Question

Anna suspected that when people exercise longer, their body temperatures change. She randomly assigned people to exercise for 30 or 60 minutes, then measured their temperatures. The 18 people who exercised for 30 minutes had a mean temperature of 38.3º C with a standard deviation of 0.27º C. The 24 people who exercised 60 minutes had a mean temperature of 38.9º C with a standard deviation of 0.29º C.

Assume Anna will use the conservative degrees of freedom from the smaller sample size.

Calculate the 90% confidence interval for the difference in mean body temperature after exercising for the two amounts of time.

What is the upper and lower limit?

Answer 1

Sample #1   ---->   30 minutes
mean of sample 1,    x̅1=   38.30
standard deviation of sample 1,   s1 =    0.27
size of sample 1,    n1=   18
      
Sample #2   ---->   60 minutes
mean of sample 2,    x̅2=   38.900
standard deviation of sample 2,   s2 =    0.29
size of sample 2,    n2=   24

DF = min(n1-1 , n2-1 )=   17              
t-critical value , t* =    1.7396 (excel formula =t.inv(α/2,df)          
                  
                  
                  
std error , SE =    √(s1²/n1+s2²/n2) =    0.087          
margin of error, E = t*SE =    1.7396   *   0.087   =   0.1512
                  
difference of means = x̅1-x̅2 =    38.3000   -   38.900   =   -0.6000
confidence interval is                   
Interval Lower Limit = (x̅1-x̅2) - E =    -0.6000   -   0.1512   =   -0.7512
Interval Upper Limit = (x̅1-x̅2) + E =    -0.6000   -   0.1512   =   -0.4488