22.5 please! Thanks introduced, the aphids and jelly helped land-ing or not). Each aphid this only once.timeters, cm) a with jelly. a wasdiagF-cigarettes and smoking cessation. Electronic cigarettes (e-cigarettes) arehortery-operated devices that can deliver nicotine and other chemicals in waterinstead of smoke. They are a relatively recent invention with little scientificvaporevaluation, yet some people believe that e-cigarettes could help smokers who wantre quit smoking. Researchers in New Zealand randomly assigned 657 adult smokerswanting to quit to three treatment groups: nicotine patches, nicotine e-cigarettes,and placebo e-cigarettes. Smoking abstinence at the end of the study was assessedbiochemically for all participants with a breath analyzer. The published researchreports that, "at 6 months, verified abstinence was 7.3% (21 of 289) with nicotinee-cigarettes, 5.8% (17 of 295) with patches, and 4.1% (three of 73) with placeboe-cigarettes."2EXT(a) Organize the study findings in the following two-way table of counts. Add thegroup totals to your table......Smoking status at six monthsTreatment groupVerified abstinenceNon-abstinentTotalNicotine e-cigarettesPatchesPlacebo e-cigarettes...(b) Now organize the cited percents into a similar two-way table. Are these per-cents part of the conditional distribution of abstinence, given treatment group, orbecukbm are they part of the conditional distribution of treatment, given abstinence sta-tus? What do they suggest about the relationship between treatment assigned andabstinence outcome?Aphid righting behavior. Pea aphids, Acyrthosiphon pisum, are small, winglesssap-sucking insects that live on plants. They evade predators (such as ladvbugeby dropping off the plants. A study examined the mechanism of anhid dreResearchers placed aphids on a leat positioned at four different heights (in centimeters, cm) above a surface covered with petroleum jelly. When a ladyhueintroduced, the aphids dropped, and the petroleum jelly helped capture their landHere are the findings: od calculator and three statistical programs. All four display the expected countsthe value of the test statistic, X2 = 21.268. The chi-square statistic is a nd8 terms, the chi-square components, one for each cell in the table. The thorstatisticalthough each uses a different terminology: "Cell Chi^ 2" for JPM, "Contriburions,Chi-square" for Minitab, and "Chi-square Components" (or "residuals^ 2") f oCommander.output forAPPLY YOUR KNOWLEDGE22.5 E-cigarettes and smoking cessation, continued. Exercise 22.1 describes oexperiment comparing three approaches for quitting smoking. Figure 22.3 shthe Minitab output for a chi-square test with a null hypothesis of no associatiobetween treatment received and abstinence outcome in the population of smokerwho want to quit.(a) Show how to obtain the expected counts of smokers with or withour verife iabstinence at six months among those using a nicotine e-cigarette. Your resulteshould agree, up to rounding error, with Figure 22.3. Verify that these two expectedcounts add to the row total for the observed counts in the nicotine e-cigarettegroup.(b) Compare the observed and expected counts in Figure 22.3. What do thersuggest?(c) Show how the first chi-square component in the Minitab output (0.48745) wascomputed.(d) What is the value of the chi-square statistic reported by Minitab? Verify that,up to rounding error, this value is the sum of all six chi-square components shownin the Minitab output.lMinitabChi-Square Test for Association: Treatment, Worksheet columnsRows: TreatmentColumns: Worksheet columnsVerifiedabstinenceNon-abstinent AllNicotine e-cigarette2126868718.04270.960.487450.03244PatchLT18.41278 295276.590.107910.00718Placebo e-cigaretteOL68.443734.560.531170.0353541616 657Cell Contents:CountExpected countContribution to Chi-square FIGURE 22.3Minitab chi-square output for thetwo-way table of e-cigarettes andsmoking cessation.Pearson Chi-Square = 1.202, DF = 2, P-Value = 0.548Likelihood Ratio Chi-Square = 1.253, DF = 2, P-Value = 0.534%3D%3D* NOTE * 1 cells with expected counts less than 5

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22.5 please! Thanks introduced, the aphids and jelly helped land-ing or not). Each aphid this only once.timeters, cm) a with jelly. a wasdiagF-cigarettes and smoking cessation. Electronic cigarettes (e-cigarettes) arehortery-operated devices that can deliver nicotine and other chemicals in waterinstead of smoke. They are a relatively recent invention with little scientificvaporevaluation, yet some people believe that e-cigarettes could help smokers who wantre quit smoking. Researchers in New Zealand randomly assigned 657 adult smokerswanting to quit to three treatment groups: nicotine patches, nicotine e-cigarettes,and placebo e-cigarettes. Smoking abstinence at the end of the study was assessedbiochemically for all participants with a breath analyzer. The published researchreports that, &#34;at 6 months, verified abstinence was 7.3% (21 of 289) with nicotinee-cigarettes, 5.8% (17 of 295) with patches, and 4.1% (three of 73) with placeboe-cigarettes.&#34;2EXT(a) Organize the study findings in the following two-way table of counts. Add thegroup totals to your table......Smoking status at six monthsTreatment groupVerified abstinenceNon-abstinentTotalNicotine e-cigarettesPatchesPlacebo e-cigarettes...(b) Now organize the cited percents into a similar two-way table. Are these per-cents part of the conditional distribution of abstinence, given treatment group, orbecukbm are they part of the conditional distribution of treatment, given abstinence sta-tus? What do they suggest about the relationship between treatment assigned andabstinence outcome?Aphid righting behavior. Pea aphids, Acyrthosiphon pisum, are small, winglesssap-sucking insects that live on plants. They evade predators (such as ladvbugeby dropping off the plants. A study examined the mechanism of anhid dreResearchers placed aphids on a leat positioned at four different heights (in centimeters, cm) above a surface covered with petroleum jelly. When a ladyhueintroduced, the aphids dropped, and the petroleum jelly helped capture their landHere are the findings: od calculator and three statistical programs. All four display the expected countsthe value of the test statistic, X2 = 21.268. The chi-square statistic is a nd8 terms, the chi-square components, one for each cell in the table. The thorstatisticalthough each uses a different terminology: &#34;Cell Chi^ 2&#34; for JPM, &#34;Contriburions,Chi-square&#34; for Minitab, and &#34;Chi-square Components&#34; (or &#34;residuals^ 2&#34;) f oCommander.output forAPPLY YOUR KNOWLEDGE22.5 E-cigarettes and smoking cessation, continued. Exercise 22.1 describes oexperiment comparing three approaches for quitting smoking. Figure 22.3 shthe Minitab output for a chi-square test with a null hypothesis of no associatiobetween treatment received and abstinence outcome in the population of smokerwho want to quit.(a) Show how to obtain the expected counts of smokers with or withour verife iabstinence at six months among those using a nicotine e-cigarette. Your resulteshould agree, up to rounding error, with Figure 22.3. Verify that these two expectedcounts add to the row total for the observed counts in the nicotine e-cigarettegroup.(b) Compare the observed and expected counts in Figure 22.3. What do thersuggest?(c) Show how the first chi-square component in the Minitab output (0.48745) wascomputed.(d) What is the value of the chi-square statistic reported by Minitab? Verify that,up to rounding error, this value is the sum of all six chi-square components shownin the Minitab output.lMinitabChi-Square Test for Association: Treatment, Worksheet columnsRows: TreatmentColumns: Worksheet columnsVerifiedabstinenceNon-abstinent AllNicotine e-cigarette2126868718.04270.960.487450.03244PatchLT18.41278 295276.590.107910.00718Placebo e-cigaretteOL68.443734.560.531170.0353541616 657Cell Contents:CountExpected countContribution to Chi-square> FIGURE 22.3Minitab chi-square output for thetwo-way table of e-cigarettes andsmoking cessation.Pearson Chi-Square = 1.202, DF = 2, P-Value = 0.548Likelihood Ratio Chi-Square = 1.253, DF = 2, P-Value = 0.534%3D%3D* NOTE * 1 cells with expected counts less than 5

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