Question

Let the following sample of 8 observations be drawn from a normal population with unknown mean and standard deviation:

Answer 1

Answer:

a

   $\= x = 18.5$  ,  $\sigma = 5.15$

b

 $15.505 < \mu < 21.495$

c

 $14.93 < \mu < 22.069$

Step-by-step explanation:

From the question we are are told that

    The  sample data is  21, 14, 13, 24, 17, 22, 25, 12

     The sample size is  n  = 8

Generally the ample mean is evaluated as

        $\= x = \frac{\sum x }{n}$

        $\= x = \frac{ 21 + 14 + 13 + 24 + 17 + 22+ 25 + 12 }{8}$

         $\= x = 18.5$

Generally the standard deviation is mathematically evaluated as

         $\sigma = \sqrt{\frac{\sum (x- \=x )^2}{n}}$

$\sigma = \sqrt{\frac{\sum ((21 - 18.5)^2 + (14-18.5)^2+ (13-18.5)^2+ (24-18.5)^2+ (17-18.5)^2+ (22-18.5)^2+ (25-18.5)^2+ (12 -18.5)^2 )}{8}}$

$\sigma = 5.15$

considering part b

Given that the confidence level is  90% then the significance level is 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  

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

The margin of error is mathematically represented as

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

=>    $E =1.645 * \frac{5.15 }{\sqrt{8} }$

=>     $E = 2.995$

The 90% confidence interval is evaluated as

       $\= x - E < \mu < \= x + E$

substituting values

       $18.5 - 2.995 < \mu < 18.5 + 2.995$

       $15.505 < \mu < 21.495$

considering part c

Given that the confidence level is  95% then the significance level is 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

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

=>    $E =1.96 * \frac{5.15 }{\sqrt{8} }$

=>     $E = 3.569$

The 95% confidence interval is evaluated as

       $\= x - E < \mu < \= x + E$

substituting values

       $18.5 - 3.569 < \mu < 18.5 + 3.569$

       $14.93 < \mu < 22.069$