We observe a sample proprotion p-hat = 0.3. If the standard deviation of the sampling distribution of p-hat is 0.04, what is the 95% confidence interval for p?

CI=point estimate +- (z)(standard deviation)

CI= 0.3 +- (1.96)(0.04)
CI= 0.3 +- 0.0784
CI= 0.3 - 0.0784
CI= 0.3 + 0.0784

CI= (0.2216, 0.3784)

To calculate the 95% confidence interval for p, we can use the formula:

p-hat ± Z * sqrt(p-hat * (1 - p-hat) / n)

where p-hat is the sample proportion, Z is the Z-score corresponding to the desired confidence level, sqrt represents the square root, and n is the sample size.

In this case, we are given p-hat = 0.3 and the standard deviation of the sampling distribution of p-hat is 0.04. Assuming that we have a large enough sample size, we can use the standard normal distribution and the Z-score associated with a 95% confidence level, which is approximately 1.96.

Thus, the formula becomes:

0.3 ± 1.96 * sqrt(0.3 * (1 - 0.3) / n)

Please note that to proceed further and find the exact confidence interval, we need to know the sample size (n).