Question

A 99% Confidence interval for mu using the sample results

x bar = 90.9 s = 34.3 n =15

Point Estimate = ______

Margin of Error = _____

99% CI is ____ to ____

Answer 1

Given that,

= 90.9

s =34.3

n = 15

Degrees of freedom = df = n - 1 = 15- 1 = 14

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t /2  df = t0.005,14 = 2.977 ( using student t table)

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

= 2.977* ( 34.3/ 15) = 26.3650

The 99% confidence interval estimate of the population mean is,

- E < < + E

90.9- 26.3650 < <90.9 + 26.3650

64.5350< < 117.2650

64.5350, 117.2650

Point Estimate = __90.9____

Margin of Error = _26.3650____

99% CI is ____= 64.5350, 117.2650__