Question

Dr. Roh was interested in whether there was any disparity in sentencing based on the race of the defendant. He selected at random 19 drug convictions for cocaine distribution and completed the prison terms given to the 10 whites and 10 blacks sampled. The sentence lengths (in years) are shown below for the white and black offenders

White	Black
3	4
5	8
2	7
7	7
4	5
5	9
5	7
4	4
3	5
2	8

FYI,

White	Black
s2/1= 2.20	s2/2=2.84

1. Please write down the null and Dr. Roh’s research hypotheses. (5 points)
2. What is the critical t – value from t-table based on df and an alpha level of .05? (5 points)
3. What is the obtained t – value based on your calculation? (10 points)
4. State your final conclusion regarding the results from this test. (10 points)

Answer 1

solution:

Given that

n1 = 10       

n2 = 10       

x1-bar = 4       

x2-bar = 6.4       

s1 = √2.2 = 1.483239697       

s2 = √2.84 = 1.685229955       

(a) Hypotheses:        

Ho: μ1 = μ2        

Ha: μ1 ≠ μ2        

Decision Rule:        

α = 0.05       

Degrees of freedom = 10 + 10 - 2 = 18      

(b) Lower Critical t- score = -2.100922037       

Upper Critical t- score = 2.100922037       

Reject Ho if |t| > 2.100922037       

(c) Test Statistic:        

Pooled SD, s = √[{(n1 - 1) s1^2 + (n2 - 1) s2^2} / (n1 + n2 - 2)] =   √(((10 - 1) * 1.48323969741913^2 + (10 - 1) * 1.68522995463527^2)/(10 + 10 -2)) = 1.587

SE = s * √{(1 /n1) + (1 /n2)} = 1.58745078663875 * √((1/10) + (1/10)) = 0.709929574      

t = (x1-bar -x2-bar)/SE = -3.380617019       

p- value = 0.003331454       

(d) Decision (in terms of the hypotheses):        

Since 3.380617019 > 2.100922037 we fail to reject Ho    

Conclusion (in terms of the problem):     

There is sufficient evidence of disparity in sentencing based on the race of the defendant.