Question

A developer collects a random sample of the ages of houses from two neighborhoods and finds that the summary statistics for each are as shown. Assume that the data come from a distribution that is Normally distributed. Complete parts a through d below.

Neighborhood 1 Neighborhood 2

n_1 40 n_2 35

y1 52.1 y2 41.6

s_1 7.71 s_2 7.17

a) Find a​ 95% confidence interval for the mean​ difference,μ1−μ2​,in ages of houses in the two neighborhoods df=72.7

A​ 95% confidence interval for the mean difference in ages of houses in the two neighborhoods when

n1equals=n2equals=6

is

left parenthesis negative 0.76 comma 20.14 right parenthesis(−0.76,20.14).

Why is the confidence interval found in part a narrower than this​ one?

Answer 1

SE = sqrt(7.71^2/40 + 7.17^2/35) = 1.7190

t-value for 95% CI= 1.9930

ME = 1.993*1.719 = 3.426

CI = (52.1 - 41.6 - 3.4260, 52.1 - 41.6 + 3.4260)
= (7.0740, 13.9260)

b)
because the sample sizes of 6 reduces the SE which in turn reduce ME Hence narrower CI