Question

1. Four different designs for a digital computer circuit are being studied to compare the amount of noise present. The following data were obtained using a completely randomized design:

         Circuit design                                            Noise observed

1	39	20	19	30	8
2	80	61	73	56	80
3	47	26	25	35	50
4	95	56	83	78	97

1. Calculate by hand (using formulas and a calculator) the sums of squares for this experiment and then construct the ANOVA table. Show your work.

2. Using a significance level of 0.01, give the rejection region of the statistical test for this experiment (write it as an interval): ________________  

3. Is there enough evidence to conclude that the mean amount of noise is not the same among the four circuit designs? _______ (use a significance level of 0.01)

4. Find the observed residual of the first observation (39) for circuit design #1 _____.

5. Use Fisher’s least significant difference method to perform pairwise comparisons among treatment means (α=0.05). Complete the calculations by hand and show your work. Write down the least significant difference _______. Which circuit designs have equal mean amounts of noise? _____________________________

6. Construct a 90% confidence interval for the difference in mean amount of noise between circuit designs #1 and #2: ___________________ Show your work.

Answer 1

	A	B	C	D
count, ni =	5	5	5	5
mean , x̅ i =	23.200	70.00	36.60	81.80
std. dev., si =	11.777	11.023	11.589	16.48
sample variances, si^2 =	138.700	121.500	134.300	271.700
total sum	116	350	183	409	1058	(grand sum)
grand mean , x̅̅ =	Σni*x̅i/Σni =	52.90
square of deviation of sample mean from grand mean,( x̅ - x̅̅)²	882.090	292.410	265.690	835.210
					TOTAL
SS(between)= SSB = Σn( x̅ - x̅̅)² =	4410.450	1462.050	1328.450	4176.050	11377
SS(within ) = SSW = Σ(n-1)s² =	554.800	486.000	537.200	1086.800	2664.8000

no. of treatment , k =   4
df between = k-1 =    3
N = Σn =   20
df within = N-k =   16

SST=Σ( x - x̅̅ )² =    14041.800
  
mean square between groups , MSB = SSB/k-1 =    3792.3333
  
mean square within groups , MSW = SSW/N-k =    166.5500
  
F-stat = MSB/MSW =    22.7699
P value =   0.0000
  

anova table
	SS	df	MS	F	p-value	F-critical
Between:	11377.00	3	3792.33	22.77	0.0000	5.29
Within:	2664.80	16	166.55
Total:	14041.80	19

...........................

critical value = 5.29

rejection region : reject Ho if F stat >5.29

...............

F stat =22.77

f stat >5.29, reject Ho

the mean amount of noise is not the same among the four circuit designs

..................

Level of significance	0.05
no. of treatments,k	4
DF error =N-k=	16
MSE	166.550
t-critical value,t(α/2,df)	2.120

Fishers LSD critical value=tα/2,df √(MSE(1/ni+1/nj))
=17.3029

			confidence interval
population mean difference	critical value	lower limit	upper limit	result
µ1-µ2	-46.800	17.3029	-64.10	-29.50	means are different
µ1-µ3	-13.400	17.3029	-30.70	3.90	means are not different
µ1-µ4	-58.600	17.3029	-75.90	-41.30	means are different
µ2-µ3	33.400	17.3029	16.10	50.70	means are different
µ2-µ4	-11.800	17.3029	-29.10	5.50	means are not different
µ3-µ4	-45.200	17.3029	-62.50	-27.90	means are different

....................

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