Question

AL LLD Records, some of the market research of college students is done during promotions on college campuses, while other market research of college students is done through anonymous mail, phone, internet, and record store questionnaires. In all cases, for each new CD the company solicits an "intent-to- purchase" score from the student, with being the lowest score ("no intent to purchase") and 100 being the highest score ("full intent to purchase"). The manager finds the following information for 145 intent to purchase scores for a soon-to-be-released CD: Group On campus By mail By phone By internet In a store Sample Sample Sample size mean variance 29 63.5 162.8 29 63.6 118.2 29 60.4 131.5 29 137.4 99.6 The manager's next step is to conduct a one-way, independent-samples ANOVA test to decide if there is a difference in the mean intent-to-purchase score for this CD depending on the method of collecting the scores Answer the following, carrying your intermediate computations to at least three decimal places and rounding your responses to at least one decimal place. 0 What is the value of the within groups" mean square that would be reported in the ANOVA test? What is the value of the "between groups' mean square that would be reported in the ANOVA test?

Answer 1

count, ni =	29	29	29	29	29
mean , x̅ i =	63.500	63.60	60.40	67.40	63.600
std. dev., si =	12.8	10.9	11.5	11.7	10.0
sample variances, si^2 =	162.800	118.200	131.500	137.400	99.600
total sum	1841.5	1844.4	1751.6	1954.6	1844.4		9236.5
grand mean , x̅̅ =	Σni*x̅i/Σni =	63.70
( x̅ - x̅̅ )²	0.040	0.010	10.890	13.690	0.010
							TOTAL
SS(between)= SSB = Σn( x̅ - x̅̅)² =	1.160	0.290	315.810	397.010	0.290		714.56
SS(within ) = SSW = Σ(n-1)s² =	4558.400	3309.600	3682.000	3847.200	2788.800		18186.0000

no. of treatment , k =   5  
df between = k-1 =    4  
N = Σn =   145  
df within = N-k =   140  
      
mean square between groups , MSB = SSB/k-1 =    714.56/4=   178.64
mean square within groups , MSW = SSW/N-k =    18186/140=   129.9
      
F-stat = MSB/MSW =    178.64/129.9=   1.3752
      
P value =   0.2456  

	SS	df	MS	F	p-value
Between:	714.56	4.00	178.64	1.38	0.246
Within:	18186.00	140.00	129.90
Total:	18900.56	144.00

Please let me know in case of any doubt.

Thanks in advance!

Please upvote!