- Question 12

In a recent poll, 560 people were asked if they liked dogs, and $82 \%$ said they did. Find the margin of error of this poll, at the $90 \%$ confidence level.

Give your answer to three decimals

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Step 1 ：The problem is asking for the margin of error of a poll, at the 90% confidence level. The margin of error for a proportion in a population is given by the formula: \(E = Z \times \sqrt{\frac{p(1-p)}{n}}\), where E is the margin of error, Z is the z-score, p is the proportion of the population giving a certain response, and n is the size of the sample.

Step 2 ：In this case, the size of the sample, n, is 560. The proportion of the population giving a certain response, p, is 0.82. The z-score, Z, for a 90% confidence level is approximately 1.645.

Step 3 ：Substitute these values into the formula to find the margin of error: \(E = 1.645 \times \sqrt{\frac{0.82(1-0.82)}{560}}\).

Step 4 ：Calculate the expression inside the square root first: \(0.82(1-0.82) = 0.1476\). Then divide this by 560: \(\frac{0.1476}{560} = 0.00026357142857142857\).

Step 5 ：Take the square root of this result: \(\sqrt{0.00026357142857142857} = 0.016232751416907282\).

Step 6 ：Finally, multiply this result by the z-score: \(1.645 \times 0.016232751416907282 = 0.026706382664074896\).

Step 7 ：Round this result to three decimal places to get the final answer: \(\boxed{0.027}\).