Question

We want to compare the weights of two independent groups of mice. Group 1 consists of 14 mice that were fed only cheese. Group 2 consists of 18 mice that were fed only walnuts. Group 1 information: sample mean x1-bar = 18 and sample standard deviation s1 = 4. Group 2 information: sample mean x2-bar = 15 and sample standard deviation s2 = 7. Perform a 2-sided hypothesis test of H0: mu1 = mu2 against H1: mu1 not equal to mu2. Do not assume the two samples share the same variance. Find the P-value. [Hint: the distribution you use will have 27 degrees of freedom.] Answer to three decimal places.

Answer 1

not equal variance                  
Ho :   µ1 - µ2 =   0          
Ha :   µ1-µ2 ╪   0          
                  
Level of Significance ,    α =    0.05          
                  
Sample #1   ---->   sample 1          
mean of sample 1,    x̅1=   18.00          
standard deviation of sample 1,   s1 =    4          
size of sample 1,    n1=   14          
                  
Sample #2   ---->   sample 2          
mean of sample 2,    x̅2=   15.000          
standard deviation of sample 2,   s2 =    7.00          
size of sample 2,    n2=   18          
                  
difference in sample means = x̅1-x̅2 =    18.000   -   15.0000   =   3.0000
                  
std error , SE =    √(s1²/n1+s2²/n2) =    1.9660          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   3.0000   /   1.9660   ) =   1.5260

Degree of freedom, DF=       27
p-value =        0.1387 (excel function: =T.DIST.2T(t stat,df) )  
Conclusion:     p-value>α , Do not reject null hypothesis