MATH 110 Module 8 Exam

M8: Exam- Requires Respondus LockDown Browser  Due No due date  Points 50  Questions 5  Time Limit 90 Minutes  Requires Respondus LockD own Browser This quiz is currently locked. Attempt History Attempt Time Score LATEST Attempt 1 80 minutes 37 out of 50 Score for this quiz: 37 out of 50 Submitted Jul 25 at 1:39pm This attempt took 80 minutes. Question 1 10 / 10 pts You may find the following files helpful throughout the exam: Statistics_Equation_Sheet (Links to an external site.) Standard Normal Table (Links to an external site.) T-Table (Links to an external site.) 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.Answer the following questions: 1. Multiple choice: Which equation would you use to solve this problem? A. B. C. D. 2. List the values you would insert into that equation. 3. State the final answer to the problem Your Answer: 1. The answer is B. 2. The values listed are: n1=420, n2=510, p1=.38, p2=.43, and 99% confidence relate to z=2.58, sample size is greater than 30. 3. When using the formula above (B): Therefore the interval will be : .-0.1333,-0.03326 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:B. So, the interval is (.-.1333,.03326). Question 2 7 / 10 pts You may find the following files helpful throughout the exam: Statistics_Equation_Sheet (Links to an external site.) standard normal table (Links to an external site.) t-table (Links to an external site.) 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 89 day shift nurses and finds that they complete an average (a mean) of 36 rounds per shift with a standard deviation of 6.3 rounds per shift. The nursing supervisor also checks the records of 70 night shift nurses and finds that they complete an average (a mean) of 31 rounds per shift with a standard deviation of 4.2 rounds per shift. a) Find the 99% confidence interval for estimating the difference in the population means (µ1 - µ2). b) Can you be 99% confident that there is a difference in the means of the two populations? Answer the following questions: 1. Multiple choice: Which equation would you use to solve this problem?A. B. C. D. 2. List the values you would insert into that equation. 3. State the final answer to the problem Your Answer: 1. The unswer is A 2. The values are: n1=89; n2=70;s1=6.3;s2=4.2; x1=36;x2=31 3. while using the formula mentioned above (A): After calculation I found: 90% of confidence interval between two population means: Correction number(7,1785 shuld read 7.1788) 99% cobfidence is: (2.8218, 7.1788 since 0 does not lies between 2.8212 the hypothesis will be rejected. However, there is 99% confident that there is a difference in the mean of the two population. Note to professor: Please undesrtand due to limited time sametimes I was not able to use proper elements for example I used (U1) instead of ( ) thank you for understanding. When we look back at table 6.1, we see that 99% confidence corresponds to z=2.58.If we say that the day shift nurses corresponds to population 1 and the night shift nurses corresponds to population 2, then: n1=89, n2=70, s1=6.3, s2=4.2, x̄1=36, x̄2=31 We will use eqn. 8.1: A. b) Since the entire confidence interval is positive, we can be 99% sure that there is a difference in the means of the two populations. Incorrect value for z. This is not a hypothesis test. Question 3 10 / 10 pts You may find the following files helpful in throughout the exam: Statistics_Equation_Sheet (Links to an external site.) standard normal table (Links to an external site.) t-table (Links to an external site.) 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 40 full-time library workers and finds that they re-shelve an average of 185 books per hour with a standard deviation of 17.1 books per hour. The records of 40 part-time library show that they re-shelve an average of 190 books per hour with a standard deviation of 9.2 books per hour.Using a level of significance of α=.10, 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? Answer the following questions: 1. Multiple choice: Which equation would you use to solve this problem? A. B. C. D. 2. List the values you would insert into that equation. 3. State the final answer to the problem Your Answer: 1. The unswer is: C 2. n1=40;n2=40;x1=185;x2=190;sd1=17.1;sd2=9.2 Z, will be found, becouse it is a two=tailed test: p(Zz)= =.05 While utilizing the table: z=-1.645 & z=1.645. 3. While using the formula above (C):The null hypothesis will not be rejected becouse the z score falls between -1.645 and 1.645 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: P(Z z)=.1/2=.05. In the standard normal table, z=-1.645 and z=1.645. We will reject the null hypothesis if the z-score is less than -1.645 or the z-score is greater than 1.645. We now find the z-score: Since the z-score is between -1.645 and 1.645, we do not reject the null hypothesis. Question 4 0 / 10 pts You may find the following files helpful throughout the exam: Statistics_Equation_Sheet (Links to an external site.) standard normal table (Links to an external site.) t-table (Links to an external site.) Consider the following dependent random samples Observations 1 2 3 4 5 6 x-values 205 y-values 219 a) Determine the difference between each set of points, xi - yi b) Do hypothesis testing to see if µd < 0 at the α = .01.Your Answer: a. The diference between each set of points X1-X2 is: X1-y1: 209-210=-1 X2-y2=220-230=-10 x3-y3=214-230=-16 x4-y4=229-238=-9 x5-y5=200-215=-15 x6-y6=205-219=-14 b. Ho:Ud=U1-U2=0 & H1: U1-U2<0, wheras; d=0.01 t= -20.83-0/2.2718 =-4.7671 t=-4.7671; t table value =3.3649, therefore we reject the hypothesis 0.01 less. Since we are testing whether or not µd < 0, then our null and alternate hypothesis will be set as follows: H0: µd = 0 H1: µd < 0 There are 6 data points. So, n=6. This is a left-tailed test. Note that for t.01 = -3.365 for 6-1 = 5. We find the mean in the usual way: The sample standard deviation is given by: Then using the mean, d = -10.833, and the standard deviation, sd= 5.565, that we found above: Since t < t.01, we reject the null hypothesis.What equation are you using? What is your value for d*? What is your calculation of Sd? You are using a t-value that is beyond the level of accuracy in the t-table provided in the exam. Use of outside resources is an academic integrity violation, and is reported to the Academic Review Committee. The full value is deducted on this page. A second violation will result in a zero exam score. A third will result in dismissal from the course. Question 5 10 / 10 pts You may find the following files helpful throughout the exam: Statistics_Equation_Sheet (Links to an external site.) standard normal table (Links to an external site.) t-table (Links to an external site.) A new energy drink is supposed to improve a person’s time in the one mile run. The times, in seconds, of eight runners with and without the drink are given below: Runner 1 2 3 4 5 6 7 8 x-time (before) y-time (after) Find the 95 % confidence interval for mean of the differences, µd. Answer the following questions: 1. Multiple choice: Which equation would you use to solve this problem? A. B.C. D. 2. List the values you would insert into that equation. 3. State the final answer to the problem Your Answer: 1. The answer is: D 2. The values are: n=8 d= -1.125 and Sd=8.7413; t-chart of 95$ is: t=2.365 3While using the formula above (D): 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 D. Quiz Score: 37 out of 50