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: 86 70 80 76 70 67

y: 57 43 46 50 50 41

Given that Se ≈ 4.075, a ≈ 16.319, b ≈ 0.435, and x-bar ≈ 74.833, ∑y = 449, ∑x² = 33,861, and ∑y² = 13,895, find a 99% confidence interval for y when x = 72

​Select one:

a. between 25.3 and 66.8

b. between 27.0 and 65.1

c. between 25.6 and 66.6

d. between 25.8 and 66.3

e. between 24.9 and 67.2

Answer 1

here,

Sum of X = 449
Sum of Y = 287
Mean X = 74.8333
Mean Y = 47.8333
Sum of squares (SS_X) = 260.8333
Sum of products (SP) = 161.8333

Regression Equation = ŷ = bX + a

b = SP/SS_X = 161.83/260.83 = 0.62045

a = M_Y - bM_X = 47.83 - (0.62*74.83) = 1.40319

ŷ = 0.62045X + 1.40319

at, x = 72

Y = 46.08

for 99% of CI, alpha = 0.01

degree of freedom, df = n-2 = 4

therefore, t = 4.604

S_xx = 33861 - (449)² / 6 = 260.8333

margin of error, E = t*S_e*sqrt(1+1/n + (x - Xbar)²/S_xx ) = 4.604* 4.075*sqrt(1+1/6 + (72-74.833)² /260.8333 ) = 20.53

lower limit = 46.08 - 20.53 = 25.55 = ~ 25.6

upper limit = 46.08 + 20.53 = 66.61 = ~66.6

between 25.6 and 66.6 (option - C)