Question

An experiment was done to compare three different methods to inoculate entozoic amoebae. The response variable was number amoebae yield per culture.

Method 1	Method 2	Method 3
224	187	215
257	225	192
259	247	211
273	202	225
272	220	198
246	238	194
255	239	182
205	204	222
	255	179
	236

Can you conclude, at the 5% level that the mean yield per culture differs among the inoculation methods (yes or no)?

What is the test statistic?

What is the p-value?

Find a 95% confidence interval for the individual μᵢ for Method 1.

When would you conclude at the 5% level that the mean yield for Method 1 is different than the mean yield for Method 2?

A. When the p-value for the F-test of Methods is less than .05 B. When the confidence interval for the mean yield from Method 1 does not overlap the confidence interval for the mean yield of Method 2 C. When the p-value for the F-test of Methods is less than .05, and the confidence interval for the mean yield from Method 1 does not overlap the confidence interval for the mean yield of Method 2 D. When the p-value for the F-test of Methods is less than .05, and the sample mean yield for Method 1 is different than the sample Mean Yield for Method 2 All of the above

If the 95% confidence Method 2 mean yield was (211.672, 238.928) and the 95% confidence interval for the Method 3 mean yield was (187.63, 216.365) answer the following TRUE/FALSE questions.

· TRUE or FALSE: Method 1 and Method 2 have significantly different mean yields. · TRUE or FALSE: Method 1 and Method 3 have significantly different mean yields. · TRUE or FALSE: Method 2 and Method 3 have significantly different mean yields. · TRUE or FALSE: None of the methods have significantly different mean yields.

Answer 1

Solution

Let xij = number amoebae yield per culture of the jth sample under Method i.

Part (a)

To test if the mean yield per culture differs among the inoculation methods

The test employed is: ANOVA (Analysis of Variance) – one way with unequal number of observations per treatment.

Final answers are given below. Back-up Theory and Details of Calculations follow at the end.

Sub-part (i)

yes Answer 1

Sub-part (ii)

Test statistic = 10.6761 Answer 2

Sub-part (iii)

p-value = 0.000482 Answer 3

Sub-part (iv)

95% confidence interval for the individual μᵢ for Method 1 is: [205.765, 291.985] Answer 4

[CI = X₁bar ± (t_{24, 0.025} x s), where

X₁bar = mean yield for Method 1 = 1991/8 = 248.875

t_{24, 0.025} = upper 2.5% point of t-distribution with 24 degrees of freedom = 2.0639

s = standard error = sqrt(MSE of ANOVA) sqrt(436.2906) = 20.8876]

Sub-part (v)

One would conclude at the 5% level that the mean yield for Method 1 is different than the mean yield for Method

When the p-value for the F-test of Methods is less than .05, and the confidence interval for the mean yield from Method 1 does not overlap the confidence interval for the mean yield of Method 2. Option C Answer 5

Part (b)

If the 95% confidence Method 2 mean yield was (211.672, 238.928) and the 95% confidence interval for the Method 3 mean yield was (187.63, 216.365),

Method 1 and Method 2 have significantly different mean yields. FALSE Answer 6 [given CI’s concern Method 2 and Method 3] Method 1 and Method 3 have significantly different mean yields. FALSE: Answer 7 [given CI’s concern Method 2 and Method 3] Method 2 and Method 3 have significantly different mean yields. FALSE: Answer 8 [given CI’s overlap] None of the methods have significantly different mean yields. FALSE: Answer 9 [given CI’s concern only Method 2 and Method 3 and Method 1 is not considered]

Back-up Theory and Details of Calculations for ANOVA

1. 1-WAY CLASSIFICATION EQUAL # OBSNS PER CELL

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, j = 1,2,…,n; i = 1,2,……,r

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

Hypotheses:

Null hypothesis: H₀: α₁ = α₂ = ….. = α_r = 0 Vs Alternative: H₁: at least one αi is different from other αi’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

alpha	0.05
#treat	3
n1	8
n2	10
n3	9
N	27
x1.	1991
x2.	2253
x3.	1818
G = x..	6062
C	1361031.2593
Sx1j^2	499385
Sx2j^2	511909
Sx3j^2	369524
Sxij^2	1380818
Sxi.^2/ni	1370347.0250
SST	19786.74074
SSR	9315.765741
SSE	10470.975

ANOVA	Table	alpha	0.05
Source	df	SS	MS	F	Fcrit	p-value
Shift	2	9315.76574	4657.88287	10.6761	3.402826	0.0004823
Error	24	10470.98	436.290625
Total	26	19786.7407	761.02849

Since F > Fcrit or equivalently, since p-value < alpha, the null hypothesis is rejected.

Hence, we conclude that

the mean yield per culture differs among the inoculation methods

DONE