A recent poll of concert attendees that would pay 200.00 for a "Rain,A Tribute to the Beatles" concert ticket has a margin of error of +3.5%.How many attendees were polled?
To determine the number of attendees who were polled, we need the margin of error and the level of confidence. However, the level of confidence is not provided in the question. In survey research, it is common to use a confidence level of 95%. Assuming a 95% confidence level, we can calculate the sample size using the formula:
n = (Z^2 * p * (1-p)) / (E^2)
Where:
n = sample size
Z = Z-score (corresponding to the confidence level)
p = estimated proportion of the population
E = margin of error
In this case, the margin of error is given as 3.5%, which can be written as 0.035.
The Z-score for a 95% confidence level is 1.96.
Since we don't know the estimated proportion of the population, we will assume a worst-case scenario where p is 0.5 (maximizing the sample size).
Let's calculate the sample size:
n = (1.96^2 * 0.5 * (1-0.5)) / (0.035^2)
n = 384.16
The calculated sample size is approximately 384.16. However, since you cannot have a fraction of a person, we need to round it up to the nearest whole number. Therefore, the estimated number of attendees polled would be 385.