Question

Gain (V/V) R Setting Totals Averages Sample 1 Sample 2 Sample 3 4 ап 7.8 8.1 7.9 3 5.2 6.0 4.3 = 359.3 i=1 j=1 2 4.4 6.9 3.8 1 2.0 1.7 0.8 This is actual data from one of Joe Tritschler's audio engineering experiments. Use Analysis of Variance (ANOVA) to test the null hypothesis that the treatment means are equal at the a = 0.05 level of significance. Fill in the ANOVA table. Source of Variation Sum of Squares Degrees of Freedom Mean Square fo Treatments Error Total

Answer 1

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
4	3	23.8	7.933333	0.023333
3	3	15.5	5.166667	0.723333
2	3	15.1	5.033333	2.703333
1	3	4.5	1.5	0.39
ANOVA
Source of Variation	SS	df	MS	F
Between Groups	62.54917	3	20.84972	21.71846
Within Groups	7.68	8	0.96
Total	70.22917	11

$F_{crit}=F_{005;3,8}=4.07$

F> F_crit, so we reject null hypothesis of equal treatments.

$LSD=t_{\frac{0.05}{2};9}\sqrt{\frac{2MSE}{3}}=2.262*\sqrt{\frac{2*0.96}{3}}=1.8096$

$|\bar{x}_{1}-\bar{x}_{2}|=$ 7.933-5.167 = 2.766 > LSD -> Significant

$|\bar{x}_{1}-\bar{x}_{3}|=7.933-5.033=2.9>LSD\Rightarrow Significant$

$|\bar{x}_{2}-\bar{x}_{3}|=0.134<LSD\Rightarrow not \:significant$