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.

a

b

Ca  

Cb

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

=>    

=>    

Generally the margin of error is mathematically represented as

=>      

=>      

Considering question C a  

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

=>  

The  sample size is  n =  22

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

Generally the degree of freedom is mathematically represented as

=>        

=>        

Generally from the student  t  distribution table the critical value  of   at a degree of freedom  of  21 is  

Considering question C b

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

=>  

Generally from the normal distribution table the critical value  of   is  

14 sq. ft

Step-by-step explanation:

First, we know that the area of a circle is , where r is the radius of the circle, and is the ratio of the length of a circle's circumference to its diameter, that is, =3.1415926535 ... . But we have here a semicircle.

So, the area of a semicircle is .

We also know that 1 feet = 12 inches, so 36 inches are:

Then, the area of the semicircular window is:

≈ 14.1372.

Well, rounding the answer to the nearest square foot, we have that the semicircular window allows light through an area of 14 sq. ft .