Question

1. Anyone who plays or watches sports has heard of the “home field advantage.” Tournaments in many sports are designed to try to neutralize the advantage of the home team or player. Most people believe that teams tend to win more often when they play at home. But do they? If there were no home field advantage, the home teams would win about half of all games played. To test this, we’ll use the games in the Major league Baseball 2017 season. That year, there were 2430 regular-season games. It turns out that the home team won 1312 of the 2430 games, or 53.99% of the time. Is there strong evidence that there exists a “home field advantage?” Use a significance level of 5% to test your hypothesis.

a) Define the parameter of interest.

b) State the hypotheses statements.

c) Check that the conditions have been met and name the significance test to be used. (Note: Though we are not interested in just the 2017 season, and this is not randomly selected, it is reasonable that the sample is representative of all Major League Baseball games in the recent past and near future.)

d) Perform the test and show all work.

e) Interpret the P-value and state your conclusion.

f) Calculate and interpret a 95% confidence interval.

g) Describe what a Type I and Type II error would be in the context of this problem.

Answer 1

n = 2430

x = 1312

p̄ = x/n = 0.5399

a) Parameter of interest: Population parameter

b) Null and Alternative hypothesis:

Ho : p ≤ 0.5

H1 : p > 0.5

c) α = 0.05

The assumptions are satisfied.

Sample is random and independent.

n*p0 = 2430*0.5 = 1215 is greater than 10.

d) Test statistic:

z = (p̄ -p)/√(p*(1-p)/n) = (0.5399 - 0.5)/√(0.5 * 0.5/2430) = 3.9355

e) p-value = 1- NORM.S.DIST(3.9355, 1) = 0.0000

Decision:

p-value < α, Reject the null hypothesis

f) 95% Confidence interval :  

At α = 0.05, two tailed critical value, z_c = NORM.S.INV(0.05/2) = 1.960

Lower Bound = p̄ - z_c*√( p̄ *(1- p̄ )/n) = 0.5399 - 1.96 *√(0.5399*0.4601/2430) = 0.5201

Upper Bound = p̄ + z_c*√( p̄ *(1- p̄ )/n) = 0.5399 + 1.96 *√(0.5399*0.4601/2430) = 0.5597

There is 95% confidence that the true proportion is between 0.5201 and 0.5597

g) Type I error is rejecting the null hypothesis when it is true.

Type I error in this context will be to conclude that there exists a “home field advantage. when in reality it does not.

Type II error is failing to reject the null hypothesis when it is not true.

Type II error in this context will be to conclude that there does not exists a “home field advantage. when in reality it does.