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.