Question

4. The National Assessment of Education Progress (NAEP) tested a sample of students who had used a computer in their mathematics classes and another sample of students who had not used a computer. The sample mean score for students using the computer was 309 with a sample standard deviation 29. For students not using computer, the sample mean was 303 and sample standard deviation was 32. Assume there were 60 students in the computer sample and 40 students in the sample that hadn’t used a computer. Can you conclude that the population mean scores differ? Use a α=0.05 significance level.

a. Calculate the Margin of error for 99% confidence level.

b. Construct a 99% confidence interval for the difference in means

c. Do the Hypothesis test.

Answer 1

a)

α=0.01

Degree of freedom, DF=   n1+n2-2 =    98              
t-critical value =    t α/2 =    2.6269   (excel formula =t.inv(α/2,df)          
                      
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    30.2296              
                      
std error , SE =    Sp*√(1/n1+1/n2) =    6.1706              
margin of error, E = t*SE =    2.6269   *   6.17   =   16.21  
                      

ii)

difference of means =    x̅1-x̅2 =    309.0000   -   303.000   =   6.0000
confidence interval is                       
Interval Lower Limit=   (x̅1-x̅2) - E =    6.0000   -   16.2097   =   -10.210
Interval Upper Limit=   (x̅1-x̅2) + E =    6.0000   +   16.2097   =   22.210

iii)

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 ╪   0                  
                          
Level of Significance ,    α =    0.05                  
                          
Sample #1   ---->   1                  
mean of sample 1,    x̅1=   309.000                  
standard deviation of sample 1,   s1 =    29.0000                  
size of sample 1,    n1=   60                  
                          
Sample #2   ---->   2                  
mean of sample 2,    x̅2=   303.000                  
standard deviation of sample 2,   s2 =    32.0000                  
size of sample 2,    n2=   40                  
                          
difference in sample means =    x̅1-x̅2 =    309.0000   -   303.0   =   6.00  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    30.2296                  
std error , SE =    Sp*√(1/n1+1/n2) =    6.1706                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   6.0000   -   0   ) /    6.17   =   0.972
                          
Degree of freedom, DF=   n1+n2-2 =    98                  
  
p-value =        0.3333   (excel function: =T.DIST.2T(t stat,df) )              
Conclusion:     p-value>α , Do not reject null hypothesis                      
                          
There is not enough evidence of significant difference