1. **Stating the problem:** We want to test if there is a significant difference in consumer opinions on three aspects (Harga, Kebersihan, Cita Rasa) of a restaurant based on the survey data. 2. **Hypothesis:** - Null hypothesis $H_0$: There is no difference in opinions across the three aspects. - Alternative hypothesis $H_a$: There is a difference in opinions. 3. **Test used:** We use the Chi-square test for independence because we have categorical data in a contingency table. 4. **Observed frequencies ($O$):** | Aspect | Baik | Cukup | Kurang | |-------------|-------|-------|--------| | Harga | 20 | 15 | 2 | | Kebersihan | 17 | 18 | 2 | | Cita Rasa | 15 | 14 | 3 | 5. **Calculate row totals, column totals, and grand total:** - Row totals: Harga = $20+15+2=37$, Kebersihan = $17+18+2=37$, Cita Rasa = $15+14+3=32$ - Column totals: Baik = $20+17+15=52$, Cukup = $15+18+14=47$, Kurang = $2+2+3=7$ - Grand total = $37+37+32=106$ 6. **Calculate expected frequencies ($E$) using:** $$E = \frac{(\text{row total})(\text{column total})}{\text{grand total}}$$ For example, for Harga-Baik: $$E = \frac{37 \times 52}{106} \approx 18.15$$ Calculate all expected values: - Harga: Baik $18.15$, Cukup $16.42$, Kurang $2.44$ - Kebersihan: Baik $18.15$, Cukup $16.42$, Kurang $2.44$ - Cita Rasa: Baik $15.70$, Cukup $14.15$, Kurang $2.11$ 7. **Calculate Chi-square statistic:** $$\chi^2 = \sum \frac{(O - E)^2}{E}$$ Calculate each cell contribution and sum: - Harga-Baik: $\frac{(20-18.15)^2}{18.15} = 0.19$ - Harga-Cukup: $\frac{(15-16.42)^2}{16.42} = 0.12$ - Harga-Kurang: $\frac{(2-2.44)^2}{2.44} = 0.08$ - Kebersihan-Baik: $\frac{(17-18.15)^2}{18.15} = 0.07$ - Kebersihan-Cukup: $\frac{(18-16.42)^2}{16.42} = 0.15$ - Kebersihan-Kurang: $\frac{(2-2.44)^2}{2.44} = 0.08$ - Cita Rasa-Baik: $\frac{(15-15.70)^2}{15.70} = 0.03$ - Cita Rasa-Cukup: $\frac{(14-14.15)^2}{14.15} = 0.002$ - Cita Rasa-Kurang: $\frac{(3-2.11)^2}{2.11} = 0.37$ Sum: $\chi^2 = 0.19+0.12+0.08+0.07+0.15+0.08+0.03+0.002+0.37 = 1.10$ 8. **Degrees of freedom:** $$df = (\text{rows} - 1)(\text{columns} - 1) = (3-1)(3-1) = 4$$ 9. **Critical value at 95% confidence:** From Chi-square table, $\chi^2_{0.05,4} = 9.488$ 10. **Decision:** Since $1.10 < 9.488$, we fail to reject $H_0$. **Conclusion:** There is no significant difference in consumer opinions on the three aspects at 95% confidence. --- "slug": "chi-square-opinions", "subject": "statistics", "desmos": {"latex": "", "features": {"intercepts": false, "extrema": false}}, "q_count": 2