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 87 88 70 84 78 76 y 53 57 50 51 46 50 Given that Se ≈ 3.054, a ≈ 16.547, b ≈ 0.425, and , find the predicted percentage of successful field goals for a player with x = 73% successful free throws. Select one: a. 47.6% b. 5.1% c. 14.5% d. 28.0% e. 31.0%

Answer 1

From given result we get, the least-squares line is,

y-hat = a + bx

y-hat = 16.547 + 0.425x

We want to predict y-hat for x = 73%

=> y-hat = 16.547 + (0.425 * 73) =16.547 + 31.025 = 47.572

=> y-hat = 47.6%

Answer: a) 47.6%,