Test the claim that the proportion of people who own cats is larger than $60 \%$ at the 0.10 significance level. The null and alternative hypothesis would be:
\[
\begin{array}{l}
H_{0}: p \leq 0.6 H_{0}: \mu \leq 0.6 \quad H_{0}: \mu=0.6 \quad H_{0}: p=0.6 \quad H_{0}: \mu \geq 0.6 \quad H_{0}: p \geq 0.6 \text { os } \\
H_{1}: p> 0.6 H_{1}: \mu> 0.6 H_{1}: \mu \neq 0.6 \quad H_{1}: p \neq 0.6 \quad H_{1}: \mu< 0.6 \quad H_{1}: p< 0.6 \\
\end{array}
\]
The test is:
two-tailed right-tailed left-tailed $\infty^{s}$
Based on a sample of 200 people, $68 \%$ owned cats
The test statistic is:
(to 2 decimals)
The $p$-value is: (to 2 decimals)

Step 1 ：We are testing the claim that the proportion of people who own cats is larger than 60% at the 0.10 significance level. The null and alternative hypothesis would be \(H_{0}: p \leq 0.6\) and \(H_{1}: p>0.6\). This is a right-tailed test.

Step 2 ：The sample size is 200 and the sample proportion is 68%.

Step 3 ：The test statistic for a proportion is calculated as: \[Z = \frac{\hat{p} - p_{0}}{\sqrt{\frac{p_{0}(1-p_{0})}{n}}}\] where \(\hat{p}\) is the sample proportion, \(p_{0}\) is the hypothesized proportion under the null hypothesis, and \(n\) is the sample size.

Step 4 ：The p-value is the probability of observing a test statistic as extreme as the one calculated, assuming the null hypothesis is true. For a right-tailed test, the p-value is calculated as \(P(Z > z)\), where \(Z\) follows a standard normal distribution and \(z\) is the calculated test statistic.

Step 5 ：Substituting the given values into the formula, we get the test statistic as approximately 2.31 and the p-value as approximately 0.01.

Step 6 ：Since the p-value is less than the significance level of 0.10, we reject the null hypothesis. This suggests that the proportion of people who own cats is significantly larger than 60%.

Step 7 ：The final answer is: The test statistic is approximately \(\boxed{2.31}\) and the p-value is approximately \(\boxed{0.01}\).