Question

im trying to estimate the difference in reported time to first failure for a product that is sold in africa and canada. They take a random sample of 500 from each area. The means obtained are 4.1 years for african areas and 3.8 years for canadian areas. Find the 95% confidence interval for the true difference in the population means. The population standard deviation for both areas is 0.4 years.

Answer 1

95% confidence interval for 1 - 2 is

Margin of error E = Z/2 * sqrt [ 1 / n1 + 2 / n2 ]

= 1.96 * sqrt [ 0.4² / 500 + 0.4² / 500 ]

= 0.0496

95% CI is

(1 - 2) - E < 1 - 2 < (1 - 2) + E

( 4.1 - 3.8) - 0.0496 < 1 - 2 < ( 4.1 - 3.8) + 0.0496

0.2504 < 1 - 2 < 0.3496

95% CI is ( 0.2504 , 0.3496)