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  

solution:

→c= a b , that is product of a and b is possible only when when a is m × n matrix and b is n × p matrix.

that is, number of columns of a= number of rows of b.

so, size of matrix , c= a b is , if a = (m × n)matrix

b= (n × p)matrix

then a b = [ (m × n)matrix] ×[   (n × p)matrix ]

→→c= a b= [m × p ] matrix

→→size of product a b will be equal to = number of rows of a × number of columns of b

y > 1/3x

y < = -x + 4

hope this : )