employee pay forty percent of employers said ...
employee pay forty percent of employers said that their employees are fairly paid, while only 15% of the employees agreed. if 50 employers and 200 employees were surveyed, find the 95% confidence level of the proportions for each group. round your answers to at least three decimal places. part: 0 / 2
Answer
# Explanation:
## Step1: Identify the formula for confidence - interval of a proportion
The formula for the confidence - interval of a proportion is $\hat{p}\pm z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$, where $\hat{p}$ is the sample proportion, $n$ is the sample size, and $z_{\alpha/2}$ is the z - value corresponding to the level of confidence. For a 95% confidence interval, $\alpha=1 - 0.95 = 0.05$, and $z_{\alpha/2}=z_{0.025}=1.96$.
## Step2: Calculate the confidence - interval for employers
For employers, $\hat{p}_1 = 0.4$ and $n_1 = 50$.
First, calculate the margin of error $E_1$:
$E_1=z_{\alpha/2}\sqrt{\frac{\hat{p}_1(1 - \hat{p}_1)}{n_1}}=1.96\sqrt{\frac{0.4\times(1 - 0.4)}{50}}=1.96\sqrt{\frac{0.4\times0.6}{50}}=1.96\sqrt{\frac{0.24}{50}}\approx1.96\times0.0693 = 0.136$.
The confidence - interval is $\hat{p}_1\pm E_1$, so $0.4-0.136 = 0.264$ and $0.4 + 0.136=0.536$.
## Step3: Calculate the confidence - interval for employees
For employees, $\hat{p}_2 = 0.15$ and $n_2 = 200$.
First, calculate the margin of error $E_2$:
$E_2=z_{\alpha/2}\sqrt{\frac{\hat{p}_2(1 - \hat{p}_2)}{n_2}}=1.96\sqrt{\frac{0.15\times(1 - 0.15)}{200}}=1.96\sqrt{\frac{0.15\times0.85}{200}}=1.96\sqrt{\frac{0.1275}{200}}\approx1.96\times0.0252=0.049$.
The confidence - interval is $\hat{p}_2\pm E_2$, so $0.15-0.049 = 0.101$ and $0.15 + 0.049 = 0.199$.
# Answer:
Employers: $(0.264,0.536)$; Employees: $(0.101,0.199)$