Question

A random sample of 12 chemists from Washington stats shows an average salary of $39,420 with a standard deviation of $1659. A random sample of 26 chemists from New Mexico has an average salary of $30,215 with a standard deviation of $4116. Find the 98% confidence interval of the difference of means.

Answer 1

Explanation:

Given that :

μ1 = 39420 ; σ1 = 1659 ; s1 = 12

μ2 = 30215 ; σ2 = 4116 ; s2 = 26

df1 = 12 - 1 = 11

df2 = 26 - 1 = 25

98 % confidence interval of the difference in sample means :

Pooled Variance :

S²p = ((df1 * s1²) + (df2 * s2²)) / (df1 + df2)

((11×1659^2)+(25×4116^2))÷(11+25) = 12605874.75

Standard Error :

√(S²p/n1) + (S²p/n2)

√(12605874.75/12) + (12605874.75/26)

= 1239.0847