Question

In a sample of 15 cell phones the average radiation emitted 0.941 W/kg with a sample standard deviation of 0.435 W/kg. Find the 99% confidence interval for the population mean.

Assume the radiation emissions are normally distributed. Round off the margin of error and the limits to three decimal places.

Answer 1

Given that,

= 0.941

s =0.435

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 * ( 0.435/ 15) = 0.334