Question

If n = 12, x bar = 43, and s = 9, construct a confidence interval at a 98% confidence level. Assume the data came from a normally distributed population. Give your answers to three decimal places.

____ < μ < ____

Answer 1

Solution :

Given that,

= 43

s = 9

n = 12

Degrees of freedom = df = n - 1 = 12 - 1 = 11

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

t _/2,df = t_0.01,11 =2.326

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

= 2.326 * ( 9 / 12)

= 6.043

Margin of error = 6.043

The 98% confidence interval estimate of the population mean is,

- E < < + E

43 - 6.043< < 43 + 6.043

36.957 < <  49.043