Question

A randomly selected sample of college basketball players has the following heights in inches. See Attached Excel for Data. Compute a 95% confidence interval for the population mean height of college basketball players based on this sample and fill in the blanks appropriately. < μ < (round to 3 decimal places)

Answer 1

Complete Question

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Answer:

The  confidence interval is  $64.86$

Step-by-step explanation:

From the question we are given the following data

   The following heights are

66, 65, 67, 62, 62, 65, 61, 70, 66, 66, 71, 63, 69, 65, 71, 66, 66, 69, 68, 62, 65, 67, 65, 71, 65, 70, 62, 62, 63, 64, 67, 67      

 The  sample size is n  =32

  The confidence level is $k = 95$% = 0.95

The mean is evaluated as

          $\= x = 66+ 65+ 67+ 62+ 62+ 65+ 61+ 70+ 66+ 66+ 71+63+ 69+ 65+ 71+ 66+ 66+ 69+ 68+ 62+ 65+ 67,+\\65+ 71+ 65+ 70+ 62+ 62+ 63+ 64+ 67+ 67 / 32$

=>   $\= x = \frac{2108}{32}$

=>     $\= x = 65.875$

The standard deviation is evaluated as

           $\sigma = \sqrt{ v}$

Now  

   $v = ( 66-65.875 )^2+(65-65.875)^2+( 67-65.875)^2+ (62-65.875)^2+ (62-65.875)^2+ (65-65.875)^2+( 61-65.875)^2+ (70-65.875)^2+ (66-65.875)^2+ (66-65.875)^2+ (71+63-65.875)^2+ (69-65.875)^2+ (65-65.875)^2+ (71-65.875)^2+( 66-65.875)^2+ (66-65.875)^2+ (69-65.875)^2+ (68-65.875)^2+ (62-65.875)^2+ (65-65.875)^2+ (67-65.875)^2,+\\(65-65.875)^2+ (71-65.875)^2+ (65-65.875)^2+ (70-65.875)^2+( 62-65.875)^2+( 62-65.875)^2+ (63-65.875)^2+ (64-65.875)^2+ (67-65.875)^2+ (67-65.875)^2 / 32$

=>$v= 8.567329$

=>   $\sigma = \sqrt{8.567329}$

=>   $\sigma = 2.927$

The level of significance is evaluated as

        $\alpha = 1 - 0.95$

        $\alpha = 0.05$

The degree of freedom is  evaluated as

    $Df = n- 1 \equiv Df = 32 -1 = 31$

The critical values for the level of significance is obtained from the z -table as

      $t_c = t_{\alpha/2 } , Df = t _{0.05/2}, 31 =\pm 1.96$

The confidence interval is evaluated as

       $\mu = \= x \pm t_c * \frac{\sigma }{\sqrt{n} }$

substituting values

        $\mu =65.875 \pm 1.96* \frac{2.927}{\sqrt{32} }$

       $\mu =65.875 \pm 1.01415$

=>    $64.86$