Question

Assume that a sample is used to estimate a population mean μ. Find the 99% confidence interval for a sample of size 641 with a mean of 21.3 and a standard deviation of 19.7. Enter your answer as a tri-linear inequality accurate to 3 decimal places.
____ < μ < ____

Answer 1

Solution:

Confidence interval for population mean() using t distribution  

Given that,

= 21.3 ....... Sample mean

s = 19.7 ...Sample standard deviation

n = 641 ....... Sample size

Note that, Population standard deviation() is unknown..So we use t distribution.

Our aim is to construct 99% confidence interval.

c = 0.99

= 1- c = 1- 0.99 = 0.01

  /2 = 0.01 2 = 0.005

Also, n = 641

d.f= n-1 = 640

     =    = = 2.584

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

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

  

21.3 - 2.584*(19.7/ 641)    21.3 + 2.584*(19.7/ 641)

19.289 < < 23.311

is the required 99% confidence interval for mean....