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

Answer 1

According to the information given, it is found that:

• The hypothesis tested are:

C. $H_0: p = 0.5$, $H_1: p > 0.5$

• The test statistic is z = 1.66.

• The p-value of the test is of 0.0485.

• The conclusion is:

Fail to reject $H_0$, as there is not sufficient evidence to conclude that the probability that the home team wins is greater than one half.

At the null hypothesis, it is tested if home teams win half their games, that is, if the proportion is of 0.5, hence:

$H_0: p = 0.5$

At the alternative hypothesis, it is tested if home teams win more than half their games, that is, if the proportion is of more than 0.5, hence:

$H_1: p > 0.5$

The test statistic is given by:

$z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$

In which:

• $\overline{p}$ is the sample proportion.
• p is the proportion tested at the null hypothesis.
• n is the sample size.

For this problem, the parameters are:

$p = 0.5, n = 52, \overline{p} = \frac{32}{52} = 0.6154$

Hence, the value of the test statistic is:

$z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$

$z = \frac{0.6154 - 0.5}{\sqrt{\frac{0.5(0.5)}{52}}}$

$z = 1.66$

The test statistic is z = 1.66.

The p-value is the probability of finding a sample proportion above 0.6154, which is 1 subtracted by the p-value of z = 1.66.

• Looking at the z-table, z = 1.66 has a p-value of 0.9515.

1 - 0.9515 = 0.0485.

The p-value of the test is of 0.0485.

Since the p-value is greater than 0.01, the conclusion is:

Fail to reject $H_0$, as there is not sufficient evidence to conclude that the probability that the home team wins is greater than one half.

A similar problem is given at https://brainly.com/question/25600854