In 2001, a Gallup poll surveyed 1016 households in the U.S. about their pets. Of those surveyed, 590 said they had at least one dog or cat as a pet. Construct and interpret a $98 \%$ confidence interval for the proportion of households who had at least one dog or cat. In your work, give the Margin of Error, the confidence interval and the interpretation of the interval. Edit View Insert Format Tools Table

Step 1 ：First, calculate the sample proportion (p̂), which is the number of households with at least one dog or cat divided by the total number of households surveyed. In this case, p̂ = \(\frac{590}{1016} = 0.581\).

Step 2 ：Next, calculate the standard error of the proportion. The formula for this is \(\sqrt{\frac{p̂ * (1 - p̂)}{n}}\), where n is the total number of households surveyed. Substituting the given values, we get \(\sqrt{\frac{0.581 * (1 - 0.581)}{1016}} = 0.0155\).

Step 3 ：Then, find the z-score for a 98% confidence interval. From the z-table or using a calculator, this value is approximately 2.33.

Step 4 ：Calculate the margin of error (ME) by multiplying the z-score by the standard error. This gives us \(2.33 * 0.0155 = 0.0361\).

Step 5 ：Finally, calculate the confidence interval by subtracting the margin of error from the sample proportion for the lower limit and adding the margin of error to the sample proportion for the upper limit. This gives us \(0.581 - 0.0361 = 0.545\) for the lower limit and \(0.581 + 0.0361 = 0.617\) for the upper limit.

Step 6 ：\(\boxed{\text{Final Answer: The margin of error is approximately 0.036. The 98% confidence interval for the proportion of households who had at least one dog or cat is approximately (0.545, 0.617). This means that we are 98% confident that the true proportion of households in the U.S. who have at least one dog or cat falls within this interval.}}\)