Question

A clinical psychologist wished to compare three methods for reducing hostility levels in university students using a certain psychological test (HLT). High scores on this test were taken to indicate great hostility, and 11 students who got high and nearly equal scores were used in the experiment. Five were selected at random from among the 11 students and treated by method A, three were taken at random from the remaining six students and treated by method B, and the other three students were treated by method C. All treatments continued throughout a semester, when the HLT test was given again. The results are shown in the table.

Method	Scores on the HLT Test
A	75	83	77	67	81
B	54	73	73
C	78	95	88

Let μ_A and μ_B, respectively, denote the mean scores at the end of the semester for the populations of extremely hostile students who were treated throughout that semester by method A and method B.

(a) Find a 95% confidence interval for μ_A.

(b) Find a 95% confidence interval for μ_B.

(c) Find a 95% confidence interval for (μ_A − μ_B).

Answer 1

ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	620.3	2	310.2	4.579	0.047	4.46
Within Groups	541.9	8	67.7
Total	1162.2	10

a)

Xbar=   76.600          
MSE=   67.73333333          
n=   5          
df error=   8          
t(α/2,df error) =    2.306
Confidence interval = Xbar ± t(α/2,dfe) *√(MSE/n)=              
(   68.11   ,   85.09   )

b)

Xbar=   66.667          
MSE=   67.73333333          
n=   3          
df error=   8          
t(α/2,df error) =    2.306
Confidence interval = Xbar ± t(α/2,dfe) *√(MSE/n)=              
(   55.71   ,   77.62   )

c)

t-critical value,t(α/2,df)=   2.3060

Confidence interval = mean difference ± t*√(MSE(1/ni+1/nj)) = 9.93 ± 2.306*√(67.73*(1/5+1/3)) = (-3.93 , 23.79)