--- title: "Chi-square: fidelity and selectivity of receivers" subject: "Probability and Statistics" chapter: 7 key: "semester_5/probability_and_statistics:chi-square-fidelity-selectivity-190" source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/chi-square-fidelity-selectivity-190 site: BCT Engineering --- # Chi-square: fidelity and selectivity of receivers Probability and Statistics · Chapter 7: Inference concerning proportion - Asked in: 2076 Asoj, 6 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/chi-square-fidelity-selectivity-190 - Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-7 ## Question Test of the fidelity and selectivity of 190 digital radio receivers produced the results shown in the following table. | Selectivity \ Fidelity | Low | Average | High | |---|---|---|---| | Low | 6 | 12 | 32 | | Average | 33 | 61 | 18 | | High | 13 | 15 | 0 | Use α = 0.05 and χ² = 5.991 to test whether there is relationship between fidelity and selectivity. ## Short answer Chi-square test of independence gives chi-square = 54.328 with df = 4; critical value at 0.05 is 9.488. Since 54.328 > 9.488, reject H0 and conclude fidelity and selectivity are associated. The stated 5.991 corresponds to df = 2 and is not the correct table value. ## Answer 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: ```text 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: ```text 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: ```text 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does H0 say? A. They are associated B. Fidelity and selectivity are independent C. Fidelity is high D. Means are equal Answer: B. H0 for the independence test: no relationship. (Key point: Hypotheses) 2. What are the d.f. for the 3 × 3 table? A. 9 B. 2 C. 4 D. 6 Answer: C. (3 − 1)(3 − 1) = 4, so the 5% value is 9.488. (Key point: Degrees of freedom) 3. Why is 5.991 not the right critical value? A. It is for 1 d.f. B. It is correct C. It is for 1% D. It is for 2 d.f., not 4 Answer: D. 5.991 is the 5% value for 2 d.f. For 4 d.f. use 9.488. (Key point: Question's table value) 4. Calculated chi-square is 54.328. What is concluded? A. Reject H0; there is a relationship B. Fail to reject H0 C. Test invalid since E < 5 D. Independent Answer: A. 54.328 > 9.488 (and > 5.991), so H0 is rejected. 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