Question

You may need to use the appropriate technology to answer this question.

Scores in the first and fourth (final) rounds for a sample of 20 golfers who competed in golf tournaments are shown in the following table.

Player	First Round	Final Round
Golfer 1	70	72
Golfer 2	71	72
Golfer 3	70	73
Golfer 4	72	71
Golfer 5	70	69
Golfer 6	67	67
Golfer 7	71	67
Golfer 8	68	74
Golfer 9	67	72
Golfer 10	70	69

Player	First Round	Final Round
Golfer 11	72	72
Golfer 12	72	70
Golfer 13	70	73
Golfer 14	70	75
Golfer 15	68	70
Golfer 16	68	65
Golfer 17	71	70
Golfer 18	70	68
Golfer 19	69	68
Golfer 20	67	71

Calculate the value of the test statistic. (Round your answer to three decimal places.)

=

Calculate the p-value. (Round your answer to four decimal places.)

p-value =

(b)

What is the point estimate of the difference between the two population means? (Use mean score first round − mean score fourth round.)

=

(c)

What is the margin of error for a 90% confidence interval estimate for the difference between the population means? (Round your answer to two decimal places.)

=

Answer 1

Sample #1	Sample #2	difference , Di =sample1-sample2	(Di - Dbar)²
70	72	-2	1.56
71	72	-1	0.06
70	73	-3	5.06
72	71	1	3.06
70	69	1	3.06
67	67	0	0.56
71	67	4	22.56
68	74	-6	27.56
67	72	-5	18.06
70	69	1	3.06
72	72	0	0.56
72	70	2	7.56
70	73	-3	5.06
70	75	-5	18.06
68	70	-2	1.56
68	65	3	14.06
71	70	1	3.06
70	68	2	7.56
69	68	1	3.06
67	71	-4	10.56

	sample 1	sample 2	Di	(Di - Dbar)²
sum =	1393	1408	-15	155.750

mean of difference ,    D̅ =ΣDi / n =   -0.750
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    2.863

a)

Level of Significance ,    α =    0.05                  
                          
sample size ,    n =    20                  
                          
  
  
     
std error , SE = Sd / √n =    2.8631   / √   20   =   0.6402      
                          
t-statistic = (D̅ - µd)/SE = (   -0.7500   -   0   ) /    0.6402   =   -1.171
                          
Degree of freedom, DF=   n - 1 =    19                  
  
p-value =       0.2559 [excel function: =t.dist.2t(t-stat,df) ]               

b)

mean of difference ,    D̅ =ΣDi / n =   -0.75

c)

Level of Significance ,    α =    0.1
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) =    2.8631          
                  
std error , SE = Sd / √n =    2.8631   / √   20   =   0.6402
margin of error, E = t*SE =    1.7291   *   0.6402   =   1.1070
                  
mean of difference ,    D̅ =   -0.750          
confidence interval is                   
Interval Lower Limit= D̅ - E =   -0.750   -   1.1070   =   -1.8570
Interval Upper Limit= D̅ + E =   -0.750   +   1.1070   =   0.3570
                  
so, confidence interval is (   -1.86 < Dbar <   0.36 )