Step 1 :Test the claim that the proportion of people who own cats is larger than 10% at the 0.05 significance level. The null hypothesis is that the proportion is equal to or less than 10%, and the alternative hypothesis is that the proportion is greater than 10%. The test is right-tailed.
Step 2 :We are given a sample of 400 people, and 15% of them own cats. We can use this information to calculate the test statistic and the p-value.
Step 3 :The test statistic is calculated as the difference between the sample proportion and the hypothesized proportion, divided by the standard error of the proportion. The p-value is the probability of observing a test statistic as extreme as, or more extreme than, the observed test statistic, under the null hypothesis.
Step 4 :We can then compare the p-value to the significance level to decide whether to reject or fail to reject the null hypothesis. If the p-value is less than the significance level, we reject the null hypothesis. If the p-value is greater than the significance level, we fail to reject the null hypothesis.
Step 5 :The test statistic is approximately \(3.33\) and the p-value is approximately \(0.00043\).
Step 6 :Since the p-value is less than the significance level of 0.05, we reject the null hypothesis. This means that we have enough evidence to support the claim that the proportion of people who own cats is larger than 10%.
Step 7 :Final Answer: The test statistic is approximately \(\boxed{3.33}\) and the p-value is approximately \(\boxed{0.00043}\). Based on this, we \(\boxed{reject the null hypothesis}\).