Question

A simple random sample of size nequals17 is drawn from a population that is normally distributed. The sample mean is found to be x overbar equals 54 and the sample standard deviation is found to be sequals16. Construct a 95​% confidence interval about the population mean.

The lower bound is =.

The upper bound is =.

​(Round to two decimal places as​ needed.)

Answer 1

Solution :

Given that,

= 54

s =16

n =17

Degrees of freedom = df = n - 1 = 17 - 1 = 16

a ) At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  / 2= 0.05 / 2 = 0.025

t /2,df = t0.025,16 = 2.120 ( using student t table)

Margin of error = E = t/2,df * (s /n)

=2.120 * ( 16/ 17)

= 8.23

The 95% confidence interval is,

- E < < + E

54 - 8.23 < < 54+8.23

45.77< < 62.23

The lower bound is =.45.77

The upper bound is =62.23