Question

1. a political candidate has asked you to conduct a poll to determine what percentage of people support her. if the candidate wants a 6% margin of error at a 90% confidence level, what sample size is needed? (use technology to find the z - score and do not round between steps. make sure you use the correct rounding rule for samples size, and give your answer in whole people.)

Explanation:

Step1: Find the z - score

For a 90% confidence level, using technology (e.g., a standard normal distribution table or calculator), the z - score \(z = 1.645\).

Step2: Use the formula for sample size in proportion

The formula for sample size \(n\) when estimating a proportion is \(n=\frac{z^{2}\cdot p(1 - p)}{E^{2}}\). When no prior estimate of the proportion \(p\) is given, we use \(p=0.5\) (this maximizes the value of \(p(1 - p)\)). The margin of error \(E = 0.06\).
Substitute the values into the formula:

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Step3: Round the sample size

Since we cannot have a fraction of a person and we use the rounding rule for sample size (we round up to the next whole number), we round \(187.9184\) up.

Answer:

\(188\)