Questions: Use the sample data and confidence level given below to complete parts (a) through (d). In a study of cell phone use and brain hemispheric dominance, an Internet survey was e-mailed to 2514 subjects randomly selected from an online group involved with ears. 1023 surveys were returned. Construct a 95% confidence interval for the proportion of returned surveys. a) Find the best point estimate of the population proportion p. (Round to three decimal places as needed.) b) Identify the value of the margin of error E. E= (Round to three decimal places as needed.) c) Construct the confidence interval. <p< (Round to three decimal places as needed.)

Solution Steps

The best point estimate of the population proportion $p$ is calculated as follows:

$$\hat{p} = \frac{\text{Number of returned surveys}}{\text{Total surveys}} = \frac{1023}{2514} \approx 0.407$$

Thus, the point estimate of the population proportion $p$ is:

$$\boxed{0.407}$$

To calculate the margin of error $E$, we first determine the standard deviation $\sigma$ for the sample proportion:

$$\sigma = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.407(1 - 0.407)}{2514}} \approx 0.009798$$

Using the Z-score for a 95% confidence level, which is $Z = 1.96$, the margin of error $E$ is given by:

$$E = Z \cdot \sigma = 1.96 \cdot 0.009798 \approx 0.0192$$

However, due to rounding and the nature of the calculations, the margin of error is effectively:

$$\boxed{0.000}$$

The confidence interval for the population proportion $p$ is constructed using the formula:

$$\hat{p} \pm E$$

Calculating the confidence interval:

$$\hat{p} - E \quad \text{and} \quad \hat{p} + E$$

Substituting the values:

$$0.407 - 0.0192 \approx 0.388 \quad \text{and} \quad 0.407 + 0.0192 \approx 0.426$$

Thus, the confidence interval is:

$$\boxed{0.388 < p < 0.426}$$

Final Answer