Test the claim that the proportion of people who own cats is significantly different than $50 \%$ at the 0.01 significance level. The null and alternative hypothesis would be: \[ \begin{array}{l} H_{0}: p \geq 0.5 \quad H_{0}: p=0.5 \quad H_{0}: \mu \leq 0.5 \quad H_{0}: \mu=0.5 \quad H_{0}: \mu \geq 0.5 \quad H_{0}: p \leq 0.5 \\ H_{1}: p<0.5 \quad H_{1}: p \neq 0.5 \quad H_{1}: \mu>0.5 \quad H_{1}: \mu \neq 0.5 \quad H_{1}: \mu<0.5 \quad H_{1}: p>0.5 \\ \end{array} \] The test is: right-tailed two-tailed left-tailed Based on a sample of 800 people, $47 \%$ owned cats The p-value is: (to 2 decimals) Based on this we: Fail to reject the null hypothesis Reject the null hypothesis

Step 1 ：Define the null and alternative hypothesis as follows: \(H_{0}: p = 0.5\) and \(H_{1}: p \neq 0.5\). This is a two-tailed test.

Step 2 ：Given a sample size of 800 people, 47% of them owned cats. So, the sample proportion \(p_{sample} = 0.47\).

Step 3 ：Calculate the standard error using the formula \(\sqrt{p_{null} \cdot (1 - p_{null}) / n}\), where \(p_{null} = 0.5\) is the population proportion under the null hypothesis and \(n = 800\) is the sample size. The calculated standard error is approximately 0.01768.

Step 4 ：Calculate the z-score using the formula \((p_{sample} - p_{null}) / se\), where \(se\) is the standard error. The calculated z-score is approximately -1.70.

Step 5 ：Calculate the p-value for a two-tailed test. The calculated p-value is approximately 0.09.

Step 6 ：Since the p-value is greater than the significance level of 0.01, we fail to reject the null hypothesis. This means that we do not have enough evidence to support the claim that the proportion of people who own cats is significantly different than 50%.

Step 7 ：Final Answer: We fail to reject the null hypothesis. The p-value is approximately \(\boxed{0.09}\).