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Please sign (or type) your name below the following honor statement: I understand that it is a violation of the UMUC Academic Dishonesty and Plagiarism policy to use unauthorized materials or work from others. I promise that I did not discuss any aspect of this exam with anyone other than my instructor. I further promise that I neither gave nor received any unauthorized assistance on this exam, and that the work presented herein is entirely my own. Name Date STAT 200 Introduction to Statistics Name: Final Examination: Fall 2019 OL2 Instructor: Dr. John Wulu Answer Sheet Instructions: This is an open-book exam. You may refer to your text and other course materials as you work on the exam, and you may use a calculator. Record your answers and work in this document. Answer all 20 questions. Make sure your answers are as complete as possible. Show all of your work and reasoning. In particular, when there are calculations involved, you must show how you come up with your answers with critical work and/or necessary tables. Answers that come straight from calculators, programs or software packages without explanation will not be accepted. If you need to use technology to aid in your calculation, you have to cite the source and explain how you get the results. For example, state the Excel function along with the required parameters when using Excel; describe the detailed steps when using a hand-held calculator; or provide the URL and detailed steps when using an online calculator, and so on. Show all supporting work and write all answers in the spaces allotted on the following pages. You may type your work using plain-text formatting or an equation editor, or you may hand- write your work and scan it. In either case, show work neatly and correctly, following standard mathematical conventions. Each step should follow clearly and completely from the previous step. If necessary, you may attach extra pages. You must complete the exam individually. Neither collaboration nor consultation with
others is allowed. It is a violation of the UMUC Academic Dishonesty and Plagiarism
policy to use unauthorized materials or work from others. Your exam will receive a
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Record your answers and work. Problem Number Solution 1 Answer: (a) ii. Density of people per square kilometer. (b) i. All countries in the world. Justifications: (a) A variable is any characteristics, number, or quantity that can be measured or counted. In this instance, we are estimating the mean density of people per square kilometer and is therefore the variable. (b) Population refers to the entire pool from which statistical information can be drawn. In this instance, the World Health Organization selected a sample of 56 countries out of all the countries in the world to estimate the mean density of people per square kilometer. Therefore, all of the countries in the world is the population. 2 Answer: (a) iii. ordinal (b) iv. sample Justifications: (a) Ordinal scales are typically used for non-numeric concepts. In other words, ordinal scales serve as a way to place things such as a scale of satisfaction in order. In this instance, 1 is poor and 5 is excellent. The numbers service as indicators of an order and therefore, the level of measurement is ordinal. (b). In this instance, it is discovered that the 100 surveyed STAT 200 students averaged 12.25 hours a week. The population is all STAT 200 students. This number is a sample because it is a sample mean. 3 Answer: (a) Frequency is found by counting the total number of scores within the given classes. Relative frequency is found by dividing the frequency by the total number
Duplicate question. (a) x-bar = 74. n = 10 x-bar = (∑xi / n) = (95+67+76+47+85+70+87+80+67+72)/ = 746/ (b) s = 13. 7 = sqrt of [(95-74.6)^2 +(67-74.6)^2 +(76-74.6)^2 +(47-74.6)^2 +(85-74.6)^2 +(70-74.6)^2 +(87-74.6)^2 +(80-74.6)^2 +(67- 74.6)^2 +(72-74.6)^2 ]/ 9 (c) I used a handheld calculator and the designated formulas for these types of problem. I also created the standard deviation formula in a separate word document and then inserted it here. I’m unable to add equations to this document and I’m not sure why. 8 Answer: (a) 32,760 ways (b) To find this answer, we multiply: 151413*12 = 32,760 ways First, a chairperson is selected from 15 people = 15 ways Second, a first vice chairperson is selected from 14 people = 14 ways Third, a second vice chairperson is selected from 13 people = 13 ways Last, a secretary is selected from 12 people = 12 ways To find the number of different ways when order selection matters, we multiply the number of possibilities by one another.
This answer was found using the formula shown and a handheld calculator. 9 (a) 15 ways (b) Because the order of book selection does not matter, the combination method is used. # of ways = !/(!(−)!) where n is the sample size (6) and r is the number of books selected (2).
of ways = 6!/(2! * (6-2)!) = 720/(2 * 24) = 720/48 = 15
This answer was found using the formula show and a handheld calculator. 10 (a) P(heart or diamond) = ½ P (hd) = (13 hearts + 13 diamonds)/52 cards = 26 / 52 = 0.5 or ½ (b) P(not a spade) = ¾ P(nas) = (13 hearts + 13 diamonds + 13 clovers)/52 cards = 0.75 or ¾ These answers were found using a handheld calculator. 11 Answer: (a) mean of x = 1. x P(x) xP(x) 0 0.1 0 1 0.27 0. 2 0.4 0. 3 0.18 0. 4 0.05 0.
(a) P(9 < x < 12) = P (9-10)/2 < z < (12-10)/ = P (-0.5 < z < 1) = P (z < 1) – P (z < - 0.5) = 0.8413 - 0. = .5328 probability that a randomly selected pecan tree is between 9 and 12 feet tall. 13 (b) The 75 th^ percentile of the pecan tree height distribution is 11.35 feet. (a) Answer was found using a handheld calculator, designated formula, and a standard normal distribution table. (b) Answer was found with excel function =NORMINV(.75, 10, 2) = 11. (a) SE = 2/(sqrt 64) = 2/8 = 0. (b) P (9.5 < x < 10) = P (9.5-10)/0.25 < z < (10-10)/0. 14 = P (-2 < z < 0) = P (z < 0) – P (z < - 2) = 0.50 – (1 – 0.9772) = 0.50 – 0. =. These answers were found using a handheld calculator, designated formula, and (for part b) a standard normal distribution table. 15 (a) confidence interval estimate = (0.723, 0.777) p = 0. n = 1000 z = 1.96 @ 95% CI CI = p ± z *sqrt of ((p * (1-p)/n) = 0.75 ± 1.96 * sqrt of ((.75 * .25)/1000) = 0.75 ± 1.96 *. = 0.75 ±. (b) We can be 95% confident that the true population portion of adults who believe in global warming are within the provided range (0.723, 0.777).
These answers were found using a handheld calculator, z score table confidence level value for 95%, and designated formula. (a) 90% confidence interval (1917.75, 2082.25) n = 64 x = 2000 ơ = 400 X ± 2 * (ơ/sqrt of n) = 2000 ± 1.645 * (400/sqrt of 64) 16
(b) We can be 90% confident that the mean number of daily parking fees collected in within the range of $1917.75 - $2082.25. Answers found using handheld calculator, z score table confidence level value for 90%, and designated formula.
(a) The appropriate test to use for this analysis is t-test for two dependent samples (matched pairs). The reason it is the correct choice because it allows for discovery of whether the mean difference between the two dependent samples is zero. (b) ii. μ 1 – μ 2 = 0 (μd = 0) (c) a. μ 1 – μ 2 > 0 (μd > 0) (d) t = 1. md = 5 (found using =AVERAGE excel function) n = 5 sd = 7.07 (found using =ST.DEV excel function) t = (md)/(sd/sqrt(n)) = (5)/(7.07 / 2.236) 18 = 5/3. = 1. (e) p-value = 0. Using excel for p-value: =T.DIST(1.5811, 5 - 1, TRUE) (f) With t = 1.581, p > 0.05, fail to reject null hypothesis at 5% level of significance (g) Because the null hypothesis is not rejected, there is not sufficient evidence to support the claim that regular exercise helps weight loss. Answers found using a combination of handheld calculator and excel functions.
Answers found using a combination of handheld calculator and excel functions. (f) With the null hypothesis not rejected, there is sufficient evidence to support the claim that bananas are the favorite fruit more than 50% of the adults. (e) Because the p-value (1.2) is greater than the significance level (0.05), the null hypothesis is not rejected. (b)H 0 : p 0 > 0. H 1 : p 0 < 0. (c) test statistic = 1. Po =. P = .56 (found by dividing 56 adults/100 adults) Z = (p-po)/(sqrt of (po(1-po))/n) = (0.56 – 0.5)/(sqrt of (.5(1-0.5))/100) = 0.06/sqrt of. = .06/. (d) p-value =. If P 0 < 0.5 then p-value = p(z < 1.2) =NORMDIST(1.2,0,1,TRUE) = 0. (a) The appropriate hypothesis test is z-test for proportions. 19
