Thirty rats were randomly assigned to three groups (10 rats per group) and all l

ID: 3364466 • Letter: T

Question

Thirty rats were randomly assigned to three groups (10 rats per group) and all learned the same maze. The number of trials required for one errorless performance was the dependent variable. In the goal box they are given a glucose solution as a reward. One group was allowed 10 licks of the solution; one was allowed 5 licks; and one, 2 licks. Summary data for the three groups are presented below. 10 Licks 60 380 10 Licks 80 875 10 2 Licks 100 1200 10 Ex 2x2 Perform an analysis of variance to test the hypothesis that mo-m-ma:05. Present your results in a summary table. Calculate HSD and effect size if necessary. Interpret your findings. a. b. c.

Explanation / Answer

Let xij = number of errorless performance of the jth mouse given ith ‘licks’, i = 1, 2 and 3 represent 10 ‘licks’, 5 ‘licks’ and 2 ‘licks’.

Model:

xij = M + mi + eij, where M = common effect, mi = effect of ith ‘lick’, eij = error, which is assumed to be N(0, 2)

Hypotheses:

Null Hypothesis H0: m1 = m2 = m3 Vs Alternative HA: H0 is false.

Calculations

Grand Total G = {(x)1 + (x)2 +(x)3} = (60 + 80 + 100) = 240

Correction Factor C = G2/30 = 1920

Total Sum of Squares SST = {(x2)1 + (x2)2 + (x2)3} – C =(380 + 875 + 1200) – 1920

= 2455 – 1920 = 535

Row (Treatment) Sum of Squares SSR = (1/10){(x)12 + (x)22 + (x)32} – C

= (1/10){602 + 802 + 1002} – 1920

= 2000 – 1920

= 80

Error Sum of Squares SSE = SST - SSR = 535 – 80 = 455.

ANOVA TABLE

Source of variation

Degrees of freedom

Sum of squares

Mean sum of squares

Fcal

F crit

Treatment

SSR =80

80/2 = 40

2.374

3.35

Error

SSE = 455

455/27 = 16.85

Total

SST = 535

535/29 = 18.448

NOTE

Degrees of freedom: Treatment: = Number of treatments – 1; Total: Total Number of observations – 1; Error: DF(Total) – DF(Treatment)

Mean sum of squares = Sum of squares/ Degrees of freedom

Fcal = MS (Treatment)/MS (Error)

Fcrit = upper 5% point of F2,27 [obtained from Standard Statistical Tables]

Decision:

Since Fcal < Fcrit, Null Hypothesis H0 is accepted at 5% level of significance. ANSWER 1

Interpretation

There is not enough evidence to suggest that number of licks has impact on the errorless performance of the mice. ANSWER