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

a

`\= x = 40.85`

b

`E = 5.85`

Ca

`t_c =  2.08`

Cb

`t_c  =   1.282`

Step-by-step 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 = (a + b )/(2)`

=>
`\= x = (47.2 + 34.5 )/(2)`

=>
`\= x = 40.85`

Generally the margin of error is mathematically represented as

`E = (b- a )/( 2)`

=>
`E = (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
`(\alpha )/(2)` at a degree of freedom of 21 is

`t_c =t_{(\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
`(\alpha )/(2)` is

`t_c  =Z_{(\alpha )/(2) } =  1.282`