Chi-Square TestChi-Square TestChi-Square TestIn-class Work on 2 Tests1. A survey is undertaken to determine the incidence of alcoholism in different professional groups. Random samples of clergy, educators, executives, and merchants were interviewed by a team of sociologists. The 2-way table with the results was given in the last in-class work. The sociologists conducting the study wish to test if any of the professions have a significantly different incidence of alcoholism than the others.(a) What are the null and alternative hypotheses being tested?(b) Fill in the following table with the EXPECTED CELL COUNTS: Alcoholic NonalcoholicClergyEducatorsExecutivesMerchants(c) Fill in the following table with the “contributions” to the test statistic X2 : Alcoholic NonalcoholicClergyEducatorsExecutivesMerchants(d) Use the table in (c) to calculate the test statistic X2 . (e) What is the distribution of the test statistic X2 ? (f) What is the p-value? (g) Make a decision at the 0.05 significance level. (h) State your conclusion in terms of this problem.(i) Describe the association in terms of this problem. That is, which professional groups have higher or lower incidences of alcoholism than the others?2. A study was conducted to explore the association between a child’s birth order and the occurrence of juvenile delinquent behavior. The subjects were a random sample of girls enrolled in public high schools in a large city. Thedata were analyzed in MINITAB and the output appears below. Note some information has been purposely deleted.Chi-Square TestExpected counts are printed below observed counts Always Sometimes Never TotalOldest 24 55 230 309 37.12 78.92 192.96 In-between 29 61 186 276 33.16 70.49 172.35 Youngest 35 89 122 246 29.55 _____ 153.62 Only child 23 31 39 93 11.17 23.75 58.07Total 111 236 577 924Chi-Sq = 4.637 + 7.251 + _____ + 0.521 + 1.279 + 1.081 + 1.004 + 10.899 + 6.507 + 12.522 + 2.211 + 6.265 = 61.289DF = ________, P-Value = 0.000 (a) What are the null and alternative hypotheses being tested? (b) Calculate the expected count for the “Youngest” and “Sometimes” cell .(c) Calculate the contribution to the test statistic for the “Oldest” and “Never” cell.(d) What is the value of the test statistic X2 ? (e) What is the distribution of the test statistic X2 ?(f) What is the p-value? (g) Make a decision at the 0.05 significance level.(h) State your conclusion in terms of this problem. (j) Describe the association in terms of this problem. 3. Use the following 2test performed in MINITAB to determine whether there is an association between a child’s gender and the child’s primary interests (grades, being popular, or sports) for middle-school children in a large city.Chi-Square TestExpected counts are printed below observed counts Grades Popular Sports Total Boys 96 32 94 222 144.19 28.40 49.42Girls 295 45 40 380 246.81 48.60 84.58Total 391 77 134 602Chi-Sq = 16.105 + 0.458 + 40.226 + 9.409 + 0.267 + 23.501 = 89.966DF = 2, P-Value = 0.000(a) What are the null and alternative hypotheses being tested? (b) Make a decision at the 0.05 significance level.(c) State your conclusion in terms of this problem. (d) Describe the association in terms of this problem. (e) Give a possible confounding variable which might lead to Simpson’s Paradox. 4. Here the 2Test is looking for an association between how many parents smoke (both, one, or none) and whether or not their child smokes.(a) What are the null and alternative hypotheses being tested? (b) Make a decision at the 0.05 significance level.(c) State your conclusion in terms of this problem. (d) Describe the association in terms of this problem. (e) Give a possible confounding variable which might lead to Simpson’s Paradox.5. The following 2Test is looking for an association between job level (1, 2, 3, 4 – where 1 represents entry-level job and 4 represents CEO of company, for example) and marital status (single, married, divorced, or widowed).Chi-Square TestExpected counts are printed below observed counts Smokes NoSmoke Total Both 400 1380 1780 332.49 1447.51 One 416 1823 2239 418.22 1820.78 None 188 1168 1356 253.29 1102.71Total 1004 4371 5375Chi-Sq = 13.709 + 3.149 + 0.012 + 0.003 + 16.829 + 3.866 = 37.566DF = 2, P-Value = 0.000Chi-Square TestExpected counts are printed below observed counts Single Married Divorced Widowed Total 1 58 874 15 8 955 39.08 896.44 14.61 4.87 2 222 3927 70 20 4239 173.47 3979.05 64.86 21.62 3 50 2396 34 10 2490 101.90 2337.30 38.10 12.70 4 7 533 7 4 551 22.55 517.21 8.43 2.81Total 337 7730 126 42 8235Chi-Sq = 9.158 + 0.562 + 0.010 + 2.011 + 13.575 + 0.681 + 0.407 + 0.121 + 26.432 + 1.474 + 0.441 + 0.574 + 10.722 + 0.482 + 0.243 + 0.504 = 67.397DF = 9, P-Value = 0.0002 cells with expected counts less than 5.0(a) What are the null and alternative hypotheses being tested? (f) Make a decision at the 0.05 significance level.(g) State your conclusion in terms of this problem. (d) Describe the association in terms of this problem. (e) Give a possible confounding variable which might lead to Simson’s