Question

Consider a sample of 52 football​ games, where 32 of them were won by the home team. Use a 0.01 significance level to test the claim that the probability that the home team wins is greater than​ one-half. Identify the null and alternative hypotheses for this test.

Answer 1

Given:

Number of favourable cases, X = 32

Sample size, n = 52

Significance level, = 0.01

Sample proportion, = X/n = 32/52 = 0.6154

Hypothesis test:

The null and alternative hypothesis is

H0 : p 0.5 (p = 0.5 also true)

Ha : p > 0.5 (This is a right tailed test)

Test statistics is

z = - p/√p(1-p/n

= 0.6154-0.5/√0.5(1-0.5)/52

= 1.6642

Test statistics is z = 1.66

P-value for right tailed test:

P-value = P(Z>1.66) = 0.0485...(from z table)

Since P-value is greater than significance level 0.01, we fail to reject null hypothesis.

Decision: Fail to reject H0.

Conclusion : At 0.01 significance level, there is not sufficient evidence to conclude that the probability that the home team wins is greater than​ one-half.