Question

g A random sample of size 16 taken from a normally distributed population revealed a sample mean of 50 and a sample variance of 36. The upper limit of a 95% confidence interval for the population mean would equal:

Answer 1

Answer:

The  upper limit is    

                   $k = 52.94$

Step-by-step explanation:

From the question we  told that

     The  sample size is $n = 16$

      The sample mean is  $\= x = 50$

      The sample variance is  $\sigma ^2 = 36$

For  a  95% confidence interval the confidence level is  95%

Given that 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(reference- math dot armstrong dot edu), the value is  

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

             

Generally the margin of error is mathematically represented as

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

 Here  $\sigma$ is the standard deviation which is mathematically evaluated as

                  $\sigma = \sqrt{\sigma^2}$

substituting values

                  $\sigma = \sqrt{36}$

=>                $\sigma = 6$

So

                    $E = 1.96 * \frac{6}{\sqrt{16} }$

                     $E = 2.94$

The 95% confidence interval is mathematically represented as

                 $\= x - E < \mu < \= x + E$

substituting values

                $50 -2.94 < \mu$

                $47.06 < \mu$

The  upper limit is    

                   $k = 52.94$