Question

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. (Round your answers into two decimal places.)

Answer 1

Answer:

The 95 confidence interval   is  $9.738  <  \mu < 10.262$

Step-by-step explanation:

The sample size is  n  =  7

 The sample data  is  9.8, 10.2, 10.4, 9.8,10.0, 10.2, and 9.6 liters

Generally the sample mean is mathematically represented as

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

=>          $\= x = \frac{ 9.8 + 10.2 + 10.4 +\cdots + 9.6}{7 }$

=>          $\= x = 10$

Generally the standard deviation is mathematically represented as

       $\sigma = \sqrt{\frac{\sum ( x_i - \= x)^2 }{n-1} }$

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

=>    $\sigma =0.283$

Note : We are making use of t distribution because n is small i.e n < 30

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  

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

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

=>    $E =  0.262$

Generally 95% confidence interval is mathematically represented as  

      $\= x -E <  \mu <  \=x  +E$

=>    $10 -0.262 <  \mu <10 + 0.262$

=>    $9.738  <  \mu < 10.262$