A researcher found a sample proportion of 0.94 for estimating the population proportion of freshmen at a university who had Facebook accounts. If the standard deviation of the sampling distribution of the sampling proportion is 0.02, what is a 95% confidence interval for the population proportion?

Formula with your data:

CI95 = 0.94 ± 1.96(0.02)

Note: ± 1.96 represents 95% confidence interval using a z-table.

I let you take it from here to calculate the interval.

To find the 95% confidence interval for the population proportion, we can use the formula:

Confidence interval = sample proportion ± (Z * standard deviation)

Where:
- The sample proportion is 0.94
- The standard deviation of the sampling distribution of the sampling proportion is 0.02
- Z is the critical value for a 95% confidence level

To find the critical value (Z), we need to refer to the standard normal distribution table or use statistical software. For a 95% confidence level, the critical value is approximately 1.96.

Now we can plug in the values into the formula:

Confidence interval = 0.94 ± (1.96 * 0.02)

Calculating this, we get:

Confidence interval = 0.94 ± 0.0392

Therefore, the 95% confidence interval for the population proportion is (0.9008, 0.9792). This means we are 95% confident that the true population proportion of freshmen at the university who have Facebook accounts falls within this range.