Question

An Article a Florida newspaper reported on the topics that teenagers most want to discuss with their parents. The findings, the results of a poll, showed that 46% would like more discussion about the family’s financial situation, 37% would like to talk about school, and 30%would like to talk about religion. These and other sampling were based on 522 teenagers. Estimate the proportion of all teenagers who want more family discussions about school. Use a 90% confidence level. Express the answer in the form P hat+- E

Answer 1

Answer:

The estimate is  $P__{hat}} \pm E = 0.37 \pm 0.0348$

Step-by-step explanation:

From the question we are told that  

    The sample size is  n =  522

    The sample proportion of students  would like to talk about school is  $\r p__{hat}} = 0.37$

  Given that the confidence level is  90 % then the level of significance can be mathematically evaluated as

                  $\alpha = 100 - 90$

                  $\alpha = 10\%$

                  $\alpha = 0.10$

Next we obtain the critical value of  $\frac{\alpha }{2}$ from the normal distribution table, the value is  

               $Z_{\frac{\alpha }{2} } =Z_{\frac{0.10}{2} } = 1.645$

Generally the margin of error can be mathematically represented as

               $E = Z_{\frac{\alpha }{2} } * \sqrt{\frac{\r P_{hat}(1- \r P_{hat} )}{n } }$

=>            $E = 1.645 * \sqrt{\frac{0.37 (1- 0.37 )}{522 } }$

=>             $E = 0.0348$

Generally the estimate the proportion of all teenagers who want more family discussions about school at 90% confidence level is  

                       $P__{hat}} \pm E$

substituting values

                     $0.37 \pm 0.0348$