Question

I only need help with (e) and (f). Thanks!

1. A study was conducted to explore the effects of ethanol on sleep time. Fifteen rats were randomized to one of three treatments. Treatment 1 got only water (control). Treatment 2 got lg of ethanol per kg of body weight, and treatment 3 got 2g/kg of ethanol. The amount of REM sleep in a 24hr period was recorded, in minutes. Data are below: Treatment 1: 63, 54, 69, 50, 72 Treatment 2: 45, 60, 40, 56 Treatment 3: 31, 40, 45, 25, 23, 28 (a) Make a preliminary graph of the data. Why did you choose the graph that you did and what does it tell you? (b) Create the complete ANOVA table for this data using the formulas provided in class. Show your work. You may use R to check your answers. (c) Evaluate the ANOVA assumptions graphically. Was ANOVA appropriate here? (d) Based on the ANOVA table, make a conclusion in the context of the problem. (e) Create 95% CIs for all pairwise comparisons of means using the Tukey method. Do this by hand and show your work. You may use R to check your answers. Summarize your results using letter codes. What do you conclude? (f) What would have been the multiplier if Fisher's LSD method had been used? Bonferroni? Use a t table. You may check your answer in R. You do not need to make the complete Cis for this question, just give the values of the multipliers.

Please show work by hand

Answer 1

(e) 1 sample 3sample n1 3 5, п2 4, пз 3 sample mean for treatment 2 50.25 mean for treatment 3 32 for treatment 1 61.6 mean 32 = 6 = of squares due to error we want test Null hypothesis, Ho(i, j): i= i vs. MSE 83.1625 = mean sum Suppose Alter native hy pothesis, H1(i, j) : HiHi where, i Test statistic tij true mean for treatment i, i= 1,2, 3 MSE 2 0.05 if ti>q0,05,2,12 3.08 (use Studentized Range Distribution table) We reject Ho at level a other wise accept it t12 = 2.6239 3.08, t3 is significantly different from others 7.5807 3.08, t3 4.3845> 3.08 so Treatment 3

Use R code:

y=c(63,54,69,50,72,45,60,40,56,31,40,45,25,23,28)
group=c("T1","T1","T1","T1","T1","T2","T2","T2","T2","T3","T3","T3","T3","T3","T3")
TukeyHSD(aov(y~group), conf.level=.95)

Output:

Tukey multiple comparisons of means
    95% family-wise confidence level

Fit: aov(formula = y ~ group)

$group
        diff       lwr        upr     p adj
T2-T1 -11.35 -27.67051   4.970507 0.1939650
T3-T1 -29.60 -44.33205 -14.867952 0.0004632
T3-T2 -18.25 -33.95442 -2.545584 0.0231619

(f Multiplier for Fisher's LSD = to.025,12=2.1788 Multiplier for Bon ferroni Code and outputs gt(1 0.025, 12) 2.178813 gt(1 0.025/3, 12) 2.779473 to.025/3,12 2.7795

Treatment 1	Treatment 2	Treatment 3		Anova: Single Factor
63	45	31
54	60	40		SUMMARY
69	40	45		Groups	Count	Sum	Average	Variance
50	56	25		Treatment 1	5	308	61.6	89.3
72		23		Treatment 2	4	201	50.25	86.91667
		28		Treatment 3	6	192	32	76
				ANOVA
				Source of Variation	SS	df	MS	F	P-value	F crit
				Between Groups	2456.983	2	1228.492	14.77218	0.000581	3.885294
				Within Groups	997.95	12	83.1625
				Total	3454.933	14