MATH 110 Module 6 Exam_Statistics (Portage learning) Module 6 Exam

MATH 110 Module 6 Exam_Statistics (Portage learning) Module 6 Exam Exam Page 1 A new drug is introduced that is supposed to reduce fevers. Tests are done with the drug. The drug is given to 60 people who have fevers. It is found that the mean time that it takes for the fever to get back to normal for this test group is 350 minutes with a standard deviation of 90 minutes. Find the 80% confidence interval for the mean time that the drug will take to reduce all fevers for all people. Case 1: large pop and large sample xbar - z(s√n) < u < xbar + z(s√n) n = 60 xbar = 350 s = 90 80% confidence interval (z) = 1.28 350 - 1.28(90√60) < u < 350 + 1.28(90√60) = 335.128, 364.872 335.13 < u < 364.87 Answer Key A new drug is introduced that is supposed to reduce fevers. Tests are done with the drug. The drug is given to 60 people who have fevers. It is found that the mean time that it takes for the fever to get back to normal for this test group is 350 minutes with a standard deviation of 90 minutes. Find the 80% confidence interval for the mean time that the drug will take to reduce all fevers for all people. The drug will ultimately sold to a very large number of people. So, we may assume a very large population. Since the sample size is greater than 30, we should use Case 1: Very large population and very large sample size. We are given the sample mean and sample standard deviation. So, we have n=60 x =¯ 350 s=90 We will use these values in the equation: Exam Page 2 A certain school has 415 male students. The school nurse would like to know how many calories the male students consume per day. So, she samples 40 male students and finds that the mean calorie consumption of the 40 is 2610 calories per day with a standard deviation of 560 calories per day. Find the 80 % confidence interval for mean calorie intake of all the male students in the school. Case 3: Finite population x* -z(s / √n) √[(N-n)/(N-1)] < < x* + z(s/ √n) √[(N-n)/(N-1)] N = 415 n = 40 xbar = 2610 s = 560 80% confidence interval (z) = 1.28 2610 - 1.28(560 / √40) √[(415-40)/(415-1)] < u < 2610 + 1.28(560 / √40) √[(415-40)/(415-1)] = 2502.13, 2717.87 2502.13 < u < 2717.87 For a 80% confidence level, we look at table 6.1 and find that z = 1.28. When we substitute these values into our equation, we get: When we do the arithmetic on the right and left hand side, we get: 335.13 < μ< 364.87. Exam Page 3 In a large city, the city supervisor wants to find the average number of aluminum cans that each family recycles per month. So, she surveys 21 families and finds that these 21 families recycle an average of 160 cans per month with a standard deviation of 45 cans per month. Find the 90 % confidence interval for the mean number of cans that all of the families in the city recycle per month. Case 2: large pop and small sample xbar ± t(s / √n) n = 21 A certain school has 415 male students. The school nurse would like to know how many calories the male students consume per day. So, she samples 40 male students and finds that the mean calorie consumption of the 40 is 2610 calories per day with a standard deviation of 560 calories per day. Find the 80 % confidence interval for mean calorie intake of all the male students in the school. The population is finite. So, we should use Case 3: Finite population. Use: In the statement of the problem, we are given: N=415 n=40 x ¯=2610 s=560 For a 80% confidence level, table 6.1 gives z=1.28 2502.13 < μ< 2217.87 Answer Key In a large city, the city supervisor wants to find the average number of aluminum cans that each family recycles per month. So, she surveys 21 families and finds that these 21 families recycle an average of 160 cans per month with a standard deviation of 45 cans per month. Find the 90 % confidence interval for the mean number of cans that all of the families in the city recycle per month. We have a very large population but the sample size is small. We should use Case 2: Very large population and small sample size. Use When we look at the student’s t chart for 90% confidence (the 90% is found along the bottom row of the chart) and df=21-1=20 (the 20 is found in the leftmost column) we find that t=1.725. So, 160±16.94 143.06<µ<176.94 xbar = 160 s = 45 df = 21-1 = 20 90% confidence interval (t) = 1.725 160 ±1.725(45 / √21) = 143.061,176.939 143.061 < u < 176.939 Answer Key A doctor has a large number of patients and would like to know if his patients prefer to fill in forms electronically or prefer to hand write their forms. So, he surveys 110 patients and finds that 58 prefer electronic forms while 52 prefer hand written forms. Find the 90% confidence limit for the proportion of all patients that prefer the electronic forms. We have an infinite population we will use Case 1: The proportion that prefer the electronic forms is 52/110 =.47 so we set P=.47. As we mentioned previously, we estimate p by P. So, p=.47.A total of 110 patients were surveyed, so n=110. Based on Exam Page 4 A doctor has a large number of patients and would like to know if his patients prefer to fill in forms electronically or prefer to hand write their forms. So, he surveys 110 patients and finds that 52 prefer electronic forms while 58 prefer hand written forms. Find the 90% confidence limit for the proportion of all patients that prefer the electronic forms. Case 1: infinite pop P ± z√(p(1-p) / n) n = 110 p = 52 P = p/n = 52/110 = 0.4727 = 0.47 P = p (estimate) 90% confidence interval (z) = 1.645 0.47 ± 1.645√(0.47(1-0.47) / 110) = 0.3917 , 0.5483 0.3917 - 0.5483 Answer Key Exam Page 5 A shipment of 450 new blood pressure monitors have arrived. Tests are done on 75 of the new monitors and it is found that 15 of the 75 give incorrect blood pressure readings. Find the 80% confidence interval for the proportion of all the monitors that give incorrect readings. case 2: finite pop Answer Key P ± z √[p(1 – p)/n] √[(N-n)/(N-1)] N = 450 n = 75 P/p = p/n = 15/75 = 0.20 P = p (estimate) P = 0.20 p = 0.20 80% confidence interval (z) = 1.28 0.20 1.28 √[0.20(1 – 0.20)/75] √[(450-75)/(450-1)] = 0.14597 , 0.25403 0.14597 - 0.25403 a confidence limit of 90 %, we find in table 6.1 that z=1.645. Now, we can substitute all of these values into our equation: .47 ± .078 So, the 90% confidence limit is: .39 to .548 A shipment of 450 new blood pressure monitors have arrived. Tests are done on 75 of the new monitors and it is found that 15 of the 75 give incorrect blood pressure readings. Find the 80% confidence interval for the proportion of all the monitors that give incorrect readings. We have a finite population, so we will use Case 2: The proportion of the sample that are defective is 15/75 = .2 so we set P=.2. As we mentioned previously, we estimate p by P. So, p=.2. A total of 75 monitors were tested, so n=75. Based on a confidence limit of 80 %, we find in table 6.1 that z=1.28. The total number of monitors is 450, so set N=450. Now, we can substitute all of these values into our equation: .2± .054