Question

Healthy subjects aged 18 to 40 participated in a study of eating habits. Subjects were given bags of potato chips and bottled water and invited to snack freely. Was there a difference between men and women in the number of potato chips consumed? Here are the data on grams of potato chips consumed.

Group	n	Mean	Standard Deviation
Males	9	38	15
Females	11	12	10

We are interested in calculating a 90% confidence interval for the difference in mean number of potato chips consumed between men and women. Without using software, what is the appropriate critical value to use in the calculation of the confidence interval? Hint: Use the approximation used in the module for determining the approximate degrees of freedom.

Answer 1

Sample #1   ---->   1
mean of sample 1,    x̅1=   38.00
standard deviation of sample 1,   s1 =    15
size of sample 1,    n1=   9
      
Sample #2   ---->   2
mean of sample 2,    x̅2=   12.000
standard deviation of sample 2,   s2 =    10.00
size of sample 2,    n2=   11

DF = min(n1-1 , n2-1 )=   8              
t-critical value , t* =    1.8595   (excel formula =t.inv(α/2,df)          
                  
                  
                  
std error , SE =    √(s1²/n1+s2²/n2) =    5.839          
margin of error, E = t*SE =    1.8595   *   5.839   =   10.8574
                  
difference of means = x̅1-x̅2 =    38.0000   -   12.000   =   26.000
confidence interval is                   
Interval Lower Limit = (x̅1-x̅2) - E =    26.0000   -   10.8574   =   15.142578621
Interval Upper Limit = (x̅1-x̅2) + E =    26.0000   -   10.8574   =   36.857421379