You may need to use the appropriate technology to answer this question. Scores in the first and fourth (final) rounds

a)

Ho :   µd=   0
Ha :   µd ╪   0

paired t test

mean of difference ,    D̅ =ΣDi / n =   -0.900
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    2.972

Level of Significance ,    α =    0.1
      
sample size ,    n =    20
      

std error , SE = Sd / √n =    2.9718   / √   20   =   0.6645      
                          
t-statistic = (D̅ - µd)/SE = (   -0.9000   -   0   ) /    0.6645   =   -1.354
                          
Degree of freedom, DF=   n - 1 =    19                  
  
p-value =        0.1915   [excel function: =t.dist.2t(t-stat,df) ]               
Conclusion:     p-value>α , Do not reject null hypothesis     

Do not reject H0. There is insufficient evidence to conclude that the mean score for the first round of a golf tournament is significantly different than the mean score for the fourth and final round.

b)

point estimate of difference ,    D̅ =ΣDi / n =   -0.900

c)

sample size ,    n =    20          
Degree of freedom, DF=   n - 1 =    19   and α =    0.1  
t-critical value =    t α/2,df =    1.7291   [excel function: =t.inv.2t(α/2,df) ]      
                  
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    2.9718          
                  
std error , SE = Sd / √n =    2.9718   / √   20   =   0.6645
margin of error, E = t*SE =    1.7291   *   0.6645   =   1.1490
                  
Yes. One could check to see if the 90% confidence interval includes a difference of zero. If the interval contains 0, the difference is not statistically significant.