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Chi-square distribution; brand-region association
Short answer
Chi-square distribution is sum of squares of independent standard normals with k d.f., positively skewed, mean k, variance 2k. For independence test, χ²=4.688 with d.f.=2, critical 5.991 at 5%. Fail to reject H0, no significant association between brand purchase and region.
Define chi-square distribution. From the following data can you conclude that there is association between the purchase of brand and geographical region? Use 5% level of significance.
| Central | Eastern | Western | |
|---|---|---|---|
| Purchase brand | 40 | 55 | 45 |
| Do not purchase brand | 60 | 45 | 55 |
Chi-square distribution: the distribution of the sum of squares of k independent standard normal variables, χ² = ΣZᵢ², with k degrees of freedom. It is positively skewed, takes only non-negative values, has mean k and variance 2k, and is used for tests of independence, goodness of fit and variance.
Test of association between brand purchase and region
Total = 300.
- H0: the two attributes are independent (no association). H1: the two attributes are associated (dependent).
- Level of significance: α = 0.05
- Test statistic: χ² = Σ (O − E)²/E, where E = (row total × column total)/grand total.
Observed frequencies with totals:
Central Eastern Western Total
Purchase 40 55 45 140
No purchase 60 45 55 160
Total 100 100 100 300
Expected frequencies:
Central Eastern Western
Purchase 46.67 46.67 46.67
No purchase 53.33 53.33 53.33
Terms (O − E)²/E:
Central Eastern Western
Purchase 0.952 1.488 0.060
No purchase 0.833 1.302 0.052
χ² = Σ (O − E)²/E = 4.688
- 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.
- Decision: calculated χ² < tabulated χ² (4.688 < 5.991, p-value = 0.0960), so fail to reject H0.
Conclusion: there is no significant association; the two attributes may be taken as independent.
✓ Answer checked by Claude Sonnet 5.5
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