Final answer:
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.