Question

You want to see if three different cafes yield different costs for a lunch meal for a week. You randomly select five measurements from trials on an automated driving machine for each cafe. At the 0.05 significance level, is there a difference among cafes in mean lunch meal cost? Test this claim showing all calculation steps clearly in the Ms World file (25P) and draw an ANOVA Table for the solution (10P). You have to use the same steps which we used in the lecture. You can find the steps from lecture slides or recorded video. Also, you have to take in a consider the HOMEWORK - PROJECT-PERFORMANCE EVALUATION FORM which is above. SEND IT IN MS WORD FORMAT NOT SCREEN SHOT
Cafe 1 Cafe 2
35 22 23 26 35 21 33 32 38 19
Cafe 3
26 22 34 34 36You want to see if three different cafes yield different costs for a lunch meal for a week. You randomly select five measurements from trials on an automated driving machine for each cafe. At the 0.05 significance level, is there a difference among cafes in mean lunch meal cost? Test this claim showing all calculation steps clearly in the Ms World file (25P) and draw an ANOVA Table for the solution (10P). You have to use the same steps which we used in the lecture. You can find the steps from lecture slides or recorded video. Also, you have to take in a consider the HOMEWORK - PROJECT-PERFORMANCE EVALUATION FORM which is above. SEND IT IN MS WORD FORMAT NOT SCREEN SHOT
Cafe 1 Cafe 2
35 22 23 26 35 21 33 32 38 19
Cafe 3
26 22 34 34 36

Answer 1

	Café 1	Café 2	Café 3
count, ni =	5	5	5
mean , x̅ i =	28.200	28.60	30.40
std. dev., si =	6.380	8.204	6.066
sample variances, si^2 =	40.700	67.300	36.800
total sum	141	143	152		436	(grand sum)
grand mean , x̅̅ =	Σni*x̅i/Σni =	29.07

square of deviation of sample mean from grand mean,( x̅ - x̅̅)²	0.751	0.218	1.778
					TOTAL
SS(between)= SSB = Σn( x̅ - x̅̅)² =	3.756	1.089	8.889		13.73333
SS(within ) = SSW = Σ(n-1)s² =	162.800	269.200	147.200		579.2000

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 =    6.8667
  
mean square within groups , MSW = SSW/N-k =    48.2667
  
F-stat = MSB/MSW =    0.1423
P value =   0.8688

	SS	df	MS	F	p-value	F-critical
Between:	13.73	2	6.87	0.14	0.8688	3.89
Within:	579.20	12	48.27
Total:	592.93	14
					α =	0.05
			conclusion :	p-value>α , do not reject null hypothesis

Ho: µ1=µ2=µ3
H1: not all means are equal

Conclusion:there is no difference among cafes in mean lunch meal cost

Thanks in advance!

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