Question

In a simple random sample of 20 residents of the city of Dallas, the mean paper products recycled per person per day was 0.95 pounds with a standard deviation of 0.32 pounds. Determine the 99% confidence interval for the mean paper products recycled per person per day for the population of Dallas, assuming the population is approximately normal

Answer 1

Answer:

The 99% confidence interval for the mean paper products recycled per person per day for the population of Dallas is

     $0.7454 < \mu < 1.1546$

Step-by-step explanation:

From the question we are told that

   The sample size is  n =  20

   The sample mean is  $\= x = 0.95 \ pounds$

   The standard deviation is  $\sigma = 0.32 \ pounds$

Generally given that the sample size is small , n<  30  we will be making use of t distribution table

Generally the degree of freedom is mathematically represented as

             $df = 20 - 1$

=>          $df = 19$

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  $\frac{\alpha }{2}$  at a degree of freedom of $df = 19$  is  

   $t_{\frac{\alpha }{2}, 19 } = 2.86$

Generally the margin of error is mathematically represented as  

      $E =t_{\frac{\alpha }{2}, 19 } * \frac{\sigma }{\sqrt{n} }$

=>   $E = 2.86 * \frac{0.32}{\sqrt{20} }$

=>   $E = 0.2046$      

Generally 99% confidence interval is mathematically represented as  

      $\= x -E < \mu < \=x +E$

=>  $0.95 -0.2046 < \mu < 0.95 + 0.2046$

=>  $0.7454 < \mu < 1.1546$