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section 10.4

The nul hypothesis is Hoi Hy Hy and the alternative hypothesis is H: * Hz. The data in the accompanying table are from a simp
The null hypothesis is Hg: Hy = Hz and the alternative hypothesis is as specified. The provided data are from a simple random
The null hypothesis is HowHy and the alternative hypothesis is He: HH2. The data in the accompanying table are from a simple

A researcher randomly selects 6 fathers who have adult sons and records the fathers and sons heights to obtain the data sho
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Answer #1

1)

sample 1 sample 2 Di (Di - Dbar)²
sum = 91 74.00 17.000 33.714

sample size ,    n =    7                  
                          
mean of sample 1,    x̅1=   13.000                  
                          
mean of sample 2,    x̅2=   10.571                  
                          
mean of difference ,    D̅ =ΣDi / n =   2.4286                  
                          
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    2.3705                  
                          
std error , SE = Sd / √n =    2.3705   / √   7   =   0.8959      
                          
t-statistic = (D̅ - µd)/SE = (   2.428571429   -   0   ) /    0.8959   =   2.71

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

2)

sample 1 sample 2 Di (Di - Dbar)²
sum = 109 108.00 1.000 112.8750

sample size ,    n =    8                  
                          
mean of sample 1,    x̅1=   13.625                  
                          
mean of sample 2,    x̅2=   13.500                  
                          
mean of difference ,    D̅ =ΣDi / n =   0.1250                  
                          
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    4.0156                  
                          
std error , SE = Sd / √n =    4.0156   / √   8   =   1.4197      
                          
t-statistic = (D̅ - µd)/SE = (   0.125   -   0   ) /    1.4197   =   0.09

......

3)

sample 1 sample 2 Di (Di - Dbar)²
sum = 161 185.00 -24.000 82.000

sample size ,    n =    9                  
                          
mean of sample 1,    x̅1=   17.889                  
                          
mean of sample 2,    x̅2=   20.556                  
                          
mean of difference ,    D̅ =ΣDi / n =   -2.6667                  
                          
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    3.2016                  
                          
std error , SE = Sd / √n =    3.2016   / √   9   =   1.0672      
                          
t-statistic = (D̅ - µd)/SE = (   -2.666666667   -   0   ) /    1.0672   =   -2.50

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

4)

sample size ,    n =    6                  
                          
mean of sample 1,    x̅1=   69.417                  
                          
mean of sample 2,    x̅2=   71.150                  
                          
mean of difference ,    D̅ =ΣDi / n =   -1.7333                  
                          
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    3.1130                  
                          
std error , SE = Sd / √n =    3.1130   / √   6   =   1.2709      
                          
t-statistic = (D̅ - µd)/SE = (   -1.733333333   -   0   ) /    1.2709   =   -1.364

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

Please let me know in case of any doubt.

Thanks in advance!


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