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	72	82	76	67	78
B	54	75	69
C	78	94	89

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. (Round your answers to two decimal places.)

(b) Find a 95% confidence interval for μ_B. (Round your answers to two decimal places.)

(c) Find a 95% confidence interval for (μ_A − μ_B). (Round your answers to two decimal places.)

Answer 1

a)

sample std dev ,    s = √(Σ(X- x̅ )²/(n-1) ) =   5.7446

Sample Mean,    x̅ = ΣX/n =    75.0000

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   4          
't value='   tα/2=   2.7764   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   5.7446   / √   5   =   2.569047
margin of error , E=t*SE =   2.7764   *   2.56905   =   7.132817
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    75.00   -   7.132817   =   67.867183
Interval Upper Limit = x̅ + E =    75.00   -   7.132817   =   82.132817
95%   confidence interval is (   67.87   < µ <   82.13   )

b)

sample std dev ,    s = √(Σ(X- x̅ )²/(n-1) ) =   10.8167

Sample Mean,    x̅ = ΣX/n =    66.0000

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   2          
't value='   tα/2=   4.3027   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   10.8167   / √   3   =   6.244998
margin of error , E=t*SE =   4.3027   *   6.24500   =   26.870058
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    66.00   -   26.870058   =   39.129942
Interval Upper Limit = x̅ + E =    66.00   -   26.870058   =   92.870058
95%   confidence interval is (   39.13   < µ <   92.87 )

c)

difference in sample means =    x̅1-x̅2 =    75.0000   -   66.0   =   9.00

Degree of freedom, DF=   n1+n2-2 =    6              
t-critical value =    t α/2 =    2.4469   (excel formula =t.inv(α/2,df)          
                      
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    7.8103              
                      
std error , SE =    Sp*√(1/n1+1/n2) =    5.7038              
margin of error, E = t*SE =    2.4469   *   5.7038   =   13.9568  
                      
difference of means =    x̅1-x̅2 =    75.0000   -   66.000   =   9.0000
confidence interval is                       
Interval Lower Limit=   (x̅1-x̅2) - E =    9.0000   -   13.9568   =   -4.96
Interval Upper Limit=   (x̅1-x̅2) + E =    9.0000   +   13.9568   =   22.96

Please revert in case of any doubt.

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