Step 1 :Define the null hypothesis as the proportion of people that own cats is 20% (\(p = 0.20\)), and the alternative hypothesis as the proportion is less than 20%.
Step 2 :Given a sample size of 200 people and a significance level of 0.025.
Step 3 :Check the condition \(h p^{\wedge}\left(1-\vec{\vartheta}^{\wedge}\right) \geq 10\), which is a rule of thumb for when the normal approximation to the binomial distribution is appropriate. Here, \(h\) is the sample size, \(p^{\wedge}\) is the sample proportion, and \(\vec{\vartheta}^{\wedge}\) is the population proportion under the null hypothesis.
Step 4 :Calculate the test statistic and the critical value, and then compare them to make a decision about the null hypothesis.
Step 5 :Given that the test statistic is 0, which is greater than the critical value of -1.96, we do not reject the null hypothesis that the proportion of people that own cats is 20%.
Step 6 :The condition for the normal approximation is also satisfied, as \(200 * 0.2 * (1 - 0.2) = 32\), which is greater than 10.
Step 7 :Therefore, the veterinarian's claim that the proportion is less than 20% is not supported by this test at the 0.025 significance level.
Step 8 :Final Answer: \(\boxed{\text{Do not reject } H_0}\). The veterinarian's claim that the proportion is less than 20% is not supported by this test at the 0.025 significance level.