Question

1. A research psychologist wishes to investigate the difference in maze test scores for a strain of laboratory mice trained under different laboratory conditions. The experiment is conducted using eighteen randomly selected mice of this strain, with six receiving no treatment at all (control group), six trained under condition 1, and six trained under condition 2. Then each of the mice is given a test score between 0 and 100, depending on its performance in a test maze. The experiment produced the following results.

Control	Condition 1	Condition 2
58	73	53
32	70	74
59	68	72
64	71	62
55	60	58
49	62	61

Is there sufficient evidence to indicate a difference among mean maze test scores for mice trained under the three different laboratory conditions. Use a=.05. Perform multiple comparisons if necessary. Please show all work. Please also show the computer software output if used.

Ho:

Ha:

level of significance = .05

value of the test statisitic

p-value

state your conclusion

Perform Tukey's procedure if necessary

Answer 1

excel data analysis tool for one factor anova,steps are: write data>menu>data>data analysis>anova :one factor>enter required labels>ok, and following o/p Is obtained,

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
Control	6	317	52.83333	128.5667
Condition 1	6	404	67.33333	27.06667
Condition 2	6	380	63.33	66.27
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	673.00	2	336.5000	4.5493	0.029	3.68
Within Groups	1109.50	15	73.967
Total	1782.50	17

Ho: µ1=µ2=µ3
H1: not all means are equal

level of significance = .05

value of the test statisitic=4.549

p-value=0.029

conclusion :    p-value<α , reject null hypothesis    
there is sufficient evidence to indicate that there is a difference among mean maze test scores for mice trained under the three different laboratory conditions at α=0.05

-------------------------

Tukey's procedure

Level of significance	0.05
no. of treatments,k	3
DF error =N-k=	15
MSE	73.967
q-statistic value(α,k,N-k)	3.6750

critical value = q*√(MSE/2*(1/ni+1/nj)) = 12.903

if absolute difference of means > critical value, means are significnantly different ,otherwise not                      
                      

population mean difference		critical value			result
µ1-µ2	14.500	12.9033			means are different
µ1-µ3	10.500	12.9033			means are not different
µ2-µ3	4.000	12.9033			means are not different