Question

In a study of the effect of glucose on insulin release, 12 identical specimens of ancreatictissue were divided into tihree groups of four specimens each. Three evels (low, medium, ihgh) of glucose concentration were randomly assigned to the three groups, and each specimen concentration of glucose. The amounts of insulin release by the tissue samples are 1. within a group was treated with its assigned s follows:(Stata code forinputting the daätais provided atthe end of this document Concentration Medium 3.36 4.01 3.49 High 3.92 Low 1.59 173 364 Т.97 387 Designate the low, medium,and high levels as treatment 1, reatment2, and treatment 3, respectively. Calculate the following: a. The overall mean 7, the treatment means 1, 2, 3, and the treatment variance si, i= 1,2,3; b. praph the boxplot for the data. Based or the plot, would it be c. Assume that the variances of populations of insulin release d. Describe the ANOVA assumptions as they pertain to these data reasonable to assume that the variances are the same for the three populations of insulin release measurements? measurements have a common value σ2. Calculate an estimate for σ2 e. Suppose that the ANOVA assumptions are satisfied by these data Calculate the values of MST and MSE and test the null hypothesis that f. Use the Bonferroni, Scheffe, and Tukey pairwise multiple comparison methods, each at the 0.05 level, to compare the mean insulin releases at the three glucose concentrations. Write your conclusions on the basis of the results from each of the multiple comparisor methods

Answer 1

Solution

[Note: Please note that solutions are NOT in the order the questions are given.]

Part (d) and (e)

Suppose we have data of a 1-way classification ANOVA, with r rows, and n observations per cell.

Let xij represent the jth observation in the ith row (treatment), j = 1,2,3,4; i = 1,2,3

Then the ANOVA model is: xij = µ + µi + εij, where µ = common effect, αi = effect of ith row (treatment), and εij is the error component which is assumed to be Normally Distributed with mean 0 and variance σ².

Hypotheses:

Null hypothesis: H₀: µ₁ = µ₂ = µ₃ = 0 Vs Alternative: H₁: at least one µi is different from other µi’s.

Now, to work out the solution,

Terminology:

Row total = xi..= sum over j of xij

Grand total = G = sum over i of xi.

Correction Factor = C = G²/N, where N = total number of observations = r x n

Total Sum of Squares: SST = (sum over i,j of xij²) – C

Row Sum of Squares: SSR = {(sum over i of xi.²)/n} – C

Error Sum of Squares: SSE = SST – SSR

Mean Sum of Squares = Sum of squares/Degrees of Freedom

Degrees of Freedom:

Total: N (i.e., rn) – 1;

Rows: (r - 1);

Error: Total - Row

F_obs: MSR/MSE;

F_crit: upper α% point of F-Distribution with degrees of freedom n1 and n2, where n1 is the DF for the numerator MS and n2 is the DF for the denominator MS of F_obs

Significance: F_obs is significant if F_obs > F_crit

Calculations:

r =	3	n =	4	N =	12
	xij
i	j				xi.	sumxij^2
Treatment 1	1.59	8.93	3.64	1.97	8.93	22.6515		G	40.68
Treatment 2	3.36	13.75	3.49	2.89	13.75	47.9019		C	137.91
Treatment 3	3.92	18	3.87	5.39	18	82.6278		SST	15.276
				Total	40.68	153.181		SSR	10.297
								SSE	4.9793

ANOVA	α	0.05
Source	DF	SS	MS	F	Fcrit	p-value
Treatment	2	10.29665	5.148325	9.305416	4.2564947	0.006445
Error	9	4.97935	0.553261
Total	11	15.276

Since F > Fcrit or equivalently, p-value < α, H₀ is rejected and hence we conclude that the effects of the three treatments are significantly different. Answer 1, 2

Part (a)

Means
Treatment 1		2.2325
Treatment 1		3.4375
Treatment 1		4.5
Overall		3.39

Answer 3

Treatment variance = 5.1483 [vide MS Treatment in ANOVA] Answer 4

Part (c)

Estimate of σ² = 0.5333 [vide MS Error in ANOVA] Answer 5

DONE