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) Question 1 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 Question 3 a) frequency of the particular class(f) total class frequency(n) Frequency (f) = Relative frequency(RF) total class frequency (n) Relative Frequency (RF) = IQ Scores Frequency 50 - 69 23 70 - 89 249 90 - 109 = 0.45 1000 = = 1000 23 249 450 25 = 25 110 - 129 130 - 149 Total 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 Relative Frequency 23 = 0.023 1000 249 = = 0.249 1000 = 0.722 0.272 = 0.45 253 = = 0.253 1000 25 = = 0.025 1000 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 = 0.278 = 27.8% Cumulative Relative Frequency 0.023 = 0.023 + 0.249 = . 0.722 = 0.722 + 0.253 = . = 0.975 + 0.025 = c) IQ Scores 50 - 69 70 - 89 90 - 109 110 - 129 130 - 149 Total Frequency 23 249 450 253 25 1000 Median position = Cumulative Frequency 23 272 722 975 1 N + 1 1000 + 1 = = 500.5 2 2 Cumulative frequency of 70 - 89 group 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 to the group Question 4 a) Range = Maximum Minimum = 100 10 = 90 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 Number of students = 100 50% 50 = 100 100 = 50 Question 5 a) P(with replacement) = b) 4 1 4 = 52 52 69 P(without replacement) = 4 3 1 = 52 51 221 Question 11 Mean, = 4 hours Standard deviation, = 3 hours a) 75th percentile P(Z > z) = 0.75 Z = 0.6745 [From Standard Normal Table] X X = + Z = 4 + (0.6745x3) = 6.0235 hours b) z= X = 7 hours z= X 74 = =1 3 P(X > 1) = P(z > 1) [From Standard Normal Table] = 0.1587 Question 12 Mean, = 150 Standard deviation, = 8.45 Sample size, n = 36 a) standard deviation of sample means = n = 8.45 36 = 1.4083 b) z= X /n For X = 148 z= 148 150 = 1.420 8.45 36 For X = 152 z= 152 150 = 1.420 8.45 36 P(148 < X < 152) = P(1.420 < z < 1.420) = P(z < 1.420) P(z < 1.420) = 0.9222 0.0778 = 0.8444 Question 13 n = 500 x = 80 x 80 p= = = 0.16 n 500 q = 1 p = 1 0.16 = 0.84 Standard error of the sample proportion pq 0.16 0.84 = = 0.01639512123 n 500 Sp = at = 0.05, Z = 1.96 2 Margin of error, E = Z Sp 2 = 1.96 0.01639512123 = 0.0321 95% CI = p E = 0.16 0.0321 = [0.1279, 0.1921] Question 14 Sample size, n = 60 = $2000 sample mean, X sample standard deviation, S = $400 S 400 Standard error of mean, SX = = = 51.6397795 n 60 at = 0.05, Z = 1.96 2 Margin of error, E = Z SX 2 = 1.96 51.6397795 = 101.21 E 95% CI = X = $2000 101.21 = [1898.79, 2101.21] Question 15 a) Null hypothesis, H0 : p 0.45 Alternative hypothesis, H1 : p > 0.45 b) n = 200 x = 100 x 100 p = = = 0.5 n 200 Standard error pq 0.45x0.55 = = 0.0351781182 n 200 SE = Test statistic p p z= SE 0.5 0.45 = 0.0351781182 = 1.4213 c) p value = P(z > 1.4213) = 1 P(z 1.4213) = 1 0.9224 = 0.0776 d) = 0.10 p = 0.776 < = 0.10, hence we reject null hypothesis 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 Question 16 a) Let d be the population mean for the differences of weights (weight after the exercise - before the exercise). Null hypothesis, H0 : d 0 Alternative hypothesis, H1 : d < 0 b) Subject Weight (pounds) Before After d 1 2 3 4 5 150 170 185 160 200 -20 -10 -5 0 -20 130 160 180 160 180 d = 55 d n 55 = 5 = 11 Mean of d, = 2 d d Standard deviation, = n1 320 51 = Test statistic d d n 11 0 = 8.9443 5 = 2.7500 t= c) = 8.9443 d d -9 1 6 11 -9 d d 81 1 36 121 81 2 2 d d = 320 df = n 1 =51 =4 t = 2.7500 The test is left tailed p value = 0.0257 d) = 0.05 p = 0.0257 < = 0.05, hence we reject null hypothesis We have sufficient evidence to support the claim that regular exercise helps weight loss at 0.05 significance level since p < 0.05 Question 17 a) Null hypothesis, H0 : 10 Alternative hypothesis, H1 : > 10 b) n = 100 S = 11 Test statistic ( 1) 2 2 = 2 (100 1) 112 = 102 = 119.79 c) Degrees of freedom, df = n 1 = 100 1 = 99 To get p value, we can use excel function P value = CHISQ. DIST. RT (119.79, 99) = 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 Question 18 a) We test this with chi square 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, H0 : p1 = 0.2, p2 = 0.2, p3 = 0.2, p4 = 0.2, p5 = 0.2 Ha : Atleast one pi do not match the claimed value b) n = 1000 The five types are equally popular hence p = Observed (O) Expected (E) 230 = np = 1000 0.2 = 200 170 = np = 1000 0.2 = 200 210 = np = 1000 0.2 = 200 180 = np = 1000 0.2 = 200 210 = np = 1000 0.2 = 200 (O E)2 Test statistic, X 2 = E = 4.5 + 4.5 + 0.5 + 2 + 0.5 = 12 c) Degrees of freedom, df = k 1 =51 =4 To get p value, we can use excel function P value = CHISQ. DIST. RT (10.65, 4) = 0.0174 1 = 0.2 5 OE 30 -30 (O E)2 E 4.5 4.5 10 0.5 -20 2 10 0.5 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 Question 19 (a) The estimated regression equation is yi = b0 + bi Xi + ei Consider the table below x 80 93 50 60 100 40 85 70 75 85 = 738 y 72 95 75 68 90 35 83 60 77 85 = 740 X2 XY 2 = 57724 = 57080 6400 8649 2500 3600 10000 1600 7225 4900 5625 7225 From the above table, Sample size, n = 10 Slope(b) = n XY - ( X)( Y) 2 n X - ( X)2 10(57080) (738)(740) = 10(57724) 7382 = 0.7571 Y b(X) n 740 0.7571(738) = 10 = 18.12602 Intercept(a) = The equation of least square regression line is: y = 0.7571 + 18.12602X (b) The predicted value of y if x = 90 y = 0.7571 + 18.12602(90) = 0.7571 + 1631.3418 = 1632.0989 5760 8835 3750 4080 9000 1400 7055 4200 5775 7225 The predicted value of y if x = 90 is 1632.0989 Question 20 a) Source of Variation Factor (Between) Error (Within) Total Sum of Squares (SS) 27.82 Degree of Mean Square (MS) Freedom (df) 9 3.0911 957.25 490 985.07 499 SSError = SSTotal SSfactor = 985.07 27.82 = . 10 weight loss: dffactor = 10 - 1 = 500 subjects: dftot = 500 - 1 = dffactor + dferror = dftot Therefore, dferror = dftot dffactor = 499 9 = MSfactor = MSError = b) SSfactor 27.82 = = . dffactor 9 SSError 957.25 = = . dfError 490 F value = c) MSfactor 3.0911 = = . MSError 1.9536 P-value is calculated using the excel function p value = fdist(2.4514, 4,245) = 0.1175 d) = 0.05 p = 0.1175 > = 0.05, hence we fail to reject null hypothesis 1.9536 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 Question 1 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 Question 3 a) frequency of the particular class(f) total class frequency(n) Frequency (f) = Relative frequency(RF) total class frequency (n) Relative Frequency (RF) = IQ Scores Frequency 50 - 69 23 70 - 89 249 90 - 109 = 0.45 1000 = = 1000 23 249 450 25 = 25 110 - 129 130 - 149 Total 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 Relative Frequency 23 = 0.023 1000 249 = = 0.249 1000 = 0.722 0.272 = 0.45 253 = = 0.253 1000 25 = = 0.025 1000 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 = 0.278 = 27.8% Cumulative Relative Frequency 0.023 = 0.023 + 0.249 = . 0.722 = 0.722 + 0.253 = . = 0.975 + 0.025 = c) IQ Scores 50 - 69 70 - 89 90 - 109 110 - 129 130 - 149 Total Frequency 23 249 450 253 25 1000 Median position = Cumulative Frequency 23 272 722 975 1 N + 1 1000 + 1 = = 500.5 2 2 Cumulative frequency of 70 - 89 group 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 to the group Question 4 a) Range = Maximum Minimum = 100 10 = 90 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 Number of students = 100 50% 50 = 100 100 = 50 Question 5 a) P(with replacement) = b) 4 1 4 = 52 52 69 P(without replacement) = 4 3 1 = 52 51 221 Question 11 Mean, = 4 hours Standard deviation, = 3 hours a) 75th percentile P(Z > z) = 0.75 Z = 0.6745 [From Standard Normal Table] X X = + Z = 4 + (0.6745x3) = 6.0235 hours b) z= X = 7 hours z= X 74 = =1 3 P(X > 1) = P(z > 1) [From Standard Normal Table] = 0.1587 Question 12 Mean, = 150 Standard deviation, = 8.45 Sample size, n = 36 a) standard deviation of sample means = n = 8.45 36 = 1.4083 b) z= X /n For X = 148 z= 148 150 = 1.420 8.45 36 For X = 152 z= 152 150 = 1.420 8.45 36 P(148 < X < 152) = P(1.420 < z < 1.420) = P(z < 1.420) P(z < 1.420) = 0.9222 0.0778 = 0.8444 Question 13 n = 500 x = 80 x 80 p= = = 0.16 n 500 q = 1 p = 1 0.16 = 0.84 Standard error of the sample proportion pq 0.16 0.84 = = 0.01639512123 n 500 Sp = at = 0.05, Z = 1.96 2 Margin of error, E = Z Sp 2 = 1.96 0.01639512123 = 0.0321 95% CI = p E = 0.16 0.0321 = [0.1279, 0.1921] Question 14 Sample size, n = 60 = $2000 sample mean, X sample standard deviation, S = $400 S 400 Standard error of mean, SX = = = 51.6397795 n 60 at = 0.05, Z = 1.96 2 Margin of error, E = Z SX 2 = 1.96 51.6397795 = 101.21 E 95% CI = X = $2000 101.21 = [1898.79, 2101.21] Question 15 a) Null hypothesis, H0 : p 0.45 Alternative hypothesis, H1 : p > 0.45 b) n = 200 x = 100 x 100 p = = = 0.5 n 200 Standard error pq 0.45x0.55 = = 0.0351781182 n 200 SE = Test statistic p p z= SE 0.5 0.45 = 0.0351781182 = 1.4213 c) p value = P(z > 1.4213) = 1 P(z 1.4213) = 1 0.9224 = 0.0776 d) = 0.10 p = 0.776 < = 0.10, hence we reject null hypothesis 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 Question 16 a) Let d be the population mean for the differences of weights (weight after the exercise - before the exercise). Null hypothesis, H0 : d 0 Alternative hypothesis, H1 : d < 0 b) Subject Weight (pounds) Before After d 1 2 3 4 5 150 170 185 160 200 -20 -10 -5 0 -20 130 160 180 160 180 d = 55 d n 55 = 5 = 11 Mean of d, = 2 d d Standard deviation, = n1 320 51 = Test statistic d d n 11 0 = 8.9443 5 = 2.7500 t= c) = 8.9443 d d -9 1 6 11 -9 d d 81 1 36 121 81 2 2 d d = 320 df = n 1 =51 =4 t = 2.7500 The test is left tailed p value = 0.0257 d) = 0.05 p = 0.0257 < = 0.05, hence we reject null hypothesis We have sufficient evidence to support the claim that regular exercise helps weight loss at 0.05 significance level since p < 0.05 Question 17 a) Null hypothesis, H0 : 10 Alternative hypothesis, H1 : > 10 b) n = 100 S = 11 Test statistic ( 1) 2 2 = 2 (100 1) 112 = 102 = 119.79 c) Degrees of freedom, df = n 1 = 100 1 = 99 To get p value, we can use excel function P value = CHISQ. DIST. RT (119.79, 99) = 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 Question 18 a) We test this with chi square 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, H0 : p1 = 0.2, p2 = 0.2, p3 = 0.2, p4 = 0.2, p5 = 0.2 Ha : Atleast one pi do not match the claimed value b) n = 1000 The five types are equally popular hence p = Observed (O) Expected (E) 230 = np = 1000 0.2 = 200 170 = np = 1000 0.2 = 200 210 = np = 1000 0.2 = 200 180 = np = 1000 0.2 = 200 210 = np = 1000 0.2 = 200 (O E)2 Test statistic, X 2 = E = 4.5 + 4.5 + 0.5 + 2 + 0.5 = 12 c) Degrees of freedom, df = k 1 =51 =4 To get p value, we can use excel function P value = CHISQ. DIST. RT (10.65, 4) = 0.0174 1 = 0.2 5 OE 30 -30 (O E)2 E 4.5 4.5 10 0.5 -20 2 10 0.5 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 Question 19 (a) The estimated regression equation is yi = b0 + bi Xi + ei Consider the table below x 80 93 50 60 100 40 85 70 75 85 = 738 y 72 95 75 68 90 35 83 60 77 85 = 740 X2 XY 2 = 57724 = 57080 6400 8649 2500 3600 10000 1600 7225 4900 5625 7225 From the above table, Sample size, n = 10 Slope(b) = n XY - ( X)( Y) 2 n X - ( X)2 10(57080) (738)(740) = 10(57724) 7382 = 0.7571 Y b(X) n 740 0.7571(738) = 10 = 18.12602 Intercept(a) = The equation of least square regression line is: y = 0.7571 + 18.12602X (b) The predicted value of y if x = 90 y = 0.7571 + 18.12602(90) = 0.7571 + 1631.3418 = 1632.0989 5760 8835 3750 4080 9000 1400 7055 4200 5775 7225 The predicted value of y if x = 90 is 1632.0989 Question 20 a) Source of Variation Factor (Between) Error (Within) Total Sum of Squares (SS) 27.82 Degree of Mean Square (MS) Freedom (df) 9 3.0911 957.25 490 985.07 499 SSError = SSTotal SSfactor = 985.07 27.82 = . 10 weight loss: dffactor = 10 - 1 = 500 subjects: dftot = 500 - 1 = dffactor + dferror = dftot Therefore, dferror = dftot dffactor = 499 9 = MSfactor = MSError = b) SSfactor 27.82 = = . dffactor 9 SSError 957.25 = = . dfError 490 F value = c) MSfactor 3.0911 = = . MSError 1.9536 P-value is calculated using the excel function p value = fdist(2.4514, 4,245) = 0.1175 d) = 0.05 p = 0.1175 > = 0.05, hence we fail to reject null hypothesis 1.9536 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