Question

A local service agency hires you to estimate the average annual income of all 700 families living in a four square block section. Since time and funds are limited, yu take a random sample of 50 families and want the estimate to have a 92% confidence level. The 50 families had an average income of $35,300. Research shows the population standard deviation for annual income in this four square block section is $1,800

Answer 1

Answer:

The true estimate of  the average annual income of all 700 families living in a four square block section lies within the 92% confidence interval below

   $34854.3  <  \mu < 35 745.7$

Step-by-step explanation:

From the question we are told that

    The  sample size is  n =  50  

    The sample mean is $\= x = \ 35300$

     The population standard deviation is  $\sigma = \ 1800$

From the question we are told the confidence level is  92% , hence the level of significance is    

      $\alpha = (100 - 92 ) \%$

=>   $\alpha = 0.08$

Generally from the normal distribution table the critical value  of  $\frac{\alpha }{2}$ is  

   $Z_{\frac{\alpha }{2} } =  1.751$

Generally the margin of error is mathematically represented as  

      $E = Z_{\frac{\alpha }{2} } *  \frac{\sigma }{\sqrt{n} }$

=>   $E = 1.751  *  \frac{1800 }{\sqrt{50} }$

=>   $E =  445.7$

Generally 92% confidence interval is mathematically represented as  

      $\= x -E <  \mu <  \=x  +E$

=> $35 300 -445.7 <  \mu < 35 300 + 445.7$

=> $34854.3  <  \mu < 35 745.7$

So the true estimate of  the average annual income of all 700 families living in a four square block section lies within the 92% confidence interval