Question

A meat inspector has randomly selected 30 packs of 95% lean beef. The sample resulted in a mean of 96.2% with a sample standard deviation of 0.8%. Calculate an upper prediction bound for the leanness of a new pack using a prediction level of 99%. Assume normality. The contents of seven similar containers of sulfuric acid are 9.8, 10.2, 10.4, 9.8,10.0, 10.2, and 9.6 liters. Find a 95% confidence interval for the mean contents of all such containers, assuming an approximately normal distribution.

Answer 1

Answer:

a

 The upper bound of the 99% prediction level is $98.2$  

b

 The  95% confidence interval is $9.7383 < \mu < 10.2617$

Step-by-step explanation:

Considering first question

From the question we are told that

   The sample size is  n  =  30  

   The sample mean is  $\= x = 96.2\%$

   The standard deviation is $s = 0.8\%$

Generally the degree of freedom is mathematically represented as

        $df = n - 1$

=>      $df = 30 - 1$

=>      $df = 29$

From the question we are told the confidence level is  99% , hence the level of significance is    

      $\alpha = (100 - 99 ) \%$

=>   $\alpha = 0.01$

Generally from the t distribution table the critical value  of   at a degree of freedom of is  

   $t_{\alpha , 29} = 2.462$

Generally the  99%  prediction level is mathematically represented as

      $\= x \pm [(t_{\alpha , df }) * s * (\sqrt{1 + \frac{1}{ n} } )}]$

Generally the upper bound of the 99%  prediction level is mathematically represented as

      $\= x + [(t_{\alpha , df }) * s * (\sqrt{1 + \frac{1}{ n} } )}]$  

=>    $96.2 + (2.462 ) * 0.8 * (\sqrt{1 + \frac{1}{ 30} } )}]$  

=>    $98.2$  

Considering second question

 Generally the sample is mathematically represented as

             $\= x = \frac{\sum x_i}{n}$

=>           $\= x = \frac{ 9.8 + 10.2 + \cdots +9.6 }{7}$  

=>           $\= x = 10$    

Generally the standard deviation  is mathematically represented as

           $\sigma = \sqrt{ \frac{ \sum ( x_ i - \= x)}{n-1} }$

=>        $\sigma = \sqrt{ \frac{ ( 9.8 -10)^2 + ( 10.2 -10)^2 + \cdots + ( 9.6 -10)^2 }{7-1} }$

=>        $\sigma = 0.283$

Generally the degree of freedom is mathematically represented as

      $df = n- 1$

=>    $df = 7- 1$

=>    $df = 6$

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 t distribution table the critical value  of   at a degree of freedom of  is  

   $t_{\frac{\alpha }{2} , 6 } = 2.447$

Generally the margin of error is mathematically represented as  

      $E = t_{\frac{\alpha }{2} , 6 } * \frac{\sigma }{\sqrt{n} }$

=>    $E =2.447* \frac{0.283 }{\sqrt{7} }$

=>    $E =0.2617$

Generally 95% confidence interval is mathematically represented as  

      $\= x -E < \mu < \=x +E$

=>   $10 -0.2617 < \mu < 10 + 0.2617$

=>   $9.7383 < \mu < 10.2617$