Angela is conducting a poll on campus to determine the next student representative. How many students does she need to sample for a confidence level of 90% with a margin of error of + or - 5%?

To determine how many students Angela needs to sample for her poll, we need to consider the confidence level and margin of error.

1. Confidence Level: The confidence level represents the level of confidence we have in the accuracy of the poll's results. It is typically expressed as a percentage, with 90% being a commonly used value. This means that there is a 90% chance that the results obtained from the sample accurately reflect the opinions of the whole population.

2. Margin of Error: The margin of error indicates the maximum amount by which the results obtained from the sample may deviate from the real population values. In this case, it is ±5%, meaning that the results from the sample could potentially be 5% higher or lower than the actual population values.

To calculate the required sample size given the confidence level and margin of error, we use the following formula:

n = (Z² * p * q) / E²

Where:
n = required sample size
Z = Z-score corresponding to the desired confidence level
p = estimated proportion of the population (0.5 for maximum sample size)
q = 1 - p
E = margin of error (expressed as a decimal)

First, let's find the Z-score for a confidence level of 90%. You can use a Z-table or use a statistical calculator for this calculation. The Z-score corresponding to a 90% confidence level is approximately 1.645.

Next, we substitute the values into the formula:

n = (1.645² * 0.5 * 0.5) / (0.05)²

Simplifying the equation:

n = (2.706 * 0.25) / 0.0025
n = 6.7656 / 0.0025
n ≈ 2706.56

Since we can't have a fractional sample size, we round up to the nearest whole number. Therefore, Angela needs to sample approximately 2707 students from the campus to achieve a confidence level of 90% with a margin of error of ±5%.