Tests for Independence/ Homogeneity

Calculations by Professor S. Gramlich

(updated 3/22/09)

Example 1

Testing Influence on Gender

(from Triola, “Elementary Statistics,” 8^th ed, C2001, p. 600)

10-3 #5

Using a 0.01 significance level, and assuming the sample sizes of 800 men and 400

women are predetermined, test the claim that proportions of agree/disagree responses

are the same for the subjects interviewed by men and the subjects interviewed by

women.

	Man Interviewer	Woman Interviewer
Women who agree	512	336
Women who disagree	288	64

H₀: proportions of agree/disagree responses are homogeneous (original claim)

H₁: proportions of agree/disagree responses are NOT homogeneous

degrees of freedom= (r – 1) (c – 1) = (2 – 1) * (2 – 1) = 1

critical value: c_c² = 6.635 (from Table A-4, df=1, alpha= 0.01)

test statistic:

Pearson X² approximation= S((O-E)²/E)

= (512 - 565.333)² /565.333 + (336 - 282.666)² /282.666 + (288 - 234.666)² /234.666 + (64 - 117.333)² /117.333

=51.458

P-Value < 0.005 (from Table A-4, df =1)

(actual p-value = 7.315 x 10^-13 from Technology)

Decision:

Traditional: Reject H₀ since X² > c_c² (test statistic in critical region)

P-value: Since P-Value < alpha, Reject Null.

Conclusion: There is sufficient evidence to warrant rejection of the claim...

Example 2

Crime and Strangers

(from Triola, “Elementary Statistics,” 8^th ed, C2001, p. 601)

10-3 #11 (10-3 #15 from 9th ed)

The accompanying table lists survey results obtained from a random sample of

different crime victims. At the 0.05 significance level, test the claim that the type

of crime is independent of whether the criminal is a stranger. How might the results

affect the strategy police officers use when they investigate crimes?

	Homicide	Robbery	Assault
Criminal was a stranger	12	379	727
Criminal was acquaintance or relative	39	106	642

H₀: crime is independent of whether criminal is stranger (original claim)

H₁: crime is dependent on whether criminal is stranger

degrees of freedom= (r – 1) (c – 1) = (2 – 1) * (3 – 1) = 2

critical value: c_c² = 5.991 (from Table A-4, df =2, alpha =.05)

test statistic:

Pearson X² approximation= S((O-E)²/E)

= (12 – 29.931)² / 29.931 + (379 – 284.64)² /284.64 + (727 – 803.43)² / 803.43

+ (39- 21.069)² / 21.069 + (106 – 200.36)² /200.36 + (642 – 565.57)² /565.57

= 119.33

P-Value < 0.005 (from Table A-4, df=2)

(actual p-value = 1.224 x 10^-26 from Technology)

Decision:

Traditional: Reject Null since X² > c_c² (test statistic in critical region)

P-Value: Since P-Value < alpha, Reject H₀

Conclusion: There is sufficient evidence to warrant rejection of the claim...