A sample of size 125 was gathered from high school seniors to estimate how many seniors at the school intended to attend an in- state university. Of the 125 seniors, 80 responded "Yes" they intend to attend an in-state university. Use the sample to estimate the mean and standard error of the sampling distribution of the sample proportions. Round answers to 4 decimal places. (a) mean \( = \) (b) standard error \( = \)

To estimate the mean and standard error of the sampling distribution of the sample proportions, we can use the following formulas: 1. **Mean of the sample proportions** (\( \hat{p} \)): \[ \hat{p} = \frac{x}{n} = \frac{80}{125} = 0.64 \] 2. **Standard Error (SE)** of the sample proportions: \[ SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.64(1 - 0.64)}{125}} = \sqrt{\frac{0.64 \times 0.36}{125}} \approx \sqrt{\frac{0.2304}{125}} \approx \sqrt{0.0018432} \approx 0.0429 \] Now rounding the results to four decimal places: (a) mean \( = 0.6400 \) (b) standard error \( = 0.0429 \)