STAT 200: Introduction to Statistics Final Examination, Spring 2017 OL1/US1 Page 1 of 8 STAT 200 OL1/US1 Sections Final Exam Spring 2017 The final exam will be posted at 12:01 am on March 3, and it is due at 11:59 pm on March 5, 2017. Eastern Time is our reference time. 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. 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. Answer all 20 questions. Make sure your answers are as complete as possible. Show all of your supporting work and reasoning. Answers that come straight from calculators, programs or software packages without any explanation will not be accepted. If you need to use technology (for example, Excel, online or handheld calculators, statistical packages) to aid in your calculation, you must cite the sources and explain how you get the results. Record your answers and work on the separate answer sheet provided. This exam has 20 questions; 5% for each question. You must include the Honor Pledge on the title page of your submitted final exam. Exams submitted without the Honor Pledge will not be accepted. STAT 200: Introduction to Statistics 1. Final Examination, Spring 2017 OL1/US1 Page 2 of 8 True or False. Justify for full credit. (a) (b) (c) (d) If A and B are any two events, then P(A AND B) = P(A) + P(B). If the variance of a data set is 0, then all the observations in this data set must be identical. A normal distribution curve is always symmetric to its mean. When plotted on the same graph, a distribution with a mean of 60 and a standard deviation of 5 will look more spread out than a distribution with a mean of 40 and standard deviation of 8. (e) In a left-tailed test, the value of the test statistic is -2. The test statistic follows a distribution with the distribution curve shown below. If we know the shaded area is 0.03, then we have sufficient evidence to reject the null hypothesis at 0.05 level of significance. 2. Choose the best answer. Justify for full credit. (a) A study was conducted at a local college to analyze the average GPA of students graduated from UMUC in 2016. 100 students graduated from UMUC in 2016 were randomly selected, and the average GPA for the group is 3.5. The value 3.5 is a (i) statistic (ii) parameter (iii) cannot be determined (b) The hotel ratings are usually on a scale from 0 star to 5 stars. The level of this measurement is (i) interval (ii) nominal (iii) ordinal (iv) ratio (c) On the day of the last presidential election, UMUC News Club organized an exit poll in which specific polling stations were randomly selected and all voters were surveyed as they left those polling stations. This type of sampling is called: (i) (ii) cluster convenience STAT 200: Introduction to Statistics (iii) (iv) 3. Final Examination, Spring 2017 OL1/US1 Page 3 of 8 systematic stratified The frequency distribution below shows the distribution for IQ scores for a random sample of 1000 adults. (Show all work. Just the answer, without supporting work, will receive no credit.) IQ Scores Frequency 50 - 69 23 70 - 89 249 Cumulative Relative Frequency 0.722 90 -109 110 - 129 (a) (b) (c) 4. 130 - 149 25 Total 1000 Complete the frequency table with frequency and cumulative relative frequency. Express the cumulative relative frequency to three decimal places. What percentage of the adults in this sample has an IQ score of at least 110? Which of the following IQ score groups does the median of this distribution belong to? 70 - 89, 90 - 109, or 110 - 129? Why? The five-number summary below shows the grade distribution of a STAT 200 quiz for a sample of 100 students. Answer each question based on the given information, and explain your answer in each case. (a) (b) What is the range in the grade distribution? Which of the following score bands has the most students? (i) 30 - 50 STAT 200: Introduction to Statistics (c) Final Examination, Spring 2017 OL1/US1 Page 4 of 8 (ii) 50 - 70 (iii) 70 - 90 (Iv) Cannot be determined How many students in the sample are in the score band between 70 and 100? 5. Consider selecting one card at a time from a 52-card deck. What is the probability that the first card is an ace and the second card is also an ace? (Note: There are 4 aces in a deck of cards) (Show all work. Just the answer, without supporting work, will receive no credit.) (a) (b) Assuming the card selection is with replacement. Assuming the card selection is without replacement. 6. There are 2000 students in a high school. Among the 2000 students, 1500 students have a laptop, and 900 students have a tablet. 500 students have a laptop and a tablet. Let L be the event that a randomly selected student has a laptop, and T be the event that a randomly selected student has a tablet. (Show all work. Just the answer, without supporting work, will receive no credit.) (a) (b) Provide a written description of the complement event of (L OR T). What is the probability of complement event of (L OR T)? 7. Consider rolling a fair 6-faced die twice. Let A be the event that the sum of the two rolls is at most 6, and B be the event that the first one is an even number. (a) What is the probability that the sum of the two rolls is at most 6 given that the first one is an even number? Show all work. Just the answer, without supporting work, will receive no credit. Are event A and event B independent? Explain. (b) 8. Answer the following two questions. (Show all work. Just the answer, without supporting work, will receive no credit). (a) The steering committee of UMUC Green Solutions Team consists of 3 committee members. 10 people are interested in serving in the committee. How many different ways can the committee be selected? A bike courier needs to make deliveries at 6 different locations. How many different routes can he take? (b) 9. Let random variable x represent the number of girls in a family of three children. (a) Construct a table describing the probability distribution. (5 pts) STAT 200: Introduction to Statistics Final Examination, Spring 2017 OL1/US1 Page 5 of 8 (b) Determine the mean and standard deviation of x. (Round the answer to two decimal places) 10. Mimi just started her tennis class three weeks ago. On average, she is able to return 20% of her opponent's serves. Assume her opponent serves 10 times. (a) Let X be the number of returns that Mimi gets. As we know, the distribution of X is a binomial probability distribution. What is the number of trials (n), probability of successes (p) and probability of failures (q), respectively? Find the probability that that she returns at least 1 of the 10 serves from her opponent. (b) 11. A research concludes that the number of hours of exercise per week for adults is normally distributed with a mean of 4 hours and a standard deviation of 3 hours. Show all work. Just the answer, without supporting work, will receive no credit. (a) Find the 75th percentile for the distribution of exercise time per week. (round the answer to 2 decimal places) What is the probability that a randomly selected adult has more than 7 hours of exercise per week? (round the answer to 4 decimal places) (b) 12. (a) (b) Based on the performance of all individuals who tested between July 1, 2012 and June 30, 2015, the GRE Verbal Reasoning scores are normally distributed with a mean of 150 and a standard deviation of 8.45. (https://www.ets.org/s/gre/pdf/gre_guide_table1a.pdf). Show all work. Just the answer, without supporting work, will receive no credit. Consider all random samples of 36 test scores. What is the standard deviation of the sample means? What is the probability that 36 randomly selected test scores will have a mean test score that is between 148 and 152? 13. An insurance company checks police records on 500 randomly selected auto accidents and notes that teenagers were at the wheel in 80 of them. Construct a 95% confidence interval estimate of the proportion of auto accidents that involve teenage drivers. Show all work. Just the answer, without supporting work, will receive no credit. 14. A city built a new parking garage in a business district. For a random sample of 60 days, daily fees collected averaged $2,000, with a standard deviation of $400. Construct a 95% confidence interval estimate of the mean daily income this parking garage generates. Show all work. Just the answer, without supporting work, will receive no credit. 15. ABC Company claims that the proportion of its employees investing in individual investment accounts is higher than national proportion of 45%. A survey of 200 employees in ABC Company indicated that 100 of them have invested in an individual investment account. Assume we want to use a 0.10 significance level to test the claim. (a) Identify the null hypothesis and the alternative hypothesis. STAT 200: Introduction to Statistics (b) (c) (d) 16. (c) (d) 17. (a) (b) (c) (d) Page 6 of 8 Determine the test statistic. Show all work; writing the correct test statistic, without supporting work, will receive no credit. Determine the P-value for this test. Show all work; writing the correct P-value, without supporting work, will receive no credit. Is there sufficient evidence to support ABC Company's claim that the proportion of its employees investing in individual investment accounts is higher than 45%? Explain. Mimi was curious if regular excise really helps weight loss, hence she decided to perform a hypothesis test. A random sample of 5 UMUC students was chosen. The students took a 30minute exercise every day for 6 months. The weight was recorded for each individual before and after the exercise regimen. Does the data below suggest that the regular exercise helps weight loss? Assume Mimi wants to use a 0.05 significance level to test the claim. Subject 1 2 3 4 5 (a) (b) Final Examination, Spring 2017 OL1/US1 Weight (pounds) Before After 150 130 170 160 185 180 160 160 200 180 Identify the null hypothesis and the alternative hypothesis. Determine the test statistic. Show all work; writing the correct test statistic, without supporting work, will receive no credit. Determine the p-value. Show all work; writing the correct critical value, without supporting work, will receive no credit. Is there sufficient evidence to support the claim that regular exercise helps weight loss? Justify your conclusion. In a pulse rate research, a simple random sample of 100 men results in a mean of 80 beats per minute, and a standard deviation of 11 beats per minute. Based on the sample results, the researcher concludes that the pulse rates of men have a standard deviation greater than 10 beats per minutes. Use a 0.05 significance level to test the researcher's claim. Identify the null hypothesis and alternative hypothesis. Determine the test statistic. Show all work; writing the correct test statistic, without supporting work, will receive no credit. Determine the P-value for this test. Show all work; writing the correct P-value, without supporting work, will receive no credit. Is there sufficient evidence to support the researcher's claim? Explain. STAT 200: Introduction to Statistics 18. (c) (d) 19. (b) 1 230 2 170 3 210 4 180 5 210 Identify the null hypothesis and the alternative hypothesis. Determine the test statistic. Show all work; writing the correct test statistic, without supporting work, will receive no credit. Determine the P-value. Show all work; writing the correct P-value, without supporting work, will receive no credit. Is there sufficient evidence to support the manager's claim that the five types of teddy bears are equally popular? Justify your answer. A STAT 200 instructor believes that the average quiz score is a good predictor of final exam score. A random sample of 10 students produced the following data where x is the average quiz score and y is the final exam score. x y (a) Page 7 of 8 The UMUC MiniMart sells five different types of teddy bears. The manager reports that the five types are equally popular. Suppose that a sample of 1000 purchases yields observed counts 230, 170, 210, 180, and 210 for types 1, 2, 3, 4, and 5, respectively. Use a 0.05 significance level to test the claim that the five types are equally popular. Show all work and justify your answer. Type Number of Teddy Bears (a) (b) Final Examination, Spring 2017 OL1/US1 80 72 93 95 50 75 60 68 100 90 40 35 85 83 70 60 75 77 85 85 Find an equation of the least squares regression line. Show all work; writing the correct equation, without supporting work, will receive no credit. Based on the equation from part (a), what is the predicted final exam score if the average quiz score is 90? Show all work and justify your answer. 20. A study of 10 different weight loss programs involved 500 subjects. Each of the 10 programs had 50 subjects in it. The subjects were followed for 12 months. Weight change for each subject was recorded. We want to test the claim that the mean weight loss is the same for the 10 programs. (a) Complete the following ANOVA table with sum of squares, degrees of freedom, and mean square (Show all work): Mean Square Sum of Squares Degrees of Freedom Source of Variation (MS) (SS) (df) Factor (Between) Error (Within) Total 27.82 985.07 499 N/A STAT 200: Introduction to Statistics Final Examination, Spring 2017 OL1/US1 Page 8 of 8 (b) Determine the test statistic. Show all work; writing the correct test statistic, without supporting work, will receive no credit. (c) Determine the P-value. Show all work; writing the correct P-value, without supporting work, will receive no credit. Is there sufficient evidence to support the claim that the mean weight loss is the same for the 10 programs at the significance level of 0.05? Explain. (d) STAT 200 Final Examination Spring 2017OL1/US1 Page 1 of 21 STAT 200 Introduction to Statistics Name______________________________ Final Examination: Spring 2017 OL1/US1 Instructor __________________________ 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 handwrite 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 zero grade unless you complete the following honor statement. 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 Final Examination Spring 2017OL1/US1 Page 2 of 21 Record your answers and work. Proble m Numbe r 1 Answer: Solution (a) FALSE (b)TRUE (c)TRUE (d)FALSE (e) TRUE Justification: a) False If A and B are independent events, then P (A and B) P (A) + P (B). Instead it should be P(A and B) = P (A) * P(B). b) True Identical numbers doesn't have any variation and variance will be equal 0 c) True A normal distribution has a symmetric bell-shaped curved centered about its mean. The standard deviation then determines its spread d) False The value of a standard deviation doesn't affect the spread of a distribution but instead z score reveals the true spread of distribution. e) True The shaded area, P(z > 2) = 0.03 is right-tailed. For left-tailed, P(z< -2) = 0.03. Here pvalue is less than at 0.05 level of significance hence we fail to reject null hypothesis STAT 200 Final Examination Spring 2017OL1/US1 Page 3 of 21 Answer: (a) Statistic (b) Ordinal (c) cluster Justification: 2 1. (a) (i) Statistic A statistic describe the sample while a parameter describe the entire population. Since only 100 students were randomly selected, they form a sample and the average GPA is a statistic. (b) (iii) Ordinal A ordinal variable is a variable which the order matters but the difference in values of the order is not meaningful. The hotel rating ranks the hotel from 0 star to 5 stars, but the different between 5 stars and 4 stars cannot be measured or expressed by the scale. (c) cluster STAT 200 Final Examination Spring 2017OL1/US1 Page 4 of 21 Answer: (a) IQ Scores Frequenc y 50 - 69 23 70 - 89 249 Cumulative Relative Frequency 90 -109 0.722 110 - 129 130 - 149 25 Total 1000 3 (b) (c) Work for (a) and (b): a) Relative Frequency ( RF )= frequency of the particular class ( f ) total class frequency ( n ) Frequency ( f ) =Relative frequency ( RF ) total class frequency (n) IQ Scores Frequency Relative Frequency 50 - 69 23 23 =0.023 1000 70 - 89 249 249 =0.249 1000 Cumulative Relative Frequency 0.023 0.023+0.249 0 .272 STAT 200 Final Examination 90 - 109 Spring 2017OL1/US1 0.45 1000 450 Page 5 of 21 0.7220.272 0.722 0.45 10002324945025 110 - 129 253 130 - 149 Total 25 1000 Hence the complete table is given as IQ Scores Frequency 50 - 69 70 - 89 90 - 109 110 - 129 130 - 149 Total 23 249 450 253 25 1000 253 =0.253 1000 0.722+0.253 25 =0.025 1000 0.975+0.025 1000 =1 1000 Cumulative Relative Frequency 0.023 0.272 0.722 0.975 1 b) Percentage of adult with IQ scores at least 110=P(IQ Scores 4 ) P( IQ Scores=110 - 129)+ P(IQ Scores=130 - 149) 0.253+0.025 c) IQ Scores 50 - 69 70 - 89 90 - 109 110 - 129 130 - 149 Total 0.278 Frequency 23 249 450 253 25 1000 27.8 Cumulative Frequency 23 272 722 975 1 0 . 975 1 STAT 200 Final Examination Median position= Spring 2017OL1/US1 Page 6 of 21 N + 1 1000+1 = =500.5 Cumulative frequency of 70 - 89 group 2 2 is 272 and Cumulative frequency of the 90 - 109 group is 722 implying that the 500.5th value lies in 90 - 109 group Hence the median belongs the 90129 group Answer: (a) 90 (b) (c) 50 Justification: a) Range=MaximumMinimum 10010 90 4 b) 30 - 50 will contain less than 25% since it's slightly more than the tail to Q1 50 - 70 will contain 25% of values since it ranges from Q1 and median (Q2) 70 - 90 contain more than 25% as it is after median to slightly more than Q3 Hence 70 - 90 score band has the most students c) Score band 70 - 100 contain 50% since it's between maximum value and median 50 100 Number of students=100 50 50 100 STAT 200 Final Examination Spring 2017OL1/US1 Page 7 of 21 Answer: 1 (a) 69 (b) 1 221 Work for (a) and (b): a) 5 P ( withreplacement )= 4 4 1 = 52 52 69 b) P ( without replacement )= 4 3 1 = 52 51 221 Answer: (a) Complement event of (L OR T) is the even that randomly selected student does not take laptop ortablet. 6 (b) 0.65 Work for (b): P(L OR T)C = 1- P(L or T) = 1- (1500+900-500)/2000 = 0.65 7 Answer: (a) 1/12 (b) No Work for (a) and (b): a) P(A|B) = P(Sum is 7| 1st one is 1) + P(Sum is 7|1st one is 3) + P(Sum is 7|1st one is 5) = P(2nd one is 6 and 1st one is 1)+P(2nd one is 4 and 1st one is 3) +P(2nd one is 2 STAT 200 Final Examination Spring 2017OL1/US1 Page 8 of 21 and 1st one is 5) = 1/36+1/36 +1/36 = 1/12 b) P(A) = P(Sum is 7) P(A|B) = 1/12 Thus the events are not independent Answer: (a) 455 (b) 720 Work for (a) and (b): 8 (a) nCr = n!/[r!(n-r)!] 15C3 = 15!/[3!(15-3)!] =455 (b) The amount of different routes available, given that the courier makes 6 stops along the way, is 6x5x4x3x2x1, or 6!, or 720 9 Answer: (a) x 0 1 2 3 Total (b) mean =1.5, s,d = 0.866 Work for (a) and (b) P(x) 1/8 3/8 3/8 1/8 1 STAT 200 Final Examination 10 Spring 2017OL1/US1 Answer: (a) 0.80 (b) 0.99999992 Work for (b) : a) n = 10, p = 0.20 and q = 1-0.20 = 0.80 b) X = number of serves that Mimi returns P (X1) = 1- P(X=0) 10! ( 0.20 )10( 0.8) =1 - 0! ( 100 ) ! =0.99999992 Page 9 of 21 STAT 200 Final Examination Spring 2017OL1/US1 Page 10 of 21 Answer: (a) 6.0235 hours (b) 0.1587 Work for (a) and (b): Mean, = 4 hours Standard deviation, = 3 hours a) 75 th percentile P ( Z > z )=0.75 11 z= X X = +Z Z =0.6745 [ Standard Normal Table ] 4 + ( 0.6745 x 3 ) 6.0235 hours b) X =7 hours z= X 74 = =1 3 P ( X >1 )=P ( z> 1 ) [ Standard Normal Table] 0.1587 STAT 200 Final Examination Spring 2017OL1/US1 Page 11 of 21 Answer: (a) 1.4083 (b) 0.8444 Work for (a) and (b): Mean, =150 Standard deviation, =8.45 a) standard deviation of sample means= b) z= 12 Sample n=36 8.45 = =1.4083 n 36 X /n For X = 148 148150 z= =1.420 8.45 36 For X = 152 152150 z= =1.420 8.45 36 P (148< X <152 )=P (1.420< z<1.420 ) 0.92220.0778 0.8444 P ( z <1.420 ) P ( z <1.420 ) STAT 200 Final Examination Spring 2017OL1/US1 Page 12 of 21 Answer: [ 0.1279,0.1921 ] Work: n=500 13 x 80 p= = =0.16 n 500 x=80 Standard error of the sample proportion S p= pq 0.16 0.84 = =0.01639512123 n 500 Margin of error , E=Z S p 95 CI = p E at =0.05, Z =1.96 2 1.96 0.01639512123 2 14 q=1 p=10.16=0.84 0.0321 [0.1279, 0.1921] 0.16 0.0321 Answer: [1898.79, 2101.21] Work: Sample n=60 sample mean, X =$ 2000 Standard error of mean, S X = sample standard deviation , S=$ 400 S 400 = =51.6397795 n 60 at =0.05, Z =1.96 2 Margin of error , E=Z S X 2 95 CI = X E 1.96 51.6397795 $ 2000 101.21 101.21 [1898.79,2101.21] STAT 200 Final Examination 15 Spring 2017OL1/US1 Page 13 of 21 Answer: . Alternative hypothesis , H 1 : p>0.45 (a) (b) 1.4213 (c) 0.0776 (d) =0.10 p=0.776< =0.10, h ence we reject null h ypot h esis We have sufficient evidence to support ABC Company's claim that the proportion of its employees investing in individual investment accounts is higher than 45% at 0.10 significance level since p<0.10 Work for (b) and (c): a) Nullhypothesis , H 0 : p 0.45 b) n=200 Test statistic c) x 100 ^p= = =0.5 n 200 x=100 Standard error Alternative hypothesis , H 1 : p>0.45 SE= z= ^p p SE pq 0.45 x 0.55 = =0.0351781182 n 200 0.50.45 0.0351781182 1.4213 STAT 200 Final Examination Spring 2017OL1/US1 pvalue=P ( z> 1.4213 ) 16 Page 14 of 21 10.9224 1P ( z 1.4213 ) 0.0776 Answer: (a) Nullhypothesis , H 0 : d 0 Alternative hypothesis , H 1 : d <0 (b) 2.7500 (c) 0.0257 (d) =0.05 p=0.0257< > 10 (b) 119.79 (c) 0.0761 (d). =0.05 p=0.0761< =0.05, hence we reject null hypothesis There is sufficient evidence to support the researcher's claim that the pulse rates of men have a standard deviation greater than 10 beats per minutes at 0.05 significance level since p<0. 05 Work for (b) and (c): 17 b) n=100 S=11 Test statistic 2 X = ( n1 ) S2 2 ( 1001 ) 112 2 10 119.79 c) Degrees of freedom, df =n1 1001 99 get pvalue , we canuse excel function Pvalue=CHISQ . DIST . RT (119.79, 99 ) 0.0761 STAT 200 Final Examination 18 Spring 2017OL1/US1 Page 17 of 21 Answer: (a) We test this with chisquare goodness of fit test . Null Hypothesis :The five types of teddy bears are equally popular . Alternative hypothesis :The five types of teddy bears are not equally equally popular Expressing Mathematically , H 0 : p1 =0.2, p2 =0.2, p3 =0.2, p 4=0.2, p5=0.2 H a : Atleast one p i do not match the claimed value (b) 12 (c) 0.0174 (d) =0.05 p=0.0174 < =0. 05, hence we reject null hypothesis There is sufficient evidence to support the manager's claim that the five types of teddy bears are equally popular at 0.05 significance level since p<0. 05 Work for (b) and (c): b) n=1000 1 The five types are equally popular hence p= =0.2 5 Observed (O) 230 Expected (E) np=1000 0.2 OE (OE)2 E 30 4.5 -30 4.5 200 170 np=1000 0.2 STAT 200 Final Examination Spring 2017OL1/US1 Page 18 of 21 200 np=1000 0.2 210 10 0.5 -20 2 10 0.5 200 np=1000 0.2 180 200 np=1000 0.2 210 200 Test statistic , X 2 = ( OE )2 E 4.5+ 4.5+0.5+2+0.5 12 c) Degrees of freedom, df =k 1 51 get pvalue , we canuse excel function 4 Pvalue=CHISQ . DIST . RT (10.65, 4 ) 0.0174 19 Answer: (a) y=0.7571+18.12602 X (b) The predicted value of y if x = 90 is 1632.0989 Work for (a) and (b): STAT 200 Final Examination Spring 2017OL1/US1 Page 19 of 21 (a) The estimated regression equation is Consider the table below x y 80 72 93 95 50 75 60 68 100 90 40 35 85 83 70 60 75 77 85 85 X = 738 Y = 740 y i=b 0+ bi X i +e i X2 6400 8649 2500 3600 10000 1600 7225 4900 5625 7225 2 X =57724 XY 5760 8835 3750 4080 9000 1400 7055 4200 5775 7225 XY = 57080 From the above table, Sample size, n = 10 Slope ( b ) = ( n XY - ( X )( Y ) ) 2 2 (n X - ( X ) ) Intercept ( a )= ( Y b ( X ) ) n 10 ( 57080 )(738)(740) 10 (57724 )7382 7400.7571 (738 ) 10 18.12602 The equation of least square regression line is: y=0.7571+18.12602 X (b) The predicted value of y if x = 90 y=0.7571+18.12602(90) 0.7571+1631.3418 The predicted value of y if x = 90 is 1632.0989 1632.0989 0.7571 STAT 200 Final Examination 20 Spring 2017OL1/US1 Page 20 of 21 Answer: (a) Source of Variation Factor (Between) Error (Within) Total Sum of Squares (SS) Mean Square (MS) 27.82 Degree of Freedom (df) 9 957.25 490 1.9536 985.07 499 3.0911 (b) 1.5823 (c) 0.1175 (d) =0.05 p=0.1175 > =0. 05, hence we fail reject null hypothesis There is no sufficient evidence to support the claim that the mean weight loss is the same for the 10 programs at the significance level of 0.05 since p>0. 05 Work for (a), (b) and (c): a) SS Error =SS Total SSfactor 985.0727.82 10 weight loss :df factor=10 - 1=9 957.25 STAT 200 Final Examination Spring 2017OL1/US1 500 subjects : df tot =500 - 1=499 df factor +df error =df tot Therefore , df error =df tot df factor =4999=490 MS factor = SS factor 27.82 = =3.0911 df factor 9 MS Error = SS Error 957.25 = =1.9536 df Error 490 b) Fvalue = MS factor 3.0911 = =1.5823 MS Error 1.9536 c) P-value is calculated using the excel function pvalue=fdist ( 2.4514, 4,245 )=0.1175 Page 21 of 21 \fAnswer wrong 5 A - change to 1/169 6b - change to 0.05 7 a- fully wrong, atmost 6 means 2,3,4,5 and 6 and first one is even not odd 8a it is 10C3 not 15C3 10b -wrong formula, 0.2^0 and 0.8^10, not other way around 14a should use t-distribution not normal Explanation wrong or missing 1A, they need not be independent 1D more spread out because of higher standard deviation 2C write definition of cluster sampling as explanation 7b Fix 7A and this will get corrected Others In 16,17,18,19,20, there arevsome symbols and numbers that are replaced by square boxes possibly due to error in copy pa Typo in answer 18a, there are two "equally" STAT 200 Final Examination Spring 2017OL1/US1 Page 1 of 22 STAT 200 Introduction to Statistics Name______________________________ Final Examination: Spring 2017 OL1/US1 Instructor __________________________ 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 handwrite 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 zero grade unless you complete the following honor statement. 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 Final Examination Spring 2017OL1/US1 Page 2 of 22 Record your answers and work. Proble m Numbe r 1 Answer: Solution (a) FALSE (b)TRUE (c)TRUE (d)FALSE (e) TRUE Justification: a) False P(AUB) =P(A)+P(B)-P(A and B) ,For two events A and B,P(A and B) =P(A) +P(B) , the union must be equal to 0 which is not possible. Therefore it is false. b) True Identical numbers doesn't have any variation and variance will be equal 0 c) True A normal distribution has a symmetric bell-shaped curved centered about its mean. The standard deviation then determines its spread d) False The value of a standard deviation reveals more spread of distribution. e) True The shaded area, P(z > 2) = 0.03 is right-tailed. For left-tailed, P(z< -2) = 0.03. Here pvalue is less than at 0.05 level of significance hence we fail to reject null hypothesis STAT 200 Final Examination Spring 2017OL1/US1 Page 3 of 22 Answer: (a) Statistic (b) Ordinal (c) cluster Justification: 1. (a) (i) Statistic A statistic describe the sample while a parameter describe the entire population. Since only 100 students were randomly selected, they form a sample and the average GPA is a statistic. 2 (b) (iii) Ordinal A ordinal variable is a variable which the order matters but the difference in values of the order is not meaningful. The hotel rating ranks the hotel from 0 star to 5 stars, but the different between 5 stars and 4 stars cannot be measured or expressed by the scale. (c) cluster It is an example of cluster sampling where we are selecting few polling station from all polling station available and then selecting all voters from the given polling station. STAT 200 Final Examination Spring 2017OL1/US1 Page 4 of 22 Answer: (a) IQ Scores Frequenc y 50 - 69 23 70 - 89 249 Cumulative Relative Frequency 90 -109 0.722 110 - 129 130 - 149 25 Total 1000 3 (b) (c) Work for (a) and (b): a) Relative Frequency ( RF )= frequency of the particular class ( f ) total class frequency ( n ) Frequency ( f ) =Relative frequency ( RF ) total class frequency (n) IQ Scores Frequency Relative Frequency 50 - 69 23 23 =0.023 1000 70 - 89 249 249 =0.249 1000 Cumulative Relative Frequency 0.023 0.023+0.249 0 .272 STAT 200 Final Examination 90 - 109 Spring 2017OL1/US1 0.45 1000 450 Page 5 of 22 0.7220.272 0.722 0.45 10002324945025 110 - 129 253 130 - 149 Total 25 1000 Hence the complete table is given as IQ Scores Frequency 50 - 69 70 - 89 90 - 109 110 - 129 130 - 149 Total 23 249 450 253 25 1000 253 =0.253 1000 0.722+0.253 25 =0.025 1000 0.975+0.025 1000 =1 1000 Cumulative Relative Frequency 0.023 0.272 0.722 0.975 1 b) Percentage of adult with IQ scores at least 110=P(IQ Scores 4 ) P( IQ Scores=110 - 129)+ P(IQ Scores=130 - 149) 0.253+0.025 c) IQ Scores 50 - 69 70 - 89 90 - 109 110 - 129 130 - 149 Total 0.278 Frequency 23 249 450 253 25 1000 27.8 Cumulative Frequency 23 272 722 975 1 0 . 975 1 STAT 200 Final Examination Median position= Spring 2017OL1/US1 Page 6 of 22 N + 1 1000+1 = =500.5 Cumulative frequency of 70 - 89 group 2 2 is 272 and Cumulative frequency of the 90 - 109 group is 722 implying that the 500.5th value lies in 90 - 109 group Hence the median belongs the 90129 group Answer: (a) 90 (b) (c) 50 Justification: a) Range=MaximumMinimum 10010 90 4 b) 30 - 50 will contain less than 25% since it's slightly more than the tail to Q1 50 - 70 will contain 25% of values since it ranges from Q1 and median (Q2) 70 - 90 contain more than 25% as it is after median to slightly more than Q3 Hence 70 - 90 score band has the most students c) Score band 70 - 100 contain 50% since it's between maximum value and median 50 100 Number of students=100 50 50 100 STAT 200 Final Examination Spring 2017OL1/US1 Page 7 of 22 Answer: 1 (a) 69 (b) 1 221 Work for (a) and (b): a) 5 P ( withreplacement )= 4 4 1 = 52 52 1 69 b) P ( without replacement )= 4 3 1 = 52 51 221 Answer: (a) Complement event of (L OR T) is the even that randomly selected student does not take laptop or tablet. 6 (b) 0.05 Work for (b): P(L OR T)C = 1- P(L or T) = 1- (1500+900-500)/2000 = 0.95 7 Answer: (a) 2/3 (b) No. Work for (a) and (b): The sample space for the sum of rolling of two die is shown below STAT 200 Final Examination Spring 2017OL1/US1 a)Event A is Sum of two die is at most 6. B: First one is an even number ={2,4,6} A and B ={(2,4) ,(4,2)} Now,we have to find P ( A|B )= P ( AB) P(B) P(A AND B) = 2/36 =1/18 P(B) =3/36 =1/12 1 /18 P ( A|B )= =2/3 1/12 b)For events to be independent P(A and B) =P(A) *P(B) P ( A|B )=P( A) P(A) =5/36 As P(A| B) is not equal to P(A),therefore the events are not independent. Page 8 of 22 STAT 200 Final Examination Spring 2017OL1/US1 Page 9 of 22 Answer: (a) 10C3 =120 ways (b) 720 Work for (a) and (b): 8 a) Total number of ways of selecting 3 member out of 10 is given by 10C3 as ordering do not matter . In the first place there are 10 ways,in the second place there are 9 ways and in the last seat there are 8 ways. Total ways = 10*9*8 /3! =120 as ordering do not matter. b)Total routes is given by the formula = n! = 6! = 6*5*4*3*2=720 9 Answer: (a) x 0 1 2 3 Total (b) mean =1.5, s,d = 0.866 Work for (a) and (b) P(x) 1/8 3/8 3/8 1/8 1 STAT 200 Final Examination Spring 2017OL1/US1 Page 10 of 22 Answer: (a) n=10,p=0.2,q=0.8 (b) Probability =0.8926 Work for (b) : 10 P(X>=1) =1 -P(X=0) =1- 11 10 0 C 0.210 0.80=10.1074=0.8926 Answer: (a) 6.0235 hours (b) 0.1587 Work for (a) and (b): Mean, = 4 hours Standard deviation, = 3 hours a) 75 th percentile P ( Z > z )=0.75 Z =0.6745 [ Standard Normal Table ] STAT 200 Final Examination z= X X = +Z Spring 2017OL1/US1 4 + ( 0.6745 x 3 ) Page 11 of 22 6.0235 hours b) X =7 hours z= X 74 = =1 3 P ( X >1 )=P ( z> 1 ) [ Standard Normal Table] 12 0.1587 Answer: (a) 1.4083 (b) 0.8444 Work for (a) and (b): Mean, =150 Standard deviation , =8.45 a) standard deviation of sample means= b) z= X /n For X = 148 148150 z= =1.420 8.45 36 For X = 152 152150 z= =1.420 8.45 36 Sample n=36 8.45 = =1.4083 n 36 STAT 200 Final Examination Spring 2017OL1/US1 P (148< X <152 )=P (1.420< z<1.420 ) 0.92220.0778 Page 12 of 22 P ( z <1.420 ) P ( z <1.420 ) 0.8444 Answer: [ 0.1279,0.1921 ] Work: n=500 13 x 80 p= = =0.16 n 500 x=80 Standard error of the sample proportion S p= pq 0.16 0.84 = =0.01639512123 n 500 Margin of error , E=Z S p 2 95 CI = p E 14 q=1 p=10.16=0.84 0.16 0.0321 at =0.05, Z =1.96 1.96 0.01639512123 [0.1279, 0.1921] Answer: 95% CI = (1896.67, 2103.33) Work: Sample mean = 2000 Sample size =60 Standard error = 400/sqrt(60) = 51.64 2 0.0321 STAT 200 Final Examination Spring 2017OL1/US1 Page 13 of 22 Critical t-value with 59 degrees of freedom = 2.001 Margin of error =Critical t-value*Standard error =2.001*51.64 = 103.33 CI =2000 103.33=(1896.67,2103 .33) 15 Answer: . Alternative hypothesis , H 1 : p>0.45 (a) (b) 1.4213 (c) 0.0776 (d) =0.10 p=0.776< =0.10, h ence we reject null h ypot h esis We have sufficient evidence to support ABC Company's claim that the proportion of its employees investing in individual investment accounts is higher than 45% at 0.10 significance level since p<0.10 Work for (b) and (c): a) Nullhypothesis , H 0 : p 0.45 b) n=200 x=100 Standard error Alternative hypothesis , H 1 : p>0.45 x 100 ^p= = =0.5 n 200 SE= pq 0.45 x 0.55 = =0.0351781182 n 200 STAT 200 Final Examination Test statistic z= Spring 2017OL1/US1 ^p p SE 0.50.45 0.0351781182 Page 14 of 22 1.4213 c) pvalue=P ( z> 1.4213 ) 16 1P ( z 1.4213 ) 10.9224 0.0776 Answer: (a) Nullhypothesis , H 0 : d 0 Alternative hypothesis , H 1 : d <0 (b) 2.7500 (c) 0.0257 (d) =0.05 p=0.02570.05, we have sufficient evidence to support the claim that regular exercise helps weight loss at 0.05 significance level since p<0.05 work for and (c): a) let d be population mean differences of weights (weight after - before exercise). nullhypothesis , h 0 : alternative hypothesis 1 <0 b) (pounds) stat 200 final examination spring 2017ol1 us1 subject 2 3 4 5 150 170 185 160 130 180 -20 -10-5 -9 6 11 n 55 standard deviation, =d c) df =n1 110 8.9443 51 ( dd ) n1 t =2.7500 test is )2 81 36 121 statistic page 15 22 pvalue =0.0257 320 16 answers: (a) 10> 10 (b) 119.79 (c) 0.0761 (d). =0.05 p=0.07610. 05,hence we reject null hypothesis There is sufficient evidence to support the researcher's claim that the pulse rates of men have a standard deviation greater than 10 beats per minutes at 0.05 significance level since p<0. 05 Work for (b) and (c): 17 b) n=100 S=11 Test statistic b ) Degree of freedom =100-1 =99 Chi-square statistic = 99*11^2/100 = 119.79 c) Degrees of freedom, df =n1 1001 99 get pvalue , we canuse excel function Pvalue=CHISQ . DIST . RT (119.79, 99 ) 0.0761 STAT 200 Final Examination 18 Spring 2017OL1/US1 Page 17 of 22 Answer: (a) We test this with chisquare goodness of fit test . Null Hypothesis :The five types of teddy bears are equally popular . Alternative hypothesis :The five types of teddy bears are not equally popular Expressing Mathematically , H 0 : p1 =0.2, p2 =0.2, p3 =0.2, p 4=0.2, p5=0.2 H a : Atleast one p i do not match the claimed value (b) 12 (c) 0.0174 (d) =0.05 p=0.0174 < 0. 05,hence we reject null hypothesis There is sufficient evidence to support the manager's claim that the five types of teddy bears are equally popular at 0.05 significance level since p<0. 05 Work for (b) and (c): b) 1 The five types are equally popular hence p= =0.2 5 Observed (O) 230 Expected (E) np=1000 0.2 OE (OE)2 E 30 4.5 -30 4.5 200 170 np=1000 0.2 STAT 200 Final Examination Spring 2017OL1/US1 Page 18 of 22 200 210 np=1000 0.2 10 0.5 -20 2 10 0.5 200 180 np=1000 0.2 200 210 np=1000 0.2 200 Test statistic , X 2 = ( OE )2 E 4.5+ 4.5+0.5+2+0.5 c) get pvalue , we canuse excel function 12 Pvalue=CHISQ . DIST . RT (12 , 4 ) STAT 200 Final Examination 19 Spring 2017OL1/US1 Page 19 of 22 Answer: (a) Equation is ,y =18.122+0.757x (b) 86.27 Work for (a) and (b): a)The regression equation was determined through excel Procedure Data Data Analysis -- > Regression The result of the excel regression equation is shown below; Regression Statistics Multiple R R Square Adjusted R Square Standard Error Observations 0.8372 0.7009 0.6635 9.9835 10 ANOVA df SS Regression 1 1868.641551 Residual 8 797.3584489 MS 1868.64 2 99.6698 1 F 18.7483 Significance F 0.0025 STAT 200 Final Examination Total Spring 2017OL1/US1 9 Coefficients 18.122 0.757 Intercept x Page 20 of 22 2666 Standard Error 13.2855 0.1749 t Stat 1.3641 4.3299 P-value 0.2097 0.0025 Lower 95% -12.5139 0.3539 b)When x= 90 y=18.122+0.757*90 = 86.27 20 Answer: (a) Source of Variation Factor (Between) Error (Within) Total Sum of Squares (SS) Mean Square (MS) 27.82 Degree of Freedom (df) 9 957.25 490 1.9536 985.07 499 3.0911 (b) 1.5823 (c) 0.1175 (d) =0.05 p=0.1175 > 0. 05, hence we fail reject null hypothesis There is no sufficient evidence to support the claim that the mean weight loss is the same for the 10 programs at the significance level of 0.05 since p>0. 05 STAT 200 Final Examination Spring 2017OL1/US1 Work for (a), (b) and (c): a) SS Error =SS Total SSfactor 985.0727.82 957.25 10 weight loss :df factor=10 - 1=9 500 subjects : df tot =500 - 1=499 df factor +df error =df tot Therefore , df error =df tot df factor =4999=490 MS factor = SS factor 27.82 = =3.0911 df factor 9 MS Error = SS Error 957.25 = =1.95 df Error 490 b) Fvalue= MS factor 3.0911 = =1.5823 MS Error 1.9536 c) P-value is calculated using the excel function pvalue=fdist ( 2.4514, 4,245 )=0.1175 Page 21 of 22 STAT 200 Final Examination Spring 2017OL1/US1 Page 22 of 22