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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.

CentralEasternWestern
Purchase brand405545
Do not purchase brand604555

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.

  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:

               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

  1. 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.
  2. 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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