A newspaper reports that the governor's approval rating stands at 60%. The article adds that the poll is based on a random sample of 1,476 adults and has a margin of error of 2.5%.... A newspaper reports that the governor's approval rating stands at 60%. The article adds that the poll is based on a random sample of 1,476 adults and has a margin of error of 2.5%. What confidence level did the pollsters use? Read more

Steps to Solve

1. Identify the given information

We are given:

• Sample size $n = 1476$
• Margin of Error $E = 2.5% = 0.025$
• Sample proportion $\hat{p} = 60% = 0.60$

2. Formula for Margin of Error

The margin of error for a proportion is given by the formula: $ E = z \cdot \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} $ where $E$ is the margin of error $z$ is the z-score corresponding to the desired confidence level $\hat{p}$ is the sample proportion $n$ is the sample size

3. Solve for the z-score

We need to find the z-score first. Plug in the given values into the margin of error formula:

$ 0.025 = z \cdot \sqrt{\frac{0.60(1 - 0.60)}{1476}} $

$ 0.025 = z \cdot \sqrt{\frac{0.60(0.40)}{1476}} $

$ 0.025 = z \cdot \sqrt{\frac{0.24}{1476}} $

$ 0.025 = z \cdot \sqrt{0.0001626} $

$ 0.025 = z \cdot 0.01275 $

$ z = \frac{0.025}{0.01275} $

$ z \approx 1.96 $

4. Find the confidence level corresponding to the z-score

A z-score of 1.96 corresponds to a confidence level of 95%. We can find this using a z-table or calculator. The area between $-1.96$ and $1.96$ is approximately 0.95.