Question

A random sample of 1003 adult Americans was asked, "Do you think televisions are a necessity or a luxury you could do without?" Of the 1003 adults surveyed, 521 indicated that televisions are a luxury they could do without. Construct and interpret a 95% confidence interval for the population proportion of adult Americans who believe that televisions are a luxury they could do without out.

Answer 1

Answer:

The  95% confidence interval is  $0.503 < p < 0.535$

The  interpretation is that there is 95% confidence that the true population proportion lie within the confidence interval

Step-by-step explanation:

From the question we are told that

    The  sample size is n  =  1003

     The number that indicated television are a luxury is  k  =  521

Generally the sample mean is mathematically represented as

           $\r p = \frac{k}{n}$

          $\r p = \frac{521}{1003}$

         $\r p = 0.519$

Given the confidence level is  95% then the level of significance is mathematically evaluated as

       $\alpha = 100 - 95$

       $\alpha = 5\%$

       $\alpha = 0.05$

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

          $Z_{\frac{\alpha }{2} } = 1.96$

The  margin of error is mathematically represented as

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

=>       $E = 1.96 * \sqrt{ \frac{ 0.519 (1- 0.519 )}{1003} }$

=>       $E = 0.016$

The  95%  confidence interval is mathematically represented as

       $\r p -E < p < \r p +E$

=>   $0.519 - 0.016 < p < 0.519 + 0.016$

=>    $0.503 < p < 0.535$