Chi-square: fidelity and selectivity of receivers

Chi-square test of independence between selectivity and fidelity (3 × 3 table, N = 190).

1. H0: the two attributes are independent (no association). H1: the two attributes are associated (dependent).
2. Level of significance: α = 0.05
3. Test statistic: χ² = Σ (O − E)²/E, where E = (row total × column total)/grand total.

Observed frequencies with totals:

                   Low Average    High   Total
Sel Low              6      12      32      50
Sel Avg             33      61      18     112
Sel High            13      15       0      28
Total               52      88      50     190

Expected frequencies:

                   Low Average    High   
Sel Low          13.68   23.16   13.16
Sel Avg          30.65   51.87   29.47
Sel High          7.66   12.97    7.37

Terms (O − E)²/E:

                   Low Average    High   
Sel Low          4.315   5.376  26.982
Sel Avg          0.180   1.606   4.467
Sel High         3.717   0.318   7.368

χ² = Σ (O − E)²/E = 54.328

4. Degrees of freedom = (r − 1)(c − 1) = (3 − 1)(3 − 1) = 4. Critical value: χ²(0.05; 4) = 9.488. Reject H0 if χ² > 9.488.
5. Decision: calculated χ² > tabulated χ² (54.328 > 9.488, p-value < 0.0001), so reject H0.

Conclusion: there is a significant association between the two attributes; they are not independent.

Note: the value 5.991 given in the question is the χ² value for 2 d.f. For a 3 × 3 table, d.f. = (3−1)(3−1) = 4, so the correct table value is 9.488. Since the calculated χ² = 54.33 is larger than both, H0 is rejected either way. All expected frequencies are above 5 (the smallest is 7.37), so the chi-square test is valid.