Question

A Pew Research study conducted in 2017 found that approximately 75% of Americans believe that robots and computers might one day do many of the jobs currently done by people.† Suppose we have the following data collected from nurses, tax auditors, and fast-food workers in which a higher score means the person feels his or her job is more likely to be automated.

Nurse	Tax Auditor	Fast-Food Worker
4	4	6
5	5	7
5	4	6
2	3	8
2	6	5
3	4	7
4	5	5
5	3	4

(a)

Use α = 0.05 to test for differences in the belief that a person's job is likely to be automated for the three professions.

State the null and alternative hypotheses.

H₀: At least two of the population means are equal.
H_a: At least two of the population means are different.H₀: μ_Nurse = μ_{Tax auditor} = μ_{Fast-food worker}
H_a: Not all the population means are equal.     H₀: Not all the population means are equal.
H_a: μ_Nurse = μ_{Tax auditor} = μ_{Fast-food worker}H₀: μ_Nurse = μ_{Tax auditor} = μ_{Fast-food worker}
H_a: μ_Nurse ≠ μ_{Tax auditor} ≠ μ_{Fast-food worker}H₀: μ_Nurse ≠ μ_{Tax auditor} ≠ μ_{Fast-food worker}
H_a: μ_Nurse = μ_{Tax auditor} = μ_{Fast-food worker}

Find the value of the test statistic. (Round your answer to two decimal places.)

Find the p-value. (Round your answer to three decimal places.)

p-value =

State your conclusion.

Reject H₀. There is sufficient evidence to conclude that there is a difference in mean scores for the three professions.Do not reject H₀. There is not sufficient evidence to conclude that there is a difference in mean scores for the three professions.     Do not reject H₀. There is sufficient evidence to conclude that there is a difference in mean scores for the three professions.Reject H₀. There is not sufficient evidence to conclude that there is a difference in mean scores for the three professions.

(b)

Use Fisher's LSD procedure to compare the belief that a person's job will be automated for nurses and tax auditors. (Use α = 0.05.)

Find the value of LSD. (Round your answer to two decimal places.)

LSD =

Find the pairwise absolute difference between sample means for nurses and tax auditors.

x_Nurse − x_{Tax auditor}

=

What conclusion can you draw after carrying out this test?

There is a significant difference between the means for nurses and tax auditors.There is not a significant difference between the means for nurses and tax auditors.     

Answer 1

	Nurse	Tax Auditor	Fast-Food Worker
	4	4	6
	5	5	7
	5	4	6
	2	3	8
	2	6	5
	3	4	7
	4	5	5
	5	3	4
Total(yi)	30	34	48
Averages y̅i	3.75	4.25	6
Treatment Effect	3.75 - 4.6667 = -0.9167	4.25 - 4.6667 = -0.4167	6 - 4.6667 = 1.3333

	Sum of Squares	Degree of freedom	Mean Square	F0 = MST / MSE	P value
Treatment	22.3333	2	11.1666	7.5644	0.0034
Error	31	21	1.4762
Total	53.3333	23

Overall total = 112
Overall mean Y̅.. = 112 / 24 = 4.6667

Y̅ .. is overall mean
y̅i . is treatment mean
SS total = ΣΣ(Yij - & Y̅..)2 = 53.3333
SS treatment = ΣΣ(Yij - & y̅i.)2 = 22.3333
SS error = Σ(y̅i. - & Y̅..)2 = 31

MS treatment = ΣΣ(Yij - & y̅i.)2 / a - 1 = 11.1666
MS error = Σ(y̅i. - & Y̅..)2 / N - a = 1.4762

Part a)

H0: μNurse = μTax auditor = μFast-food worker

Ha: Not all the population means are equal.  

Test Statistic :-
f = MS treatment / MS error = 7.5644

P value = 0.003

Decision based on P value
Reject null hypothesis if P value < α = 0.05
Since P value = 0.0034 < 0.05, hence we reject the null hypothesis
Conclusion :- Treatment means differs  

Reject H0. There is sufficient evidence to conclude that there is a difference in mean scores for the three professions.

Part b)

Fisher's Least significant Difference ( LSD Method )
The pair of means µi. and µj. would be declared significantly different if > LSD
t(α/2 , N-a ) = 2.0796 ( Critical value from t table )

LSD = = 1.26 Since the design is balanced n1 = n1 = n3 = n

= | 3.75 - 4.25 | = 0.5

Since | 3.75 - 4.25 | < 1.2634, we conclude that the population means µ1. and µ2. does not differ.

There is not a significant difference between the means for nurses and tax auditors.