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Apr 3, 2024

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2/27/24, 9:42 PM Quizzes - MGMT 650 9046 Statistics for Managerial Decision Making (2242) - UMGC Learning Management System Attempt Score 100/ 100 -100% Overall Grade (Highest Attempt) 100 / 100 - 100 % Question 1 7 |/ 7 points The report of a sample survey of 1014 adults says, with 95% confidence level, between 9% and 15% of all Americans expect to spend more on gifts this year than last year. What does "95% confidence level” mean? v The method used to get the interval, when used over and over again, produces intervals which include the true population proportion 95% of the time. () 9% to 15% of all Americans will spend 95% of what they spent last year. ) 95% of all Americans will spend between 9% and 15% more than what they spent last year. O There is a 95% chance that the percent who expect to spend more is between 9% and 15%. Question 2 7 | 7 points What is the critical value of t for a 90% confidence interval with df = 14? (Assume a 2- tailed test) () 2.14479 O None of the answers match my calculation. () 2.62449 v ) 1.761310 ~1.345030 W Hide question 2 feedback =T.INV.2T(0.1,14) = 1.761 Question 3 7 | 7 points What is the critical value for a 95% confidence interval for the population mean for a sample size of 20? (Population standard deviation is unknown). (Assume a 2-tailed test) () None of the answers match my calculation. v ) 2.09302 () 2.08596 () 1.729133 ) 1.724718 https://learn.umgc.edu/d2l/Ims/quizzing/user/attempt/quiz_start_frame_auto.d21?o0u=930725&isprv=&qi=2119147&cfql=0&dnb=0&fromQB=0&inProgre... 1/6

2/27/24, 9:42 PM Quizzes - MGMT 650 9046 Statistics for Managerial Decision Making (2242) - UMGC Learning Management System W Hide question 3 feedback =T.INV.2T(0.05,19) = 2.09302 Question 4 7 | 7 points A survey of 30 random shoppers found that shoppers will spend between $24.50 and $28.30 at confidence level 95%. A second survey surveyed 25 random shoppers. Which of the following statements is true? () The second interval will be narrower because the sample size is smaller. O The second interval will be wider because it is likely that there are more outliers. v ) The second interval will be wider because the sample size is smaller. () The second interval will be narrower because it is likely that there are less outliers. W Hide question 4 feedback The margin of error depends on confidence level and sample size. In the second sample the sample size decreased. The margin of error of the second sample = tcritical value x sd/SQRT(25) is greater than that of the first sample = tcritical value x sd/SQRT(30). Question 5 7/ | 7 points A survey of 507 citizens found that 364 of them favor a new bill introduced by the city. We want to find a 95% confidence interval for the true proportion of the population who favor the bill. What is the lower limit of the interval? (Round to 3 decimal digits) Answer: 0.679 v W Hide question 5 feedback With p the sample proportion and n the sample size, the lower limit of the confidence interval for the population proportion = p - (Z x SQRT(p(1-p)/n)) Z = NORM.S.INV(.975) = 1.960 Question 6 6 / 6 points Compute a 95% confidence interval for the population mean, based on the sample numbers 21, 28, 33, 34, 25, 26, and 135. https://learn.umgc.edu/d2l/Ims/quizzing/user/attempt/quiz_start_frame_auto.d21?o0u=930725&isprv=&qi=2119147&cfql=0&dnb=0&fromQB=0&inProgre... 2/6

2/27/24, 9:42 PM Quizzes - MGMT 650 9046 Statistics for Managerial Decision Making (2242) - UMGC Learning Management System Which distribution do we use to find the critical value for the margin of error computation? O G CF ) Chi-squared 02 vit W Hide question 6 feedback The t distribution applies to calculation of a confidence interval for the population mean when the population standard deviation is unknown and must be estimated by the sample standard deviation. Incorporating the additional variability of the sample standard deviation results in a flatter probability distribution than the standard normal, called the t distribution with parameter DF (Degrees of Freedom) = sample size minus 1. As sample size n becomes larger the approximation to the standard normal improves (Law of Large Numbers). This distribution is derived in higher level statistics courses. Question 7 6 / 6 points Compute a 95% confidence interval for the population mean, based on the sample numbers 21, 28, 33, 34, 25, 26, and 135. What is the Critical value? (Get 4 decimal digits). () 2.3646 ) 3.7074 () None of the answers match my calculation. ) 2.5706 v ) 2.4469 W Hide question 7 feedback DF = sample size - 1 =T.INV.2T(0.05,6) = 2.4469 Question 8 6 / 6 points Compute a 95% confidence interval for the population mean, based on the sample numbers 21, 28, 33, 34, 25, 26, and 135. What is the margin of error? (Round to two decimal digits) ) 41.83 ) 29.57 https://learn.umgc.edu/d2l/Ims/quizzing/user/attempt/quiz_start_frame_auto.d21?o0u=930725&isprv=&qi=2119147&cfql=0&dnb=0&fromQB=0&inProgre... 3/6

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2/27/24, 9:42 PM Quizzes - MGMT 650 9046 Statistics for Managerial Decision Making (2242) - UMGC Learning Management System ) 38.80 v ) 37.69 . ) None of the answers match my calculation. W Hide question 8 feedback =T.INV.2T(0.05,6)*STDEV.S(21,28,33,34,25,26,135)/SQRT(7) = 37.69 Question 9 7 |/ 7 points Compute a 95% confidence interval for the population mean, based on the sample numbers 21, 28, 33, 34, 25, 26, and 135. Find the lower and upper limits of the interval. () Lower - 18.925; Upper - 50.741 . ) None of the answers match my calculation. v ) Lower - 5.449; Upper - 80.837 () Lower - 20.492; Upper - 40.876 () Lower - 15.449; Upper - 67.174 W Hide question 9 feedback EXAMPLE Question 10 7 |/ 7 points Change the last value to 27 and re-compute the confidence interval. The numbers now are: 21, 28, 33, 34, 25, 26, and 27 g_‘) Lower - 22.32; Upper - 35.861 . ) None of the answers match my calculation. v ) Lower - 23.52; Upper - 31.909 () Lower - 3.52; Upper - 41.929 () Lower - 19.73; Upper - 51.994 W Hide question 10 feedback EXAMPLE Question 11 7 | 7 points An outlier is a value significantly outside the body of data. How does it affect the confidence interval? https://learn.umgc.edu/d2l/Ims/quizzing/user/attempt/quiz_start_frame_auto.d21?o0u=930725&isprv=&qi=2119147&cfql=0&dnb=0&fromQB=0&inProgre... 4/6

2/27/24, 9:42 PM Quizzes - MGMT 650 9046 Statistics for Managerial Decision Making (2242) - UMGC Learning Management System ) An outlier stretches the interval because it decreases the standard deviation () An outlier compacts the interval because it decreases the standard deviation () An outlier compacts the interval because it increases the standard deviation v ) An outlier stretches the interval because it increases the standard deviation. W Hide question 11 feedback Compare the standard deviation of Question 9 with outlier 135 to the standard deviation of Question 10 with replacement 27. The outlier 135 increases the standard deviation. The width of the confidence interval = upper limit - lower limit = 2 x margin of error = 2 x the critical value x standard deviation/sqgrt(n). Increasing the standard deviation increases the width of the confidence interval. Question 12 6 / 6 points The director of admissions at the University of Maryland, University College is concerned about the high cost of textbooks for the students each semester. A sample of 25 students enrolled in the university indicates that x(bar) = $315.40 and s = $43.20. Using a significance level of .1, what are the boundaries of the confidence interval for the population mean? () 297.485 - 327.820 () 282.871 - 338.475 . ) None of the answers match my calculation. ()302.438 - 326.184 v ) 300.618 - 330.182 W Hide question 12 feedback EXAMPLE Question 13 6 / 6 points Does the confidence interval in Question 12 reject mean $3007? () Yes, because 300 is in the confidence interval v ) Yes, because 300 is below the lower limit of the confidence interval. ) No, because 300 is below the lower limit of the confidence interval . ) No, because 300 is in the confidence interval Question 14 7 | 7 points The director of admissions at the University of Maryland, University College is concerned about the high cost of textbooks for the students each semester. A sample https://learn.umgc.edu/d2l/Ims/quizzing/user/attempt/quiz_start_frame_auto.d21?o0u=930725&isprv=&qi=2119147&cfql=0&dnb=0&fromQB=0&inProgre... 5/6

2/27/24, 9:42 PM Quizzes - MGMT 650 9046 Statistics for Managerial Decision Making (2242) - UMGC Learning Management System of 25 students enrolled in the university indicates that x(bar) = $315.40 and s = $75. (Note the sample standard deviation is different from the first part). At the 0.05 level of significance, does the new confidence interval reject the mean $3007? v ) No, because 300 is in the confidence interval () Yes, because 300 is in the confidence interval () Yes, because 300 is below the lower limit of the confidence interval O No, because 300 is below the lower limit of the confidence interval W Hide question 14 feedback EXAMPLE Question 15 7 |/ 7 points Explain the difference between questions 12 and 14. In questions we are asked whether or not there is evidence against a population mean of $300. In Question 12: x(bar) = $315.40; s = $43.2; n = 25 In Question 14: x(bar) = $315.40; s = $75.0; n = 25 v ) The larger standard deviation in Question 14 reflects greater spread in the sample data indicating that the sample may have had some outliers, increasing the width of the confidence interval. O The smaller standard error indicated the sample may have had some outliers and increases the width of the confidence interval. O The larger mean adds uncertainty in the sample. The sample may have had some outliers. () The smaller mean adds uncertainty in the sample. The sample may have had some outliers. W Hide question 15 feedback The larger standard deviation in Question 14 increases the standard error, increasing the margin of error, and increasing the width of the confidence interval such that 300 fell inside the interval. Done https://learn.umgc.edu/d2l/Ims/quizzing/user/attempt/quiz_start_frame_auto.d21?ou=930725&isprv=&qi=2119147&cfql=0&dnb=0&fromQB=0&inProgre... 6/6

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