MATH 110 Module 8 Exam (New, 2023-2024) / MATH110 Module 8 Exam: Portage Learning, Exams of Mathematics

Download MATH 110 Module 8 Exam (New, 2023-2024) / MATH110 Module 8 Exam: Portage Learning and more Exams Mathematics in PDF only on Docsity! MATH 110 Module 8 Exam Exam Page 1 Suppose we have independent random samples of size n1 = 420 and n2 = 510. The proportions of success in the two samples are p1= .38 and p2 = .43. Find the 99% confidence interval for the difference in the two population proportions. n1=420 n2=510 p1=.38 p2=.43 z=2.58 P1 - P2 ± z √p1 (1 - p1) + p2 (1 - p2) n1 n2 .38 - .43 ± 2.58 √.38 (1 - .38) + .43 (1 - .43) 420 510 -.05 ± 2.58(.03227) So the interval is ( -.1332566 .0332566 ) Instructor Comments Suppose we have independent random samples of size n1 = 420 and n2 = 510. The proportions of success in the two samples are p1= .38 and p2 = .43. Find the 99% confidence interval for the difference in the two population proportions. From table 6.1, we see that 99% confidence corresponds to z=2.58. Notice that the sample sizes are each greater than 30, so we may use eqn. 8.2: So, the interval is (.-.1333,.03326). Very good. Answer Key Exam Page 2 In certain hospital, nurses are required to constantly make rounds to check in on all of the patients. The nursing supervisor would like to know if there is a difference between the number of rounds completed per shift by the nurses on the day shift compared to the nurses on the night shift. So, the nursing supervisor checks the records of 61 day shift nurses and finds that they complete an average (a mean) of 39 rounds per shift with a standard deviation of 6.1 rounds per shift. The nursing supervisor also checks the records of 49 night shift nurses and finds that they complete an average (a mean) of 29 rounds per shift with a standard deviation of 5.2 rounds per shift. a) Find the 98% confidence interval for estimating the difference in the population means (µ1 - µ2). b) Can you be 98% confident that there is a difference in the means of the two populations? a) z=2.33 H0: u1-u2=0 H1:u1-u2≠0 Two tailed test: a=.05 P(Z<z)= a/2 P(Z>z)=a/2 P(Z<z)=.05/2 = .025 P(Z>z)=a/2 = .05/2= .025 z=-1.96 and z=1.96 Z-Score: z=(xbar1 - xbar2) - 0 √s1 2 + s2 2 √n1 n2 z = (113- 110) - 0 A head librarian supervises a number of libraries in a large county. He wants to know if full-time library workers and part-time library workers re-shelve books at the same rate. So, he checks the records of 42 full-time library workers and finds that they re-shelve an average of 113 books per hour with a standard deviation of 8.1 books per hour. The records of 42 part-time library show that they re- shelve an average of 110 books per hour with a standard deviation of 10.1 books per hour. Using a level of significance of α=.05, is there enough evidence to indicate a difference in the mean number of books re-shelved by full-time workers compared to part-time workers? The null hypothesis is that there is no difference between the mean number of books re-shelved by the full-time and part-time workers: H0 : µ1 - µ2 = 0 H1 : µ1 - µ2 ≠0. Since this is a two-tailed test, we must find the z that satisfies: √8.1 2 + 10.1 2 √42 42 z=3/1.9978 z=1.5017 The z-score is between -1.96 and 1.96 so we do not reject the null hypothesis, there is no difference in the mean number of books re-shelved by full time and part time workers. Instructor Comments Very good. Answer Key Exam Page 4 Consider the following dependent random samples Observations 1 2 3 4 5 6 x-values 22.0 21.0 18.1 19.6 13.2 17.5 y-values 23.1 21.7 18.7 20.7 14.7 16.4 a) Determine the difference between each set of points, xi - yi b) Do hypothesis testing to see if µd < 0 at the α = .05. H0: ud=0 H1: ud <0 n=6 and this is a left tailed test and note for t.05= -2.015 for 6-1=5 P(Z z)=.05/2=.025. In the standard normal table, z=-1.96 and z=1.96. We will reject the null hypothesis if the z-score is less than -1.96 or the z-score is greater than 1.96. We now find the z-score: Since the z-score is between -1.96 and 1.96, we do not reject the null hypothesis. y-time (after) 260 277 281 279 260 242 244 250 Find the 95 % confidence interval for mean of the differences, µd. n=8 d= -1.125 sd= 8.7413 DOF= 8-1=7 t=2.365 dbar- t sd <ud < dbar + t sd √n √n -1.125- 2.365 8.714 <ud < -1.125+ 2.365 8.714 √8 √8 -1.125 - 7.31 <ud < -1.125 + 7.31 The times, in y-time (after) 260 277 281 279 260 242 244 250 Find the 95 % confidence interval for mean of the differences, µd. Note that n=8. We will define , di = xi - yi. After doing the appropriate calculations, we find that d=-1.125 sd= 8.7413. When we look at the student’s t chart for 95% confidence (the 95% is found along the bottom row of the chart) and DOF=8-1=7 (the df=7 is found in the leftmost column) we find that t=2.365. Then -8.434. <ud < 6.1838 Instructor Comments Very good. Answer Key A new energy drink is supposed to improve a person’s time in the one mile run. seconds, of eight runners with and without the drink are given below: Runner 1 2 3 4 5 6 7 8 x-time (before) 267 283 270 265 261 247 250 241