Question

Based on a sample of 100 employees a 95% confidence interval is calculated for the mean age of all employees at a large firm. The interval is (34.5 years, 47.2 years).
A. What was the sample mean?
B. Find the margin of error?
C. Find the critical value tc for:
a. a 90% confidence level when the sample size is 22.
b. an 80% confidence level when the sample size is 49.

Answer 1

Answer:

a

$\= x = 40.85$

b

$E = 5.85$

Ca  

   $t_c = 2.08$

Cb

   $t_c = 1.282$

Explanation:

From the question we are told that

     The sample size is n  =  100

     The upper limit of the 95% confidence interval is  b =  47.2 years

     The lower limit of the 95% confidence interval is   a =  34.5 years

Generally the sample mean is mathematically represented as

         $\= x = \frac{a + b }{2}$

=>    $\= x = \frac{47.2 + 34.5 }{2}$

=>    $\= x = 40.85$

Generally the margin of error is mathematically represented as

         $E = \frac{b- a }{ 2}$

=>      $E = \frac{47.2- 34.5 }{ 2}$

=>      $E = 5.85$

Considering question C a  

From the question we are told the confidence level is  90% , hence the level of significance is    

      $\alpha = (100 - 90 ) \%$

=>   $\alpha = 0.10$

The  sample size is  n =  22

Given that the sample size is not sufficient enough i.e $n < 30$ we will make use of the student t distribution table  

Generally the degree of freedom is mathematically represented as

           $df = n- 1$

=>        $df = 22 - 1$

=>        $df = 21$

Generally from the student  t  distribution table the critical value  of  $\frac{\alpha }{2}$ at a degree of freedom  of  21 is  

   $t_c =t_{\frac{\alpha }{2} , 21 } = 2.08$

Considering question C b

From the question we are told the confidence level is  80% , hence the level of significance is    

      $\alpha = (100 - 80 ) \%$

=>   $\alpha = 0.20$

Generally from the normal distribution table the critical value  of  $\frac{\alpha }{2}$ is  

   $t_c =Z_{\frac{\alpha }{2} } = 1.282$