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A study was performed on a certain river to investigate how the distance from the mouth of the river (in km) relates to the ce) f) g) only plz

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e) Ho:   ß1=   0          
H1:   ß1╪   0          
n=   10              
alpha=   0.05   
estimated std error of slope =Se(ß1) = Se/√Sxx =    1.161   /√   16   =   0.2876
                  
t stat = estimated slope/std error =ß1 /Se(ß1) =    3.1779   /   0.2876   =   11.051
                  
Degree of freedom ,df = n-2=   8              
p-value =    0.0000              
decision :    p-value<α , reject Ho              

cocnlsuion: there is significant effect.........

f) confidence interval for slope                  
α=   0.05              
t critical value=   t α/2 =    2.306   [excel function: =t.inv.2t(α/2,df) ]      
estimated std error of slope = Se/√Sxx =    1.16104   /√   16.30   =   0.288
                  
margin of error ,E= t*std error =    2.306   *   0.288   =   0.663
estimated slope , ß^ =    3.1779              
                  
                  
lower confidence limit = estimated slope - margin of error =   3.1779   -   0.663   =   2.5148
upper confidence limit=estimated slope + margin of error =   3.1779   +   0.663   =   3.8411

g)

confidence interval do not contain 0, so, null hypothesis is rejected .

hence, both confidence interval and hypothesis test have same result.

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