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 \( \alpha=0.008 \), is the sample showing strong evidence that \( p \) is other than 0.65 ? Use 4 decimal places for \( \sigma_{\hat{p}} \). YES NO Note: Round any \( z \)-score to 2 decimal places before any calculation.

To determine if there's strong evidence that the proportion \( p \) is different from 0.65, we can conduct a hypothesis test. The null hypothesis is \( H_0: p = 0.65 \) and the alternative hypothesis is \( H_a: p \neq 0.65 \). First, we need to calculate the sample proportion \( \hat{p} \): \[ \hat{p} = \frac{395}{600} = 0.6583 \] Next, we calculate the standard deviation of the sampling distribution \( \sigma_{\hat{p}} \) using the null hypothesis proportion: \[ \sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.65(1-0.65)}{600}} = \sqrt{\frac{0.65 \times 0.35}{600}} \approx 0.0207 \] Now we calculate the z-score: \[ z = \frac{\hat{p} - p}{\sigma_{\hat{p}}} = \frac{0.6583 - 0.65}{0.0207} \approx 0.0153 \] Since we do a two-tailed test, we compare \( |z| \) to the critical value from the z-table at \( \alpha/2 = 0.004 \). The critical z-value for \( 0.004 \) is approximately \( \pm2.88 \). Since \( |0.0153| < 2.88 \), we fail to reject the null hypothesis and there is no strong evidence that \( p \) is different from 0.65. Thus, the answer is NO.