Question

To test the belief that sons are taller than their fathers, a student randomly selects 13 fathers who have adult male children. She records the heights of both father and son in inches and obtains the following data. A normal probability plot and boxplot of the data indicate that the differences are approximately normally distributed with no outliers. Test the claim at the α = 0.05 level of significance. Need to explain and show all work.

fathers height: 70.3 67.1 70.9 66.8 72.8 70.4 71.8 70.1 69.9 70.8 70.2 70.4 72.4

Sons Height: 74.1 69.2 66.9 69.2 68.9 70.2 70.4 69.3 75.8 72.3 69.2 68.6 73.9

Answer 1

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 <   0                  
                          
Level of Significance ,    α =    0.05                  
                          
Sample #1   ---->   sample 1                  
mean of sample 1,    x̅1=   70.30                  
standard deviation of sample 1,   s1 =    1.74                  
size of sample 1,    n1=   13                  
                          
Sample #2   ---->   sample 2                  
mean of sample 2,    x̅2=   70.62                  
standard deviation of sample 2,   s2 =    2.60                  
size of sample 2,    n2=   13                  
                          
difference in sample means =    x̅1-x̅2 =    70.3000   -   70.6   =   -0.32  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    2.2141                  
std error , SE =    Sp*√(1/n1+1/n2) =    0.8685                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -0.3154   -   0   ) /    0.87   =   -0.363
                          
Degree of freedom, DF=   n1+n2-2 =    24                  

p-value =        0.359833   [ excel function: =T.DIST(t stat,df) ]               
Conclusion:     p-value>α , Do not reject null hypothesis                      
                          
There is not enough evidence to support the claim

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