Question

Suppose the accompanying summary statistics for a measure of social marginality for samples of youths, young adults, adults, and seniors appeared in a research paper. The social marginality score measured actual and perceived social rejection, with higher scores indicating greater social rejection.

Age Group	Youths	Young Adults	Adults	Seniors
Sample Size	109	257	311	37
x	2.00	3.40	3.06	2.84
s	1.59	1.68	1.67	1.88

For purposes of this exercise, assume that it is reasonable to regard the four samples as representative of the U.S. population in the corresponding age groups and that the distributions of social marginality scores for these four groups are approximately normal with the same standard deviation.

Is there evidence that the mean social marginality scores are not the same for all four age groups? Test the relevant hypotheses using

α = 0.01.

Calculate the test statistic. (Round your answer to two decimal places.)

F =

What can be said about the P-value for this test?

P-value > 0.1000.050 < P-value < 0.100    0.010 < P-value < 0.0500.001 < P-value < 0.010P-value < 0.001

What can you conclude?

Reject H₀. There is convincing evidence that the mean social marginality scores are not the same for all four age groups.Fail to reject H₀. There is not convincing evidence that the mean social marginality scores are not the same for all four age groups.    Fail to reject H₀. There is convincing evidence that the mean social marginality scores are not the same for all four age groups.Reject H₀. There is not convincing evidence that the mean social marginality scores are not the same for all four age groups.

You may need to use the appropriate table in Appendix A to answer this question.

Answer 1

One Way Analysis of Variance (ANOVA)              

	Youth	young ault	adults	seniors
count, ni =	109	257	311	37
mean , x̅ i =	2.000	3.40	3.06	2.84
std. dev., si =	1.590	1.680	1.670	1.880
sample variances, si^2 =	2.528	2.822	2.789	3.534
total sum	218	873.8	951.66	105.08	2148.54	(grand sum)
grand mean , x̅̅ =	Σni*x̅i/Σni =				3.01
square of deviation of sample mean from grand mean,( x̅ - x̅̅)²	1.018	0.153	0.003	0.029
					TOTAL
SS(between)= SSB = Σn( x̅ - x̅̅)² =	111.006	39.258	0.804	1.059	152.1269
SS(within ) = SSW = Σ(n-1)s² =	273.035	722.534	864.559	127.238	1987.3666

no. of treatment , k =   4
df between = k-1 =    3
N = Σn =   714
df within = N-k =   710
  
mean square between groups , MSB = SSB/k-1 =    50.7090
  
mean square within groups , MSW = SSW/N-k =    2.7991
  
F-stat = MSB/MSW =    18.1161
P value =   0.0000

	SS	df	MS	F	p-value	F-critical
Between:	152.13	3	50.71	18.12	0.0000	3.81
Within:	1987.37	710	2.80
Total:	2139.49	713
					α =	0.01

test stat ,F = 18.12

P-value < 0.001

Reject H₀. There is convincing evidence that the mean social marginality scores are not the same for all four age groups.