Answer:
The estimate is  [tex]P__{hat}} \pm E = 0.37 \pm 0.0348[/tex]
Step-by-step explanation:
From the question we are told that Â
  The sample size is  n =  522
  The sample proportion of students  would like to talk about school is  [tex]\r p__{hat}} = 0.37[/tex]
 Given that the confidence level is  90 % then the level of significance can be mathematically evaluated as
         [tex]\alpha = 100 - 90[/tex]
         [tex]\alpha = 10\%[/tex]
         [tex]\alpha = 0.10[/tex]
Next we obtain the critical value of  [tex]\frac{\alpha }{2}[/tex] from the normal distribution table, the value is Â
        [tex]Z_{\frac{\alpha }{2} } =Z_{\frac{0.10}{2} } = 1.645[/tex]
Generally the margin of error can be mathematically represented as
        [tex]E = Z_{\frac{\alpha }{2} } * \sqrt{\frac{\r P_{hat}(1- \r P_{hat} )}{n } }[/tex]
=> Â Â Â Â Â Â [tex]E = 1.645 * \sqrt{\frac{0.37 (1- 0.37 )}{522 } }[/tex]
=> Â Â Â Â Â Â [tex]E = 0.0348[/tex]
Generally the estimate the proportion of all teenagers who want more family discussions about school at 90% confidence level is Â
            [tex]P__{hat}} \pm E[/tex]
substituting values
           [tex]0.37 \pm 0.0348[/tex]
