Question

You are looking at the impact of diabetes on elementary school students’ academic performance. You decide to assess this using the standardized test performance of students (units for this are percentile). You want to see how students with and w/o diabetes as a group compare on passing a standardized test. So you decide to assess whether the two different groups each have an average that is above the states minimum passing score of 65%. The school has 235 students w/o diabetes and their average test score is 75% with a standard deviation of 20%. You also have data from 20 diabetic students who have an average score of 70% and a standard deviation of 25%. To lower the risk of error, the school asks you to use a significance of 0.01. (Hint: you have to run two t-tests and you need to use the 65% in both)

Answer 1

T-TEST FOR W/O DIABETIC

Ho :   µ =   65                  
Ha :   µ >   65       (Right tail test)          
                          
Level of Significance ,    α =    0.01                  
sample std dev ,    s =    20.0000                  
Sample Size ,   n =    235                  
Sample Mean,    x̅ =   75.0000                  
                          
degree of freedom=   DF=n-1=   234                  
                          
Standard Error , SE = s/√n =   20.0000   / √    235   =   1.3047      
t-test statistic= (x̅ - µ )/SE = (   75.000   -   65   ) /    1.3047   =   7.66
                          
  
p-Value   =   0.0000   [Excel formula =t.dist(t-stat,df) ]              
Decision:   p-value<α, Reject null hypothesis

SO, MEAN IS GREATER THAN 65%

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

WITH DIABETICS

Ho :   µ =   65                  
Ha :   µ > 65       (Two tail test)          
                          
Level of Significance ,    α =    0.01                  
sample std dev ,    s =    25.0000                  
Sample Size ,   n =    20                  
Sample Mean,    x̅ =   70.0000                  
                          
degree of freedom=   DF=n-1=   19                  
                          
Standard Error , SE = s/√n =   25.0000   / √    20   =   5.5902      
t-test statistic= (x̅ - µ )/SE = (   70.000   -   65   ) /    5.5902   =   0.89
                          
  
p-Value   =   0.1911 [Excel formula =t.dist(t-stat,df) ]              
Decision:   p-value>α, Do not reject null hypothesis

MEAN IS LESS THAN 65%   

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