Question

Suppose a new standardized test is given to 97 randomly selected​ third-grade students in New Jersey. The sample average score Upper Y overbarY

on the test is 56 ​points and the sample standard​ deviation, s Subscript Upper YsY​, is 10 points. The authors plan to administer the test to all​ third-grade students in New Jersey.

1. The​ 95% confidence interval for the mean score of all New Jersey third graders is (round to two decimal places).

2. Suppose the same test is given to 190 randomly selected third graders from​ Iowa, producing a sample average of 66 points and sample standard deviation of

13 points.The​ 90% confidence interval for the difference in mean scores between Iowa and New Jersey is (round to two decimal places)

3. The p​-value of the test of no difference in means versus some difference is

Answer 1

1)

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   96          
't value='   tα/2=   1.985   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   10.0000   / √   97   =   1.0153
margin of error , E=t*SE =   1.9850   *   1.0153   =   2.0154
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    56.00   -   2.015446   =   53.9846
Interval Upper Limit = x̅ + E =    56.00   -   2.015446   =   58.0154
95%   confidence interval is (   53.98   < µ <   58.02   )

2)

Level of Significance ,    α =    0.1          
degree of freedom=   DF=n-1=   189          
't value='   tα/2=   1.653   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   13.0000   / √   190   =   0.9431
margin of error , E=t*SE =   1.6530   *   0.9431   =   1.5589
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    66.00   - 1.558934   =   64.4411
Interval Upper Limit = x̅ + E =    66.00   -   1.558934   =   67.5589
90%   confidence interval is (   64.44   < µ <   67.56   )

3)

Sample #1   ---->   1          
mean of sample 1,    x̅1=   56.00          
standard deviation of sample 1,   s1 =    10          
size of sample 1,    n1=   97          
                  
Sample #2   ---->   2          
mean of sample 2,    x̅2=   66.000          
standard deviation of sample 2,   s2 =    13.00          
size of sample 2,    n2=   190          
                  
difference in sample means = x̅1-x̅2 =    56.000   -   66.0000   =   -10.0000
                  
std error , SE =    √(s1²/n1+s2²/n2) =    1.3858          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -10.0000   /   1.3858   ) =   -7.2161
                  
Degree of freedom,     df= 241          
                  
                  
p-value =        0.0000   (excel function: =T.DIST.2T(t stat,df) )