Let's work through the steps to test the hypothesis that the proportion of home team wins is different from 0.5 (50%). This involves conducting a hypothesis test for the proportion.
### Step-by-Step Solution:
1. State the null and alternative hypotheses:
- Null Hypothesis ([tex]\(H_0\)[/tex]): The true proportion of home team wins [tex]\( p \)[/tex] is 0.5.
- Alternative Hypothesis ([tex]\(H_a\)[/tex]): The true proportion of home team wins [tex]\( p \)[/tex] is not 0.5.
2. Set the significance level:
- Significance level ([tex]\(\alpha\)[/tex]) = 0.01
3. Calculate the sample proportion:
- Number of games in the sample ([tex]\( n \)[/tex]) = 50
- Number of home wins ([tex]\( X \)[/tex]) = 30
- Sample proportion of home team wins ([tex]\( \hat{p} \)[/tex]) = [tex]\( \frac{X}{n} = \frac{30}{50} = 0.6 \)[/tex]
4. Calculate the standard error (SE) for the sample proportion under the null hypothesis:
- The proportion under the null hypothesis ([tex]\( p_0 \)[/tex]) = 0.5
- Standard Error (SE) = [tex]\( \sqrt{\frac{p_0 (1 - p_0)}{n}} = \sqrt{\frac{0.5 \times 0.5}{50}} = 0.07071067811865475 \)[/tex]
5. Calculate the z-score:
- Z-score = [tex]\( \frac{\hat{p} - p_0}{SE} = \frac{0.6 - 0.5}{0.07071067811865475} = 1.4142135623730947 \)[/tex]
6. Calculate the p-value:
- For a two-tailed test, the p-value is calculated using the standard normal distribution. Since the test is two-tailed, we consider both tails of the distribution.
- P-value = [tex]\( 2 \times (1 - \Phi(|z|)) = 2 \times (1 - \Phi(1.4142135623730947)) \approx 0.15729920705028544 \)[/tex]
7. Compare the p-value with the significance level:
- If the p-value is less than the significance level ([tex]\(\alpha\)[/tex]), we reject the null hypothesis.
- P-value = 0.15729920705028544
- Since 0.15729920705028544 > 0.01, we fail to reject the null hypothesis.
### Conclusion:
At the 0.01 significance level, we do not have enough evidence to reject the null hypothesis. Therefore, we do not reject the claim that the true proportion of home team wins is 50%. The result does not provide sufficient evidence to say that the proportion of home wins is different from 0.5.
