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 \(\boxed{0.01}\).