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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.

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.

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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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, please EXAMPLE 21.7 A random sample of 700 births from local records shows this distribution across the days of the week: 4. STEP Day Mon. Tue. Wed. Thu. Fri. Sat. Sun. Births 110 124 104 94 112 72 84 ..
Transcribed Image Text:Not on the weekend, please EXAMPLE 21.7 A random sample of 700 births from local records shows this distribution across the days of the week: 4. STEP Day Mon. Tue. Wed. Thu. Fri. Sat. Sun. Births 110 124 104 94 112 72 84 ..
Chi-Square Test for in Birth o test whether some from the sample data. For each as the probability of successD STATE: any particular day of the week, but modern births INTERPRETING SIGNIFICANT CHI-SQUARE may not ocess step by step. give the week? STATS IN YOUR WORLD on all days of WEEKEND C null hypothesis is ndows down so that they reflect tal bird strikes. Use software or You work all w get 1/7th of all births. So the rains on the we Ho: Þ1 = P2=P3 = P4 = P5 = P6 = P7 = there really be e in context. %3D behind our per %3D weather is aga 21.2 investigated a genetic model P-value of 0.397. Interpret this heat-resistant phenotypes in F2 %3D 1. on the East Cc States, the ans Going back to that Sundays precipitation t likely explanati pollution from cars and truc for raindrops- Day Mon. Tue. Wed. Thu. Fri. Sat. Sun. Probability P1 P2 p3 P4 P5 P6 P7 The alternative hypothesis simply says that days are not all equally likely for births ARE RESULTS (Ho is not true, not all p; = ÷): delay to causE %3D epartures from a distribution y significant, it is important distribution. Here are three SOLVE: Check the conditions for inference. We will need to get back to this weekend. part later in the chapter. SOLVE: Obtain the test statistic and P-value. The Minitab output in Figure 21.3 shows that the chi-square statistic is X' = 19.12 and the P-value is 0.004. as been established: %3D CONCLUDE: There is very strong evidence that the population distribution of local 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 this finding. Look at the seven X components provided by Minitab in the rightmost column of Figure 21.3. Roughly 40% of the value of X2 (7.84 out of 19.12) comes occur in percents quite dif- h outcomes have more or true? comes contribute the most Irom just one outcome. This points to the most important difference between the observations and the uniform model: a really low count of births on Saturday, rep- Tesenting the largest difference between any observed and expected counts. Most Example 21.6 was highly phids landed on their legs 10; inversely, only 1 aphid n chance alone 10 aphids Minitab eezers are more likely than nanism is at work. Contribution to Chi-Sq Test Observed Proportion Expected 0.142857 Category Monday Tuesday Wednesday Thursday Friday 100 5.76 110 0.16 124 0.142857 0.36 se 0.142857 104 1.44 across 0.142857 94 7.84 112 0.142857 100 s this distribution Saturday 2.56 72 0.142857 Sunday 84 0.142857 < FIGURE output for a : 6 19.12 0.004 Sat. Sun.
Transcribed Image Text:Chi-Square Test for in Birth o test whether some from the sample data. For each as the probability of successD STATE: any particular day of the week, but modern births INTERPRETING SIGNIFICANT CHI-SQUARE may not ocess step by step. give the week? STATS IN YOUR WORLD on all days of WEEKEND C null hypothesis is ndows down so that they reflect tal bird strikes. Use software or You work all w get 1/7th of all births. So the rains on the we Ho: Þ1 = P2=P3 = P4 = P5 = P6 = P7 = there really be e in context. %3D behind our per %3D weather is aga 21.2 investigated a genetic model P-value of 0.397. Interpret this heat-resistant phenotypes in F2 %3D 1. on the East Cc States, the ans Going back to that Sundays precipitation t likely explanati pollution from cars and truc for raindrops- Day Mon. Tue. Wed. Thu. Fri. Sat. Sun. Probability P1 P2 p3 P4 P5 P6 P7 The alternative hypothesis simply says that days are not all equally likely for births ARE RESULTS (Ho is not true, not all p; = ÷): delay to causE %3D epartures from a distribution y significant, it is important distribution. Here are three SOLVE: Check the conditions for inference. We will need to get back to this weekend. part later in the chapter. SOLVE: Obtain the test statistic and P-value. The Minitab output in Figure 21.3 shows that the chi-square statistic is X' = 19.12 and the P-value is 0.004. as been established: %3D CONCLUDE: There is very strong evidence that the population distribution of local 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 this finding. Look at the seven X components provided by Minitab in the rightmost column of Figure 21.3. Roughly 40% of the value of X2 (7.84 out of 19.12) comes occur in percents quite dif- h outcomes have more or true? comes contribute the most Irom just one outcome. This points to the most important difference between the observations and the uniform model: a really low count of births on Saturday, rep- Tesenting the largest difference between any observed and expected counts. Most Example 21.6 was highly phids landed on their legs 10; inversely, only 1 aphid n chance alone 10 aphids Minitab eezers are more likely than nanism is at work. Contribution to Chi-Sq Test Observed Proportion Expected 0.142857 Category Monday Tuesday Wednesday Thursday Friday 100 5.76 110 0.16 124 0.142857 0.36 se 0.142857 104 1.44 across 0.142857 94 7.84 112 0.142857 100 s this distribution Saturday 2.56 72 0.142857 Sunday 84 0.142857 < FIGURE output for a : 6 19.12 0.004 Sat. Sun.
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