Uuestions 15-18; The Wage Data report information on annual swages for a sample
ID: 3363671 • Letter: U
Question
Uuestions 15-18; The Wage Data report information on annual swages for a sample of 66 workers. An ANOVA table is. obtained as below to determine if there is a difference in the mean annual wages among workers with different Fcritical MS occupations. ANOVA Source of Variation SS Between Groups (Treatment) 582803 within Groups (Error) Total 2843320 Descriptive statistics Occupation Mean 28921 40074 26755 24388 21510 43002 Sample Standard Deviation 145 227 245 365 694 220 5(Service) 6 (Professional) 15 15. Using (0,05 significance level, what is your conclusion? A) There is not enough evidence to reject null hypothesis: Wa: 0, == 0, was won = 06 B) There is not enough evidence to reject null hypothesis: No: A = 4 - 5 = 4-= 6 O There is enough evidence to reject null hypothesis: He: 0, 0,2 - 05° = a = 0, D) There is enough evidence to reject null hypothesis: Ho: , -, -* % E) There is enough evidence to reject null hypothesis: M: 0, 0, , , mas ta, 16. What is the critical value of the P-test at the 0.05 level of significance (une approximation if necessary)? A) 2.37 B) 3.34 C) 4.85 D) 2.53 E) 0.001 17. From the results, derive the 95% confidence interval for the mean wase IWerence between 'Occupation 5 (Service) and 'Occupation 6 (Professional)'. Show your solution A)’(629 - 436137 B) [21411 - 21572) C) [21244 - 21739] D) (11454 - 31529] E) (21345 - 216391 18. How much of the total variation (NOT variance) in annual waves is explained by the differences between different occupations? Show your solution A) About 56% B) About 4% C) About 29% D) About 20% E) About 45%Explanation / Answer
Solution:
First of all we have to complete given ANOVA table by using following formulas:
SSW = SST – SSB
DF within = DF total – DF between
MSB = SSB / DF between
MSE = SSE / DF within
F = MSB / MSE
We have to find F critical and p-value by using F-table.
Completed ANOVA table by using above formulas is given as below:
SV
SS
df
MS
F
F crit
P-value
Between groups
582803
5
116560.6
3.093821
2.36827
0.015005
Within groups
2260517
60
37675.28
Total
2843320
65
Question 15
Correct Answer:
D) There is enough evidence to reject the null hypothesis H0: µ1 = µ2 = µ3 = µ4 = µ5 = µ6
(Because P-value = 0.015005 < = 0.05, so we reject the null hypothesis H0.)
Question 16
Correct Answer:
A) 2.37
(By using F-table)
Question 17
We have to find 95% confidence interval for mean difference between service and professional occupation.
Formula is given as below:
Confidence interval = Absolute difference -/+ t*sqrt(MSE*((1/n1)+(1/n2)))
Absolute difference = 43002 – 21510 = 21492
DF for critical value t = DF within = 60
= 0.05
So, by using t-table or excel, we get
Critical value = t = 2.000298
MSE = 37675.28 (from above ANOVA)
((1/n1)+(1/n2)) = (1/13) + (1/15) = 0.14359
MSE*((1/n1)+(1/n2)) = 37675.28*0.14359 = 5409.784
sqrt(MSE*((1/n1)+(1/n2))) = sqrt(5409.784) = 73.55124
Margin of error = t*sqrt(MSE*((1/n1)+(1/n2))) = 2.000298*73.55124 = 147.1244
Confidence interval = Absolute difference -/+ t*sqrt(MSE*((1/n1)+(1/n2)))
Lower limit = 21492 - 147.1244 = 21344.88
Upper limit = 21492 + 147.1244 = 21639.12
Confidence interval = (21345, 21639)
Correct Answer:
E) [21345, 21639]
Question 18
Here, we have to find total variation or coefficient of determination or R square.
R square = 1 – (SSW / SST) = 1 – (2260517/2843320) = 20.49727%
Correct Answer:
D) About 20%
SV
SS
df
MS
F
F crit
P-value
Between groups
582803
5
116560.6
3.093821
2.36827
0.015005
Within groups
2260517
60
37675.28
Total
2843320
65
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.