Question

Consider a sample of 54 football​ games, where 32 of them were won by the home team. Use a 0.10 significance level to test the claim that the probability that the home team wins is greater than​ one-half.

Answer 1

Solution:

Here, we have to use z test for population proportion.

H₀: p ≤ 0.5 versus H_a: p > 0.5

This is an upper tailed test.

Test statistic formula for this test is given as below:

Z = (p̂ - p)/sqrt(pq/n)

Where, p̂ = Sample proportion, p is population proportion, q = 1 - p, and n is sample size

x = number of items of interest = 32

n = sample size = 54

p̂ = x/n = 32/54 = 0.592592593

p = 0.5

q = 1 - p = 0.5

Z = (p̂ - p)/sqrt(pq/n)

Z = (0.592592593 – 0.5)/sqrt(0.5*0.5/54)

Z = 1.3608

P-value = 0.0868

(by using z-table)

α = 0.10

P-value < α

So, we reject the null hypothesis

There is sufficient evidence to conclude that the probability that the home team wins is greater than ​one-half.