Question

TABLE 12-11 The director of admissions at a state college is interested in seeing if admissions status (admitted, waiting list, denied admission) at his college is independent of the type of community in which an applicant resides. He takes a sample of recent admissions decisions and forms the following table:

Admitted Wait

	ADMITTED	WAIT LIST	DENIED	TOTAL
URBAN	45	21	17	83
RURAL	33	13	24	70
SUBURBAN	34	12	39	85
TOTAL	112	46	80	238

He will use this table to do a chi-square test of independence with a level of significance of 0.01.

Referring to Table 12-11, the value of the test statistic is

Answer 1

Observed Frequencies
	Admitted	Wait List	Denied	Total
Urban	45	21	17	83
Rural	33	13	24	70
Suburban	34	12	39	85
Total	112	46	80	238
Expected Frequencies
	Admitted	Wait List	Denied	Total
Urban	112 * 83 / 238 = 39.0588	46 * 83 / 238 = 16.042	80 * 83 / 238 = 27.8992	83
Rural	112 * 70 / 238 = 32.9412	46 * 70 / 238 = 13.5294	80 * 70 / 238 = 23.5294	70
Suburban	112 * 85 / 238 = 40	46 * 85 / 238 = 16.4286	80 * 85 / 238 = 28.5714	85
Total	112	46	80	238
	(fo-fe)²/fe
Urban	(45 - 39.0588)²/39.0588 = 0.9037	(21 - 16.042)²/16.042 = 1.5323	(17 - 27.8992)²/27.8992 = 4.2579
Rural	(33 - 32.9412)²/32.9412 = 0.0001	(13 - 13.5294)²/13.5294 = 0.0207	(24 - 23.5294)²/23.5294 = 0.0094
Suburban	(34 - 40)²/40 = 0.9	(12 - 16.4286)²/16.4286 = 1.1938	(39 - 28.5714)²/28.5714 = 3.8064

Null and Alternative hypothesis:

Ho: Factors are independent.

H1: Factor are dependent.

Test statistic:

χ² = ∑ ((fo-fe)²/fe) = 12.6244

df = (r-1)(c-1) = 4

Critical value:

χ²α = CHISQ.INV.RT(0.01, 4) = 13.2767

p-value:

p-value = CHISQ.DIST.RT(12.6244, 4) = 0.0133

Decision:

p-value > α, Do not reject the null hypothesis.