Confidence Interval problem for two population DIFFERENCES (not means or proportions) that have the SAME NUMBER OF DATA VALUES (n1 = n2). Using the same traffic count data and statistics as below, the margin of error, E, formula is E = t-critical * s_d/sqrt(n) We will assume a significance level of = 5%(95% confidence level) with n=10 data points, hence df = (10 - 1 ) = 9, so t-critical =______ ? What are the MEAN ______and SD _____for the "differences"?

E = _______ * _______ / ______ = __________ (fill in the different values and the calculated result) .

Our Confidence Intervalis (d-bar - E) <_d < (d-bar + E) = _________ <_d < __________

Real-world Conclusion: Whose median income mean is larger and by how much?

(IF this or any CI includes zero there is NO difference)

Dates	6th (x1)	13th (x2)	d = (x1 - x2)
1990, July	139246	138548
1990, July	138012	132908
1991, September	137055	136018
1991, September	140675	131843
1991, December	123552	121641
1991, December	121139	120723
1992, March	128293	125532
1992, March	124631	120249
1992, November	124609	125770
1992, November	119584	116263