Question

It has been suggusted that the highest priority of retirees is travel. Thus, a study was conducted to investigate the differences in the length of stay of a trip for pre- and post-retirees. A sample of 688 travelers were asked how long they stayed on a typical trip. The observed results of the study are found below.

Number of Nights / Pre-retirement / Post-retirement / Total

4−7 / 236 / 167 / 403

8−13 / 81 / 60 / 141

14−21 / 33 / 55 / 88

22 or more / 19 / 37 / 56 /

Total / 369 / 319 / 688

With this information, construct a table of estimated expected values.

Number of Nights / Pre-retirement / Post-retirement

4−7 / 216.1438 / 186.8561

8−13 / 75.6235 / 65.3764

14−21 / 47.1976 / 40.8023

22 or more / 30.0348 / 25.9651

Now, with that information, determine whether the length of stay is independent of retirement using α=0.05.

What is χ2=________

Answer 1

Observed Frequencies
	Pre-retirement	Post-retirement	Total
4 - 7	236	167	403
8 - 13	81	60	141
14 - 21	33	55	88
22 or more	19	37	56
Total	369	319	688
Expected Frequencies
	Pre-retirement	Post-retirement	Total
4 - 7	369 * 403 / 688 = 216.1439	319 * 403 / 688 = 186.8561	403
8 - 13	369 * 141 / 688 = 75.6235	319 * 141 / 688 = 65.3765	141
14 - 21	369 * 88 / 688 = 47.1977	319 * 88 / 688 = 40.8023	88
22 or more	369 * 56 / 688 = 30.0349	319 * 56 / 688 = 25.9651	56
Total	369	319	688
	(fo-fe)²/fe
4 - 7	(236 - 216.1439)²/216.1439 = 1.8241	(167 - 186.8561)²/186.8561 = 2.11
8 - 13	(81 - 75.6235)²/75.6235 = 0.3822	(60 - 65.3765)²/65.3765 = 0.4422
14 - 21	(33 - 47.1977)²/47.1977 = 4.2708	(55 - 40.8023)²/40.8023 = 4.9403
22 or more	(19 - 30.0349)²/30.0349 = 4.0542	(37 - 25.9651)²/25.9651 = 4.6897

Test statistic:

χ² = ∑ ((fo-fe)²/fe) = 22.7135

df = (r-1)(c-1) = 3

p-value = CHISQ.DIST.RT(22.7135, 3) = 0.000

Decision:

p-value < α, Reject the null hypothesis.

There is enough evidence to conclude that length of stay is dependent of retirement using α=0.05.