Chi-square: interest and computer ability

Chi-square distribution: if Z₁, …, Z_k are independent standard normal variables, χ² = ΣZᵢ² follows the chi-square distribution with k degrees of freedom. It is positively skewed, non-negative, with mean k and variance 2k; it is used for goodness of fit, independence of attributes and variance tests.

Test of relationship between interest in engineering and ability in computer

The third row (High ability) is cut off in the question, so the test uses the two rows given (N = 162).

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
Ability Low          6      12      32      50
Ability Avg         33      61      18     112
Total               39      73      50     162

Expected frequencies:

                   Low Average    High   
Ability Low      12.04   22.53   15.43
Ability Avg      26.96   50.47   34.57

Terms (O − E)²/E:

                   Low Average    High   
Ability Low      3.028   4.922  17.787
Ability Avg      1.352   2.197   7.941

χ² = Σ (O − E)²/E = 37.227

4. Degrees of freedom = (r − 1)(c − 1) = (2 − 1)(3 − 1) = 2. Critical value: χ²(0.05; 2) = 5.991. Reject H0 if χ² > 5.991.
5. Decision: calculated χ² > tabulated χ² (37.227 > 5.991, p-value = 0.0000), so reject H0.

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

Assumption: only the two rows shown are used. If the missing row is 13, 15, 0 (as in the identical-looking fidelity table of 2076 Asoj), the 3 × 3 test gives χ² = 54.33 with 4 d.f. (critical 9.488), which also rejects H0.