Assume that the population proportion is 0.46 . Compute the standard error of the proportion, \( \sigma-\overline{p^{\prime}} \) for sample sizes of 500,\( 000 ; 1,000,000 ; 5,000,000 ; 10,000,000 ; \) and \( 100,000,000 \). (Round your answers to five decimal places.) sample size of 500,000 sample size of \( 1,000,000 \) sample size of \( 5,000,000 \) sample size of \( 10,000,000 \) sample size of \( 100,000,000 \) What can you say about the size of the standard error of the sample proportion as the sample size is increased? The standard error of the sample proportion, \( \sigma_{\overline{p^{\prime}}}- \) Select \( -\hat{\sim} \) and becomes extremely --Select- \( \hat{v} \) as the sample size becomes huge.

The standard error of the sample proportion can be calculated using the formula: \[ \sigma_{\overline{p^{\prime}}} = \sqrt{\frac{p(1-p)}{n}} \] where \( p \) is the population proportion (0.46) and \( n \) is the sample size. Here are the calculations: 1. For \( n = 500,000 \): \[ \sigma_{\overline{p^{\prime}}} = \sqrt{\frac{0.46 \times (1-0.46)}{500000}} \approx 0.000998 \] 2. For \( n = 1,000,000 \): \[ \sigma_{\overline{p^{\prime}}} = \sqrt{\frac{0.46 \times (1-0.46)}{1000000}} \approx 0.000707 \] 3. For \( n = 5,000,000 \): \[ \sigma_{\overline{p^{\prime}}} = \sqrt{\frac{0.46 \times (1-0.46)}{5000000}} \approx 0.000316 \] 4. For \( n = 10,000,000 \): \[ \sigma_{\overline{p^{\prime}}} = \sqrt{\frac{0.46 \times (1-0.46)}{10000000}} \approx 0.000223 \] 5. For \( n = 100,000,000 \): \[ \sigma_{\overline{p^{\prime}}} = \sqrt{\frac{0.46 \times (1-0.46)}{100000000}} \approx 0.000071 \] As you can observe, the standard error of the sample proportion decreases as the sample size increases. The standard error becomes extremely small as the sample size becomes huge, indicating that larger samples provide more reliable estimates of the population proportion.