Consider the following problem that tests a hypothesis: A random sample of 600 teens with Facebook profiles found that 395 included photos of themselves. Let \( p \) be the proportion of teen Facebook users who included photos of themselves a profile pictures. At significance level \( \alpha=0.007 \), is the sample showing strong evidence that \( p \) is other than 0.8 ? Accurate to 4 decimal places, which of the following is \( \sigma_{\hat{p}} \) used for this testing hypothesis problem? 0.049 0.0427 0.0065 0.0163 0.0194

To test the hypothesis that the proportion of teen Facebook users who included photos of themselves as profile pictures is other than 0.8, we can use a one-sample z-test. Given: - Sample size (\( n \)) = 600 - Number of teens with photos of themselves (\( x \)) = 395 - Significance level (\( \alpha \)) = 0.007 - Hypothesized proportion (\( p \)) = 0.8 First, we need to calculate the standard error of the proportion (\( \sigma_{\hat{p}} \)) using the formula: \[ \sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} \] Substitute the given values into the formula to find \( \sigma_{\hat{p}} \). Calculate the value by following steps: - step0: Calculate: \(\sqrt{\frac{0.8\left(1-0.8\right)}{600}}\) - step1: Subtract the numbers: \(\sqrt{\frac{0.8\times 0.2}{600}}\) - step2: Multiply the numbers: \(\sqrt{\frac{0.16}{600}}\) - step3: Divide the terms: \(\sqrt{\frac{1}{3750}}\) - step4: Use the properties of radicals: \(\frac{\sqrt{1}}{\sqrt{3750}}\) - step5: Simplify the expression: \(\frac{1}{25\sqrt{6}}\) - step6: Multiply by the Conjugate: \(\frac{\sqrt{6}}{25\sqrt{6}\times \sqrt{6}}\) - step7: Multiply the numbers: \(\frac{\sqrt{6}}{150}\) The standard error of the proportion (\( \sigma_{\hat{p}} \)) is approximately 0.0163. Therefore, the correct answer is 0.0163.