--- title: "Chi-square: interest and computer ability" subject: "Probability and Statistics" chapter: 7 key: "semester_5/probability_and_statistics:chi-square-interest-ability-computer" source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/chi-square-interest-ability-computer site: BCT Engineering --- # Chi-square: interest and computer ability Probability and Statistics · Chapter 7: Inference concerning proportion - Asked in: 2070 Asar, 5 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-interest-ability-computer - Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-7 ## Question What do you mean by chi square distribution? The following test gives the information for the engineering students interest with ability in computer. Is there any significant relationship between interest in engineering and ability in computer? | Ability in computer \ Interest in Engineering | Low | Average | High | |---|---|---|---| | Low | 6 | 12 | 32 | | Average | 33 | 61 | 18 | [The third row (High) is cut off in the scan.] ## Short answer The chi-square distribution is the sum of squared independent standard normal variables, with k df, mean k, variance 2k. For the given table (two rows, N=162), χ² = 37.227, df=2, critical value 5.991. Reject H₀; there is a significant association between interest in engineering and ability in computer. ## Answer **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: ```text Low Average High Total Ability Low 6 12 32 50 Ability Avg 33 61 18 112 Total 39 73 50 162 ``` Expected frequencies: ```text Low Average High Ability Low 12.04 22.53 15.43 Ability Avg 26.96 50.47 34.57 ``` Terms (O − E)²/E: ```text 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the chi-square distribution with k d.f. defined? A. Sum of k independent standard normal variables B. Sum of squares of k independent standard normal variables C. Ratio of two normal variables D. Mean of k normal variables Answer: B. chi-square = Z1^2 + ... + Zk^2, with mean k and variance 2k. (Key point: Definition of chi-square) 2. For the 2 rows x 3 columns table used, how many d.f.? A. 2 B. 1 C. 3 D. 6 Answer: A. (2 - 1)(3 - 1) = 2. (Key point: Degrees of freedom) 3. Calculated chi-square is 37.227 and critical is 5.991. What is the conclusion? A. Independent; fail to reject H0 B. Significant association between interest and ability; reject H0 C. Cannot decide because a row is missing D. Reject H0 but no association Answer: B. 37.227 > 5.991, so H0 of independence is rejected. 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