Question

Consider a sample of

48

football​ games, where

26

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.

Answer 1

Solution :

This is the two tailed test .

The null and alternative hypothesis is

H₀ : p =0.50

H_a : p > 0.50

= x / n = 26/48 = 0.5417

P₀ = 0.50

1 - P₀ = 1-0.50 =0.50

Test statistic = z

= - P₀ / [P_{0 *} (1 - P₀ ) / n]

=0.5417 - 0.50 / [0.50(0.50) /48 ]

= 0.577

P(z >0.577 ) = 1 - P(z < 0.577 ) = 0.2819

P-value = 0.2819

= 0.01

p = 0.2819 ≥ 0.01, it is concluded that the null hypothesis is not rejected.

There is not enough evidence to claim that the population proportion p is greater than p₀ at theα=0.01 significance level.