Question

A summer tennis tournament sees 2425 regular games played. Each team plays roughly half their games at home, and half their games away. Home teams won 1314 out of 2425 games, or 54.2%.
Could this deviation from 50% be explained just from natural sampling variability, or is this evidence to suggest that there is a home game advantage?
State the appropriate null and alternative hypotheses for this situation.
Carry out the relevant statistical test at 5% level of significance to decide on the claim represented by the null hypothesis.

Answer 1

Solution:

Given:

Sample size = n = 2425

x = 1314

Null and alternative hypothesis:

H: p=0.5 H: p=0.5

This corresponding to a two-tailed test, for which a z-test for one population proportion needs to be used.

Sample proportion:

n p= 1314 2425 = 0.5429 Test statistic: P-P. VP.(1 - p.)/n

こ3 0.5419-0.5 10.5(1-0.5)/2425 %3D14.122

Test statistic z = 4.122

Using the P-value approach:

The p-value is p = 0 …Using excel formula, =1-NORMSDIST(4.122)

And α = 0.05

Since p < 0.05, it is concluded that the Null Hypothesis is rejected.

Conclusion: There is sufficient evidence to suggest that there is a home game advantage at 0.05 level of significance.

Done