Question

Let x be a random variable that represents the percentage of successful free throws a professional basketball player makes in a season. Let y be a random variable that represents the percentage of successful field goals a professional basketball player makes in a season. A random sample of n = 6 professional basketball players gave the following information.

x	67	64	75	86	73	73
y	44	40	48	51	44	51

(a) Verify that Σx=438, Σy=278, Σx²=32,264, Σy²=12,978, and Σxy=20,429. Find r. (Round r to three decimal places.)

Σx =
Σy =
Σx2 =
Σy2 =
Σxy =
r =

(c) Find a, b, and x. (Round your answers to four decimal places.)

a =
b =
x =

(d) Find the predicted percentage ŷ of successful field goals for a player with x = 80% successful free throws. (Round your answer to two decimal places.)
%

(f) Use a 5% level of significance to test the claim that β > 0. (Round your answers to two decimal places.) Hint 1: The standard error of b is 0.172428. Hint 2: Your answers to the t and critical t should have the same sign.

t =
critical t =

Conclusion

Reject the null hypothesis, there is sufficient evidence that β > 0.

Reject the null hypothesis, there is insufficient evidence that β > 0.

Fail to reject the null hypothesis, there is insufficient evidence that β > 0.

Fail to reject the null hypothesis, there is sufficient evidence that β > 0.

Answer 1

Part a)

ΣX = 438
ΣY = 278
ΣX * Y = 20429
ΣX2 = 32264
ΣY2 = 12978

r = 0.804

Part c)

X̅ = Σ( Xi / n ) = 438/6 = 73
Y̅ = Σ( Yi / n ) = 278/6 = 46.33

Equation of regression line is Ŷ = a + bX


b = 0.466
a =( Σ Y - ( b * Σ X) ) / n
a =( 278 - ( 0.4655 * 438 ) ) / 6
a = 12.351
Equation of regression line becomes Ŷ = 12.351 + 0.466 X

Part d)

When X = 80
Ŷ = 12.351 + 0.466 X
Ŷ = 12.351 + ( 0.466 * 80 )
Ŷ = 49.63

Part f)

Sxx =Σ (Xi - X̅ )2 = 290
Syy = Σ( Yi - Y̅ )2 = 97.3333
Sxy = Σ (Xi - X̅ ) * (Yi - Y̅) = 135

X̅ = Σ (Xi / n ) = 438/6 = 73
Y̅ = Σ (Yi / n ) = 278/6 = 46.3333

To Test :-

H0 :- ß = 0

H1 :- ß > 0

Test Statistic :-
t = ( b - β) / ( S / √(S(xx)))
t = ( 0.4655 - 0 ) / ( 2.9364 / √(290))
t = 2.6996

Test Criteria :-
Reject null hypothesis if t > t(α)
t(α,n-2) = t(0.05 , 6 - 2 ) = 2.1318
t < t(α, n-2) = 2.6996 < 2.1318
Result :- Fail to reject null hypothesis

Fail to reject the null hypothesis, there is insufficient evidence that β > 0.