Question

In a recent poll, 440 people were asked if they liked dogs, and 75 % said they did. Find the margin of error of this poll, at the 99 % confidence level. Give your answer to three decimal

Answer 1

The margin of error at the 99% confidence level for the poll where 75% of 440 people said they liked dogs is approximately 0.053 or 5.3%, which can be calculated using the margin of error formula incorporating the z-score, sample proportion, and sample size.

Step-by-step explanation:

To calculate the margin of error for a poll at a 99% confidence level when 75% of 440 people said they liked dogs, we use the formula for margin of error:

M.E. = Z * sqrt[(p*(1-p))/n]

Where 'Z' is the z-score corresponding to the confidence level, 'p' is the sample proportion, and 'n' is the sample size.

• p = 0.75, since 75% said they liked dogs.
• n = 440, the number of people polled.
• Z for 99% confidence is approximately 2.576 (from z-score tables).

Plugging the values in, the margin of error calculation is:

M.E. = 2.576 * sqrt[(0.75*(1-0.75))/440]

M.E. = 2.576 * sqrt[(0.1875)/440]

M.E. = 2.576 * sqrt[0.000426136]

M.E. = 2.576 * 0.0206422

M.E. = 0.0531667

So, the margin of error at the 99% confidence level is approximately 0.053 or 5.3%.

M.E. is a measurement of the potential discrepancy between the poll results and the actual opinion of the total population. A lower margin of error is preferable, as it indicates higher precision.