Question

1. Researchers (Hebl et al.,2009) were interested in whether European American (EA) individuals would perceive overweight EAs the same way that African American (AA) individuals would perceive overweight AAs. They asked a group of 45 EA female college students and a group of 21 AA female college students to rate photographs of overweight women of their own race; ratings indicated how positively they felt about the person (e.g., attractive, intelligent, successful, happy, etc.).

EA group (n = 45): mean rating = 5.80, SS = 28.87

AA group (n = 21); mean rating = 6.89, SS = 39.76

1. Using all of the steps of hypothesis testing, determine whether positivity ratings are different across the two groups using a .05 significance level.

2. Compute and interpret the 95% Confidence Interval of the mean difference.

3. Compute and interpret the effect size and determine whether it is small, medium, or large.

4. Compute the power of the test. How many subjects are needed for a power of .80 with a small effect size?

Answer 1

a)

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 ╪   0                  
                          
Level of Significance ,    α =    0.05                  
                          
Sample #1   ---->   sample 1                  
mean of sample 1,    x̅1=   5.80                  
standard deviation of sample 1,   s1 =    0.80                  
size of sample 1,    n1=   45                  
                          
Sample #2   ---->   sample 2                  
mean of sample 2,    x̅2=   6.89                  
standard deviation of sample 2,   s2 =    1.41                  
size of sample 2,    n2=   21                  
                          
difference in sample means =    x̅1-x̅2 =    5.8000   -   6.9   =   -1.09  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    1.0310                  
std error , SE =    Sp*√(1/n1+1/n2) =    0.2725                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -1.0900   -   0   ) /    0.27   =   -4.000
                          
Degree of freedom, DF=   n1+n2-2 =    64                  
  
p-value =        0.00017 (excel function: =T.DIST.2T(t stat,df) )              
Conclusion:     p-value <α , Reject null hypothesis

positivity ratings are different across the two groups using a .05 significance level.

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b)

  Degree of freedom, DF=   n1+n2-2 =    64              
t-critical value =    t α/2 =    1.9977   (excel formula =t.inv(α/2,df)          
                      
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    1.0310              
                      
std error , SE =    Sp*√(1/n1+1/n2) =    0.2725              
margin of error, E = t*SE =    1.9977   *   0.2725   =   0.5443  
                      
difference of means =    x̅1-x̅2 =    5.8000   -   6.890   =   -1.0900

confidence interval is                       
Interval Lower Limit=   (x̅1-x̅2) - E =    -1.0900   -   0.5443   =   -1.6343
Interval Upper Limit=   (x̅1-x̅2) + E =    -1.0900   +   0.5443   =   -0.5457

............

c)

effect size,      
cohen's d =    |( x̅1-x̅2 )/Sp | =    1.057

LARGE EFFECT

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THANKS

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