Question

A study was conducted to assess the effects that occur when children are exposed to cocaine before birth. Children were tested at age 4 for object assembly skil which was described as "a task requiring visual-spatial skills related to mathematical competence. The 193 children born to cocaine users had a mean of 74 and standard deviation of 3.0. The 193 children not exposed to cocaine had a mean score of 8.3 with a standard deviation of 3.0. Assume that the two samples are Independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Lot population 1 be children born to cocaine users. Complete parts (a) and (b) below. C. Ho: O A Hol4*12 OB. Ho: 44 =12 42 Hy: 1732 H > H₂:14-12 2 OD. Ho: 142 OE Ho-H422 Hot 12 H: H2 на: 2 <p The test statistic is -2.95 (Round to two decimal places as needed.) The P-value is .003 (Round to three decimal places as needed.) State the conclusion for the test. Reject the null hypothesis. There is sufficient evidence to support the children on the test of object assembly b. Test the claim in part (a) by using a confidence interval The 95% confidence interval estimate is -1.5 <-H2 - 3 (Round to one decimal place as needed.) claim that prenatal cocaine exposure is associated with lower scores 8 Enter your answer in the edit fields and then click Check Ans 1 remaining Clear Final Check

Answer 1

Sample #1   ---->   1                  
mean of sample 1,    x̅1=   7.400                  
standard deviation of sample 1,   s1 =    3.0000                  
size of sample 1,    n1=   193                  
                          
Sample #2   ---->   2                  
mean of sample 2,    x̅2=   8.300                  
standard deviation of sample 2,   s2 =    3.0000                  
size of sample 2,    n2=   193                  
                          
difference in sample means =    x̅1-x̅2 =    7.4000   -   8.3   =   -0.90  
                          

std error , SE =    Sp*√(1/n1+1/n2) =    0.3054                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -0.9000   -   0   ) /    0.31   =   -2.95

Degree of freedom, DF=       384
p-value =        0.002

reject , is

b)

Degree of freedom, DF=       384          
t-critical value =    t α/2 =    1.966   (excel formula =t.inv(α/2,df)      
                  
                  
                  
std error , SE =    √(s1²/n1+s2²/n2) =    0.305          
margin of error, E = t*SE =    1.966   *   0.305   =   0.60
                  
difference of means = x̅1-x̅2 =    7.4000   -   8.300   =   -0.9000
confidence interval is                   
Interval Lower Limit = (x̅1-x̅2) - E =    -0.9000   -   0.600   =   -1.50
Interval Upper Limit = (x̅1-x̅2) + E =    -0.9000   -   0.600   =   -0.30

(-1.5 , -0.3 )