Question

From a random sample of size 18, a researcher states that (11.1, 15.7) inches is a 90% confidence interval for mu, the mean length of bass caught in a small lake. A normal distribution was assumed. Using the 90% confidence interval obtain:
a. A point estimate of μ and its 90% margin of error.
b. A 95% confidence interval for μ.

Answer 1

Complete Question

From a random sample of size 18, a researcher states that (11.1, 15.7) inches is a 90% confidence interval for mu, the mean length of bass caught in a small lake. A normal distribution was assumed. Using the 90% confidence interval obtain:

a. A point estimate of $\mu$ and its 90% margin of error.

b. A 95% confidence interval for $\mu$.

Answer:

a

$\= x = 13.4$   .   $E = 2.3$

b

$10.7 < \mu < 16.1$

Step-by-step explanation:

From the question we are told that

  The sample size is  n = 18

  The 90% confidence interval is  (11.1, 15.7)

Generally the point estimate of  $\mu$ is mathematically  evaluated  as

       $\= x = \frac{11.1 + 15.7 }{2}$

=>    $\= x = 13.4$

Generally the margin of error is mathematically evaluated  as

     $E = \frac{15.7 - 11.1}{2 }$

=> $E = 2.3$

  From the question we are told the confidence level is  90% , hence the level of significance is    

      $\alpha = (100 - 90 ) \%$

=>   $\alpha = 0.10$

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

   $Z_{\frac{\alpha }{2} } = 1.645$

Generally the equation for the lower limit of the confidence interval is  

      $\= x - Z_{\frac{\alpha }{2} } * \frac{s}{\sqrt{18} } = 11.1$

=> $13.4 - 0.3877 s = 11.1$

=>  $s = 5.932$

  From the question we are told the confidence level is  95% , hence the level of significance is    

      $\alpha = (100 - 95 ) \%$

=>   $\alpha = 0.05$

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

   $Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as  

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

=>    $E = 1.96 * \frac{5.932}{\sqrt{18} }$

=>    $E = 2.7$      

Generally 95% confidence interval is mathematically represented as  

      $\= x -E < \mu < \=x +E$

=>    $13.4 - 2.7 < \mu < 13.4 + 2.7$

=>    $10.7 < \mu < 16.1$