Question

Consider a normal population with an unknown population standard deviation. A random sample results in x− = 52.15 and s2 = 21.16. [You may find it useful to reference the t table.] a. Compute the 95% confidence interval for μ if x− and s2 were obtained from a sample of 19 observations. (Round intermediate calculations to at least 4 decimal places. Round "t" value to 3 decimal places and final answers to 2 decimal places.) b. Compute the 95% confidence interval for μ if x− and s2 were obtained from a sample of 27 observations. (Round intermediate calculations to at least 4 decimal places. Round "t" value to 3 decimal places and final answers to 2 decimal places.)

Answer 1

Solution:

Given that

sample mean = 52.15

sample variance s² = 21.16

sample SD s = 21.16 = 4.6

Confidence level c = 95% = 0.95

a)

Here , n = 19

degrees of freedom df = n - 1 = 18

= 1- c = 1- 0.95 = 0.05

  /2 = 0.05 2 = 0.025

Also, d.f = n - 1 = 18

    =    =  _0.025,18 = 2.101

( use t table or t calculator to find this value..)

The margin of error is given by

E =  /2,d.f. * ( / n)

= 2.101 * (4.6 / 19 )

= 2.22

Now , confidence interval for mean() is given by:

( - E ) <   <  ( + E)

(52.15 - 2.22)   <   <  (52.15 + 2.22)

49.93 <   < 54.37

Required 95% confidence interval is (49.93 , 54.37)

b)

Here , n = 27

degrees of freedom df = n - 1 = 26

= 1- c = 1- 0.95 = 0.05

  /2 = 0.05 2 = 0.025

Also, d.f = n - 1 = 26

    =    =  _0.025,26 = 2.056

( use t table or t calculator to find this value..)

The margin of error is given by

E =  /2,d.f. * ( / n)

= 2.056* (4.6 / 27 )

= 1.82

Now , confidence interval for mean() is given by:

( - E ) <   <  ( + E)

(52.15 - 1.82)   <   <  (52.15 + 1.82)

50.33 <   < 53.97

Required 95% confidence interval is (50.33 , 53.97)