Question

A research center project involved a survey of 851 Internet users. It provided a variety of statistics on Internet users. (a) The sample survey showed that 92% of respondents said the Internet has been a good thing for them personally. Develop a 95% confidence interval for the proportion of respondents who say the Internet has been a good thing for them personally. (Round your answers to four decimal places.)

Answer 1

Answer:

The answer is "(0.9193924 , 0.9206076)".

Step-by-step explanation:

$\text{sample proportion}\ (SP) = 0.92\\\\\text{sample size}\ n = 851\\\\\text{Standard error} \ SE = \sqrt{\frac{(SP \times(1 - SP)}{n})}$

                              $= \sqrt{\frac{(0.92 \times (0.08)}{851})}\\\\= \sqrt{\frac{0.0736}{851}}\\\\= \sqrt{8.648\times 10^{-5}}\\\\=0.00031$

$\text{CI level is}\ 95\% \\\\\therefore\\\\ \alpha = 1 - 0.95 = 0.05\\\\\frac{\alpha}{2} = \frac{0.05}{2} = 0.025\\\\ Z_c = Z_{(\frac{\alpha}{2})} = 1.96$

Calculating the Margin of Error:

$ME = z_{c} \times SE$

       $= 1.96 \times 0.00031\\\\ = 0.0006076$

$CI = (SP - z*SE, SP + z*SE)$

      $= (0.92 - 1.96 * 0.00031 , 0.92 + 1.96 * 0.00031)\\\\ = (0.92 - 0.0006076 , 0.92 + 0.0006076)\\\\= (0.9193924 , 0.9206076)$