Question

For a random sample sample of size 22, one has calculated the 98% confidence interval for μ and obtained the result (18.69, 56.96)

a) What is the margin of error of this confidence interval? (2 points)

b) What is the point estimate for that sample? (3 points)

c) What is the standard deviation for that sample? (4 points)

Answer 1

Given:

The 98% confidence interval for μ and obtained the result (18.69, 56.96)

To calculate:

a) The margin of error of this confidence interval

b) The point estimate for that sample

c) The standard deviation for that sample

Now,

a) We know that the margin of error, E is half of the length of confidence interval.

So E = 56.96 - 18.69 / 2 = 38.27/2 = 19.135

b) The point estimate for that sample :

- E = 18.69 or + E = 56.96

- 19.135 = 18.69 or    + 19.135 = 56.96

= 18.69 + 19.135 or   = 56.96 - 19.135  

= 37.825 or    = 37.825

c) The standard deviation for that sample :

Sample size, = 22

Degree of freedom, df = n-1 = 22-1 = 21

At 98% confidence level, the critical value of t is

t/2,df = t0.02/2, 21 = 2.5176

We have

E = t/2,df × s/√n

19.135 = 2.5176 × s/√22

s = 19.135 × √22/ 2.5176

= 35.65

Therefore

a) The margin of error of this confidence interval is 19.135

b) The point estimate for that sample is 37.825

c) The standard deviation for that sample is 35.65