Question

Question 12): You have been assigned to test the hypothesis that the average number of cars waiting in line for the drive-thru window during lunch hour differs between Burger King, Wendy’s, and McDonald’s. The following data show the number of cars in line during randomly selected times during the lunch hour at all three chains. Perform a one-way ANOVA using α = 0.05 to determine if a difference exists in the average number of cars waiting in line at the drive-thru during the lunch hour between these chains. (. (Hint: F crit. 3.885294))

Burger King	Wendy's	McDonald's
7	7	6
10	8	7
11	5	7
8	3	7
9	2	9

Calculate :

SST		MST		SSB
MSB		SSW		MSW
F-critical		F Stat

Answer 1

treatment	Burger King	Wendy's	McDonald's
count, ni =	5	5	5
mean , x̅ i =	9.000	5.00	7.200
std. dev., si =	1.581	2.550	1.095
sample variances, si^2 =	2.500	6.500	1.200
total sum	45	25	36		106	(grand sum)
grand mean , x̅̅ =	Σni*x̅i/Σni =	7.07
square of deviation of sample mean from grand mean,( x̅ - x̅̅)²	3.738	4.271	0.018
					TOTAL
SS(between)= SSB = Σn( x̅ - x̅̅)² =	18.689	21.356	0.089		40.13333333
SS(within ) = SSW = Σ(n-1)s² =	10.000	26.000	4.800		40.8000

no. of treatment , k =   3
df between = k-1 =    2
N = Σn =   15
df within = N-k =   12
  
mean square between groups , MSB = SSB/k-1 =    20.0667
  
mean square within groups , MSW = SSW/N-k =    3.4000
  
F-stat = MSB/MSW =    5.9020

anova table
	SS	df	MS	F	p-value	F-critical
Between:	40.1333	2	20.067	5.902	0.0164	3.885
Within:	40.8000	12	3.400
Total:	80.9333	14
					α =	0.05

Ho: µ1=µ2=µ3
H1: not all means are equal
value of test stat=   5.90
p value is    0.0164
   Decision:   p-value<α , reject null hypothesis
      
conclusion :    there is enough evidence of significant mean difference among three treatments