who do aerobics and
108 people do don't do aerobics. Since 0.25*164 and (1-0.25)*164 are both larger than 10, then
the assumption for normal distribution is guaranteed.
Code:
X Record. Number Age Sex Aer Smk Caf Pr Rct Starting_Sal HRs.per.week Major
11
1 20
1 72 125
FIN
22
18
FIN
33
21
FIN
44
18
5 18 1
55
66
2
3
4
р
Xततत
0.3414634
1
1
6 18 1
SONतततत
1
2
1
2
1
1
ΝΝΝΝΝΝ
2 80 150
2 60 100
2 60 160
2 53 150
2 81 140
"1"))
> length(which(Student_data_BUS$Aer
[1] 56
> length(Student_data_BUS$Aer)
[1] 164
> prop.test(56,164, p = .25, conf.level = .95, alternative = "greater", correct = FALSE)
1-sample proportions test without continuity correction
data: 56 out of 164, null probability 0.25
X-squared = 7.3171, df = 1, p-value = 0.003415
alternative hypothesis: true p is greater than 0.25
95 percent confidence interval:
0.2835709 1.0000000
sample estimates:
-
38723
40407
39988
40166
39580
39982
58
62
55
61 ACCT
58 ACCT
57 ACCT
Conclusion & Interpretation
Our p-value of 0.003415 is less than 0.05, so we reject the null hypothesis. There is sufficient
evidence to show that the true proportion of people who do aerobics is greater than 25%.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6d5ea10b-84de-4f72-b236-1879ad8dc434%2F898a9234-04c7-4f36-b4a8-3b636461691d%2Fizbs7il_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Test the hypothesis that more than 25% of college students do aerobics at the a = 0.05 level.
Hypotheses
HO: list the null hypothesis here: μ ≤.25
Ha: list the alternative hypothesis here: u >.25
Assumptions
List assumptions here and show the tests for the assumptions.
1) Random: Sampling is assumed to be random
2)
A 22
Normal: We have to check the large counts condition. There are 56 people who do aerobics and
108 people do don't do aerobics. Since 0.25*164 and (1-0.25)*164 are both larger than 10, then
the assumption for normal distribution is guaranteed.
Code:
X Record. Number Age Sex Aer Smk Caf Pr Rct Starting_Sal HRs.per.week Major
11
1 20
1 72 125
FIN
22
18
FIN
33
21
FIN
44
18
5 18 1
55
66
2
3
4
р
Xततत
0.3414634
1
1
6 18 1
SONतततत
1
2
1
2
1
1
ΝΝΝΝΝΝ
2 80 150
2 60 100
2 60 160
2 53 150
2 81 140
"1"))
> length(which(Student_data_BUS$Aer
[1] 56
> length(Student_data_BUS$Aer)
[1] 164
> prop.test(56,164, p = .25, conf.level = .95, alternative = "greater", correct = FALSE)
1-sample proportions test without continuity correction
data: 56 out of 164, null probability 0.25
X-squared = 7.3171, df = 1, p-value = 0.003415
alternative hypothesis: true p is greater than 0.25
95 percent confidence interval:
0.2835709 1.0000000
sample estimates:
-
38723
40407
39988
40166
39580
39982
58
62
55
61 ACCT
58 ACCT
57 ACCT
Conclusion & Interpretation
Our p-value of 0.003415 is less than 0.05, so we reject the null hypothesis. There is sufficient
evidence to show that the true proportion of people who do aerobics is greater than 25%.
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