M7 Exam

M7: Exam - Requires Respondus LockDown Browser + Webcam Due No due date Points 50 Questions 4 Time Limit 90 Minutes Requires Respondus LockDown Browser Instructions You may only have the following items when taking an exam: computer, 1-2 pieces of blank scratch paper, a pen/pencil, and a calculator. You may ONLY use the equation sheets that are provided WITHIN the exam. The use of printed versions will be considered a violation of the Academic Integrity Policy. Attempt History Attempt Time Score LATEST Attempt 1 55 minutes 45 out of 50 Score for this quiz: 45 out of 50 Submitted Sep 4 at 5:36pm This attempt took 55 minutes. Question 1 15115 pts You may find the following files helpful throughout the exam: Statistics_Equation_Sheet = Answer the following questions:

a) Define the null and alternative hypothesis for the following. Also, explain what it would mean to make a Type | error and explain what it would mean to make a Type Il error. The newspaper in a certain city had a circulation of 20,000 per day in 2010. You believe that the newspaper’s circulation is less than 20,000 today. b) Define the null and alternative hypothesis for the following. Also, explain what it would mean to make a Type | error and explain what it would mean to make a Type Il error. A certain website had 4000 hits per month a year ago. You believe that the number of hits per month is different today. Your Answer: a) Null hypothesis ‘HO D= 20000‘ Alt hypothesis ‘I—Il s u < 20000‘ Type | error: Reject the null hypothesis that the circulation today is 20,000 when the circulation is at least 20,000. Type |l error: Do not reject the null hypothesis when the circulation is less tfian !UUUU g b) \Hy : p = 4000| \Hy @ p # 20000) Type | error: Reject the null hypothesis that the website gets 4000 per month when the website actually gets 4000 per month. Type |l error: Do not reject the null hypothesis when the number of hits is more than or less than 4000 per month.

Is this a right-tailed test, left-tailed test, or two-tailed test. Find the z value based on a level of significance of .03. Your Answer: This is a two tailed test P(Z<-z)=.03/2=.015 and P(Z>z)=.03/2=.015 z=-2.17 and z=2.17 Two-tailed test. P(Z<-z)=.03/2=.015 and P(Z>z)=.03/2=.015. This corresponds to z= -2.17 and z= 2.17. - 4 Question 3 12 / 15 pts You may find the following files helpful throughout the exam: Statistics_Equation_Sheet & (https://previous.nursingabc.com/upload/images/Help_file picture/Statistics Standard Normal Table = (https://previous.nursingabc.com/upload/images/Help_file picture/standardn It is recommended that pregnant women over eighteen years old get 85 milligrams of vitamin C each day. The standard deviation of the population is estimated to be 12 milligrams per day. A doctor is concerned that her pregnant patients are not getting enough vitamin C. So, she collects data on 45 of her patients and finds that the mean vitamin intake of these 45 patients is 81 milligrams per day. Based on a level of significance of a = .02, test the hypothesis. < . 4

Your Answer: \Hy @ = 85| \Hy @ p < 85 P(Z<z)=.02 = 2.236‘ Test stat=2.236 Critical value=.02=2.05 Reject null hypothesis Ho: =85 milligrams per day. H4: u<85 milligrams per day. This is a left-tailed test, so we must find a z that satisfies P(Z<z)=.02. In the standard normal table, we find z.y, = -2.05. For a left-tailed test, we will reject the null hypothesis if the z- score is less than -2.05. We find the z-score: Oz =—=—= 179 ©yn 45 _X—p 81-85 5% ‘= T 179 T~ Notice that since the z-score is less than -2.05, we reject the null hypothesis. Question 4 13 /15 pts You may find the following files helpful throughout the exam:

Suppose p represents the probability that a person is unemployed Ho: p=.05. Hq: p>.05. Since this is a right-tailed test, we must find the z that satisfies P(Z>z)=.03. In the standard normal table, we find that Z..03 =1.88. This iIs a right-tailed test, we reject the null hypothesis if the z-score is greater than 1.88. Recall that for proportions, the mean and standard deviation are found by: u=np=130(.05) = 6.5 os = np(1—p) =/130(.05)(1 —.05) = 2.48 If we find the z-score: X—u 12-65 g 248 = Z — Score = Since this is a right-tailed test, and the z-score is greater than 1.88, we reject the null hypothesis. Quiz Score: 45 out of 50