21.27 America's favorite alcoholic beverage. Are beer, wine, and liquor equally favored by Americans as the alcoholic drink they consume most often? A 2016 Gallup survey interview a random sample of American adults about their alcohol consumption and preferences. Of the 647 non-abstainers who answered the question about which beverage-beer, wine, liquor-they consume most often, 293 selected beer, 218 selected wine, and the remaining 136 selected liquor. Follow the four-step process as illustrated in Example 21.7. Not on the weekend, pleaseEXAMPLE 21.7A random sample of 700 births from local records shows this distribution acrossthe days of the week:4.STEPDayMon. Tue. Wed. Thu. Fri. Sat. Sun.Births110 124 104 94 112 72 84.. Chi-Square Test for in Birtho test whether somefrom the sample data. For eachas the probability of successDSTATE:any particular day of the week, but modern birthsINTERPRETING SIGNIFICANT CHI-SQUAREmay notocess step by step.givethe week?STATSIN YOURWORLDon all days ofWEEKEND Cnull hypothesis isndows down so that they reflecttal bird strikes. Use software orYou work all wget 1/7th of all births. So therains on the weHo: Þ1 = P2=P3 = P4 = P5 = P6 = P7 =there really bee in context.%3Dbehind our per%3Dweather is aga21.2 investigated a genetic modelP-value of 0.397. Interpret thisheat-resistant phenotypes in F2%3D1.on the East CcStates, the ansGoing back tothat Sundaysprecipitation tlikely explanatipollution fromcars and trucfor raindrops-DayMon. Tue. Wed. Thu. Fri. Sat. Sun.ProbabilityP1 P2 p3 P4 P5 P6 P7The alternative hypothesis simply says that days are not all equally likely for birthsARE RESULTS(Ho is not true, notall p; = ÷):delay to causE%3Departures from a distributiony significant, it is importantdistribution. Here are threeSOLVE: Check the conditions for inference. We will need to get back to thisweekend.part later in the chapter.SOLVE: Obtain the test statistic and P-value. The Minitab output in Figure 21.3shows that the chi-square statistic is X' = 19.12 and the P-value is 0.004.as been established:%3DCONCLUDE: There is very strong evidence that the population distribution oflocal births is not evenly (“uniformly") distributed across all 7 days of the week.The sample data are not consistent with a null hypothesis of equal proportions.Because the test is statistically significant, we can look deeper to explain thisfinding. Look at the seven X components provided by Minitab in the rightmostcolumn of Figure 21.3. Roughly 40% of the value of X2 (7.84 out of 19.12) comesoccur in percents quite dif-h outcomes have more ortrue?comes contribute the mostIrom just one outcome. This points to the most important difference between theobservations and the uniform model: a really low count of births on Saturday, rep-Tesenting the largest difference between any observed and expected counts. MostExample 21.6 was highlyphids landed on their legs10; inversely, only 1 aphidn chance alone 10 aphidsMinitabeezers are more likely thannanism is at work.Contributionto Chi-SqTestObserved Proportion Expected0.142857CategoryMondayTuesdayWednesdayThursdayFriday1005.761100.161240.1428570.36se0.1428571041.44across0.142857947.841120.142857100s this distributionSaturday2.56720.142857Sunday840.142857

Question

21.27 America's favorite alcoholic beverage. Are beer, wine, and liquor equally favored by Americans as the alcoholic drink they consume most often? A 2016 Gallup survey interview a random sample of American adults about their alcohol consumption and preferences. Of the 647 non-abstainers who answered the question about which beverage-beer, wine, liquor-they consume most often, 293 selected beer, 218 selected wine, and the remaining 136 selected liquor. Follow the four-step process as illustrated in Example 21.7. Not on the weekend, pleaseEXAMPLE 21.7A random sample of 700 births from local records shows this distribution acrossthe days of the week:4.STEPDayMon. Tue. Wed. Thu. Fri. Sat. Sun.Births110 124 104 94 112 72 84.. Chi-Square Test for in Birtho test whether somefrom the sample data. For eachas the probability of successDSTATE:any particular day of the week, but modern birthsINTERPRETING SIGNIFICANT CHI-SQUAREmay notocess step by step.givethe week?STATSIN YOURWORLDon all days ofWEEKEND Cnull hypothesis isndows down so that they reflecttal bird strikes. Use software orYou work all wget 1/7th of all births. So therains on the weHo: Þ1 = P2=P3 = P4 = P5 = P6 = P7 =there really bee in context.%3Dbehind our per%3Dweather is aga21.2 investigated a genetic modelP-value of 0.397. Interpret thisheat-resistant phenotypes in F2%3D1.on the East CcStates, the ansGoing back tothat Sundaysprecipitation tlikely explanatipollution fromcars and trucfor raindrops-DayMon. Tue. Wed. Thu. Fri. Sat. Sun.ProbabilityP1 P2 p3 P4 P5 P6 P7The alternative hypothesis simply says that days are not all equally likely for birthsARE RESULTS(Ho is not true, notall p; = ÷):delay to causE%3Departures from a distributiony significant, it is importantdistribution. Here are threeSOLVE: Check the conditions for inference. We will need to get back to thisweekend.part later in the chapter.SOLVE: Obtain the test statistic and P-value. The Minitab output in Figure 21.3shows that the chi-square statistic is X' = 19.12 and the P-value is 0.004.as been established:%3DCONCLUDE: There is very strong evidence that the population distribution oflocal births is not evenly (“uniformly&#34;) distributed across all 7 days of the week.The sample data are not consistent with a null hypothesis of equal proportions.Because the test is statistically significant, we can look deeper to explain thisfinding. Look at the seven X components provided by Minitab in the rightmostcolumn of Figure 21.3. Roughly 40% of the value of X2 (7.84 out of 19.12) comesoccur in percents quite dif-h outcomes have more ortrue?comes contribute the mostIrom just one outcome. This points to the most important difference between theobservations and the uniform model: a really low count of births on Saturday, rep-Tesenting the largest difference between any observed and expected counts. MostExample 21.6 was highlyphids landed on their legs10; inversely, only 1 aphidn chance alone 10 aphidsMinitabeezers are more likely thannanism is at work.Contributionto Chi-SqTestObserved Proportion Expected0.142857CategoryMondayTuesdayWednesdayThursdayFriday1005.761100.161240.1428570.36se0.1428571041.44across0.142857947.841120.142857100s this distributionSaturday2.56720.142857Sunday840.142857< FIGUREoutput for a :6 19.120.004Sat. Sun.

Accepted Answer

Given Information:
Total number of individuals is n=647
Of which 293 selected beer, 218 selected…