The bank has recently approved 200 loans and found that \( 9 \% \) did not make payments on time. Let \( p \) represent the proportion all loans issued by the bank that do not make payments on time. Use the sample to estimate the mean and standard error of the sampling distribution of the sample proportions. Round answers to 4 decimal places as needed. (a) mean \( = \) (b) standard error \( = \)

**Step 1: Define the Sample Proportion** The sample shows that \( 9\% \) of the 200 loans did not make payments on time. Thus, the sample proportion is \[ \hat{p} = 0.09. \] **Step 2: Mean of the Sampling Distribution** The mean of the sampling distribution of the sample proportion is the true proportion, \( p \), which we estimate by \( \hat{p} \). Therefore, the estimated mean is \[ \text{Mean} = \hat{p} = 0.09. \] **Step 3: Standard Error of the Sampling Distribution** The standard error (SE) of the sample proportion is calculated using the formula \[ \text{SE} = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}, \] where \( n = 200 \). Plug in the values: \[ \text{SE} = \sqrt{\frac{0.09 \times 0.91}{200}}. \] Calculate the numerator: \[ 0.09 \times 0.91 = 0.0819. \] Then, divide by \( n \): \[ \frac{0.0819}{200} = 0.0004095. \] Finally, take the square root: \[ \text{SE} = \sqrt{0.0004095} \approx 0.0202. \] **Final Answers:** (a) Mean \( = 0.09 \) (b) Standard Error \( = 0.0202 \)