I am conducting a study on support of ban on smartphones in class. 68/87 said yes to the ban. How large of a sample is needed to be sure the margin of error +/- 3.5% with 99% confidence. I know this is estimation...but I am not sure how to go about solving this problem.

To determine the sample size needed to achieve a specific margin of error with a certain level of confidence, you can use the following formula:

n = (Z^2 * p * (1-p)) / E^2

Where:
n = Sample size
Z = Z-score corresponding to the desired confidence level (in this case, 99% confidence level)
p = Estimated proportion in the population (in this case, the proportion who support the ban, which is 68/87)
E = Margin of error (in this case, +/- 3.5%)

First, you need to calculate the estimated proportion in the population. The proportion who support the ban is 68/87, which is approximately 0.7816.

Next, you need to find the Z-score corresponding to the 99% confidence level. The Z-score for a 99% confidence level is approximately 2.576.

Now you can plug these values into the formula and solve for the sample size:

n = (2.576^2 * 0.7816 * (1-0.7816)) / (0.035)^2

n ≈ 427.32

Since sample sizes must be whole numbers, you need to round up to the nearest whole number:

n ≈ 428

Therefore, a sample size of at least 428 is needed to have a margin of error of +/- 3.5% with 99% confidence when estimating the support for the ban on smartphones in class.