The confidence level is approximately 44.38%.
We know from the question that the sample size is 254, the proportion of bartenders who heard complaints is 0.48, and the margin of error is 0.034.
We can calculate the standard error by using the following formula:
se = sqrt(p * (1 - p) / n)
where:
se is the standard error
p is the proportion of bartenders who heard complaints
n is the sample size
Plugging in the values from the question, we get:
se = sqrt(0.48 * (1 - 0.48) / 254) = 0.028
We can then calculate the z-score by dividing the margin of error by the standard error:
z = e / se = 0.034 / 0.028 = 1.21
Finally, we can look up the confidence level in a z-table. A z-table shows the probability that a standard normal variable will be less than a certain value (z).
In this case, we want to find the probability that a standard normal variable will be less than 1.21. Using a z-table, we find that this probability is approximately 0.8830.
Since the confidence level is the probability that the true population proportion falls within the margin of error of the sample proportion, we can multiply the probability we just found by 2 to get the confidence level:
confidence level = 2 * probability (z < 1.21) = 2 * 0.8830 = 1.7660
As a percentage, the confidence level is approximately 176.60%, which is greater than 100%.
However, confidence levels cannot be greater than 100%, so we know that there must be an error in our calculation.
The error is that we used the wrong formula to calculate the confidence level. The correct formula for a margin of error is:
e = z * se
where: e is the margin of error ,z is the z-score ,se is the standard error
Solving for z, we get:
z = e / se = 0.034 / 0.028 = 1.21
We can now look up the confidence level in a z-table using a z-score of 1.21.
We find that the probability that a standard normal variable will be less than 1.21 is approximately 0.8830.
Multiplying this probability by 2 to get the confidence level, we get:
confidence level = 2 * probability (z < 1.21) = 2 * 0.8830 = 1.7660
As a percentage, the confidence level is approximately 176.60%.
However, confidence levels cannot be greater than 100%, so we know that there must be an error in our calculation.
The error is that we used the wrong formula to calculate the confidence level. The correct formula for a confidence level is:
confidence level = 100 * (1 - 2 * (1 - erf(z / sqrt(2))))
where: confidence level is the confidence level as a percentage ,erf is the error function ,z is the z-score
Plugging in the values from the question, we get:
confidence level = 100 * (1 - 2 * (1 - erf(1.21 / sqrt(2)))) = 88.76%
Therefore, the confidence level used is approximately 88.76%.