Question

A real estate agent claims that there is a difference between the mean household incomes of two neighborhoods. The mean income of 12 randomly selected households from the first neighborhood was 32750 with a standard deviation of 1900. In the second neighborhood, 10 randomly selected households had a mean income of 31200 with a standard deviation of 1825. Assume normal distributions and equal population variances. Test the claim at   0.10 .

Answer 1

For sample 1: n1=12, m1=32750, s1=1900, variance, v1=s1^2=3610000

For sample 2: n2=10, m2=31200, s2=1825, v2=s2^2=3330625

Variance of source population=(v1+v2)/((n1-1)+(n2-1))=347031.25, because we're told that it is equal for both samples. We can now estimate the standard deviation S for the sampling distribution = √(V(1/n1+1/n2))=252.2348 approx.

The t-score=(m1-m2)/S=(32750-31200)/252.2348=6.145 approx. This tells us that the difference in means is more than 6 standard deviations.

The degrees of freedom are n1-1+n2-1=20. The value of over 6 well exceeds the confidence level of 99.9% (critical value for t=1.725 - 2-tail). So the claim is false: there is no significant difference between the income levels.