How are the smoking habits of students related to their parents

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saiyansruleA · Accepted Answer

Let us assume: α=0.05=5% (a) The null hypothesis states that there is no association between the variables, while the alternative hypothesis states that there is an association between the variables. H0: There is no association between student smoking habit and parent smoking habit Hα: There is no association between student smoking habit and parent smoking habit (b) Determine the row and column totals of the given table: Student smokesStudent does not smokeTotalBoth parents smoke4001380400+1380=1780One parent smokes4161823416+1823=2239Neither parent smokes1881168188+1168=1356Total400+416+188=10041380+1823+1168=43711004+4371=5375 The expected frequencies E are the product of the column and row total, divided by the table total. E11=r1×c1n=1780×10045375≈332.4874 E12=r1×c2n=1780×43715375≈1447.5126 E21=r2×c1n=2239×10045375≈418.2244 E22=r2×c2n=2239×43715375≈1820.7756 E31=r3×c1n=1356×10045375≈253.2882 E32=r3×c2n=1356×43715375≈1102.7118 Expected counts are the counts that we expect based on the row and column totals, when there is no association between the variables. (c) The chi-square subtotals are the squared differences between the observed abd expected frequencies, divivded by the expected frequency. The value of the test-statistic is then the sum of the chi-square subtotals: X2=∑(O−E)2E =(400−322.4874)2332.4874+(1380−1447.5126)21447.5126+(416−418.2244)2418.2244+(1823−1820.7756)21820.7756+(188−253.2882)2253.2882+1168−1102.7119)21102.7118 The degrees of freedom is the product od the number of row and the number of columns, both decreased by 1. df=(r−1)(c−1)=(3−1)(2−1)=2 The P-value is the probability of obtaining the value of the test statistic, or a value more extreme. The P-value is the number (or interval) in the column title of the chi-square distribution table in the appendix containing the X2 -value in the row df=2: P&amp;lt;0.001 (d) If the P-value is less than or equal to the significance level, then the null hypothesis is rejected: P&amp;lt;0.05⇒Reject H0 There is sufficient evidence to support the claim that there is an association between student smoking habit and parent smoking habit. Result: (a) H0: There is no association between smoking habit and parent smoking habit. Hα: There is an association between student smoking habit and parent smoking habit. (b) 332.4874, 1447.5126, 418.2244, 1820.7756, 253.2882, 1102.7118 Expected counts are the counts that we expect based on the row and column totals, when there is no association between the variables. (c) X2=37.5664, degrees of freedom, P&amp;lt;0.01 (d) There is sufficient evidence to support the claim that there is an association between student smoking habit and parent smoking habit.

How are the smoking habits of students related to their parents' smoking? Here is a two-way table from a survey of student s in eight Arizona high sch

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