Question 10 3 pts To test the hypothesis that an intervention involving self-facilitated interprofessional group meetings would result in improvement in wellbeing and satisfaction in the workplace, 60 healthcare workers were interviewed to determine their feelings of satisfaction and wellbeing at work. The healthcare workers were then randomly assigned to two groups. The control group met weekly without any specific guidelines for the discussion content and the treatment group met weekly with a one-page discussion guide addressing themes relevant to challenges in the workplace. After 3 months, researchers re-interviewed subjects and recorded whether each subject felt "more satisfied", "the same" or "less satisfied" at work. The results are shown in the table below. Group More Satisfied The Same Less Satisfied Control 14 12 4 Treatment 17 10 3 Calculate the expected count for subjects in the treatment group who are less satisfied after the study. O 3.5 O 30 0 3 0 4Question 7 3 pts The Virginia Zoo in Norfolk VA keeps track of the percentage of individuals with season passes according to their home neighborhood in Norfolk. A federal grant investigator would like to show that this distribution of neighborhoods for individuals buying season passes is different from what the zoo claims. The investigator randomly sampled 150 individuals entering the zoo with a season pass and recorded the home neighborhood of each individual. Home Neighborhood Broad Creek Ghent Norview Ocean View Outside Norfolk Number of Individuals 37 52 24 21 16 The investigator will use the data to test the zoo's claim which is reflected in the following null hypothesis: Ho: PBC=0.20, PG=0.36, PN=0.18, POV=0.22, PON=0.04 Determine the expected count for each category. O Home Neighborhood Broad Creek Ghent Norview Ocean View Outside Norfolk Number of Individuals 37 152 24 21 16 Expected Count 30 54 27 33 6 O Home Neighborhood Broad Creek Ghent Norview Ocean View Outside Norfolk Number of Individuals 37 52 24 21 16 Expected Count 20 36 18 22 4 Home Neighborhood Broad Creek Ghent Norview Ocean View Outside Norfolk Number of Individuals 37 52 24 21 16 Expected Count 30 30 30 30 30Question 8 3 pts The Virginia 200 in Norfolk VA keeps track of the percentage of individuals with season passes according to their home neighborhood in Norfolk. A federal grant investigator would like to show that this distribution of neighborhoods for individuals buying season passes is different from what the zoo claims. The investigator randomly sampled 150 individuals entering the 200 with a season pass and recorded the home neighborhood of each individual. Home Neighborhood Broad Creek Outside Norfolk The investigator will use the data to test the 200's claim which is reected in the following null hypothesis: H0: ch=o.2o, pG=O.36, pN=0.18, pOV=0.22, pON=O.O4 Which of the following represents the alternative hypothesis for this test? 0 Ha: The season pass holders are not evenly distributed between neighborhoods 0 Ha: At least one of the neighborhood proportions is different from the proportion reported by the 200 0 Ha: The proportions for the different neighborhoods match those reported by the 200 0 Ha: Broad Creek represents the highest proportion of individual season pass holders Question 9 3 pts A major clothing retailer is investigating whether the distribution of the economic status of their customers has changed from last year to this year. Customers are classied as low to moderate- income, middle-income, or high-income. The company conducts a chi-square goodness of t test to investigate whether there is a change in the distribution in customer economic status. The value of the chi-square test statistic was X2 = 7,82 with a corresponding p-value of 0.02. Assuming the conditions of inference were met, which of the following is the correct interpretation of this p-value? C) If the null hypothesis were true, there is a 2 percent chance of obtaining a chi-square value of at least 7.82 C) There is a 2 percent chance that the company's claim is correct. 0 If the null hypothesis were true, there is a 2% chance that the company's claim is correct. 0 There is a 2 percent chance of obtaining a chi-square value of at least 7.82