Midterm 2 Version B - 2023

MAT 2379C Midterm Examination November 15, 2023 Instructor: Xiao Liang Time: 80 minutes Student Number: Family Name: First Name: • This is a closed book examination. • You can bring your own formula sheet (one page, one-sided). • Some statistical tables are included in the last pages of the booklet. • Only Faculty standard calculators are permitted: TI30, TI34, Casio fx-260, Casio fx-300. • You are not allowed to use any electronic device during the exam. Cell phones should be put away. • The exam consists of 6 multiple choice questions and 4 long answer questions. • Each multiple choice question is worth 5 marks and each long answer question is worth 10 marks. The total number of marks is 70. NOTE: At the end of the examination, hand in the entire booklet. ***************************************************** For professor’s use: Number of marks Total for all MC Questions Long Answer Question 1 Long Answer Question 2 Long Answer Question 3 Long Answer Question 4 Total 1

Part 1: Multiple Choice Questions Record your answer to the multiple choice questions in the table below: Question Answer 1 E 2 C 3 C 4 E 5 A 6 C 1. Our scientists study polar bears in the Beaufort Sea. Their research indicates that bears, while fasting, give birth to cubs of lower weight. We measured the birth weight (in grams) for a sample of 52 cubs, obtaining a sample mean ¯ x = 715 g and a standard deviation s = 123 g. Calculate a confidence interval of 99% for the mean birth weight μ of the cubs, and the estimated standard error of the mean. A) 715 ± 1 . 96; 17 . 057 B) 715 ± 43 . 92; 2 . 46 C) 715 ± 43 . 92; 2 . 46 D) 715 ± 34 . 09; 17 . 057 E) 715 ± 43 . 92; 17 . 057 Solution This is an interval for a large sample. A confidence interval of 99% for μ is given by : 715 ± 2 . 575 123 √ 52 = 715 ± 43 . 9218 .. The estimated standard error of the mean is s/ √ n = 123 / √ 52 = 17 . 057 . Here, since P ( Z ≤ 2 . 57) = 0 . 9949 and P ( Z ≤ 2 . 58) = 0 . 9951 , we choose z α/ 2 = 2 . 575 . The answer is E. 2. We aim to estimate the mean daily temperature on San Crist´obal Island during the dry season. Assuming that daily temperatures (during the dry season from June to December ) follow a normal distribution, we wish to determine a 90% confidence interval for this mean temperature during three weeks of the dry season, given that the daily measured mean was 21.5 ° C with a variance of 2.083. A) 21 . 5 ± 0 . 327 B) 21 . 5 ± 0 . 418 C) 21 . 5 ± 0 . 543 D) 21 . 5 ± 0 . 682 E) 21 . 5 ± 0 . 701 Solution (Section 8.1) We have ¯ x = 21 . 5 and s 2 = 2 . 083 . The confidence interval is based on a random variable T which follows a t (20) distribution. To construct this interval, we first need to find the value of t such that P ( - t ≤ T ≤ t ) = 0 . 90 , where T is a random variable following a t (20) distribution. Therefore, P ( T > t ) = (1 - 0 . 90) / 2 = 0 . 05 and, consequently, P ( T ≤ t ) = 1 - 0 . 05 = 0 . 95 . Using Table 18.4, we find t 0 . 05 , 20 = 1 . 725 . A 90% confidence interval for μ is 21 . 5 ± 1 . 725 √ 2 . 083 √ 21 ! = 21 . 5 ± 0 . 543 . The answer is C. 3. Glaucoma is an eye disease characterized by elevated intraocular pressure. Let’s assume that in the general population, intraocular pressure follows an approximately normal distribution, with a mean of 15 mm Hg and a standard deviation of 4 mm Hg. The usual range of intraocular pressure is considered to be between 10 mm Hg and 19 mm Hg. What proportion of the general population has an intraocular pressure within 2

the usual range? A) 0 . 7164 B) 0 . 0718 C) 0 . 7357 D) 0 . 7426 E) 0 . 8012 Solution (Section 5.2) We want to calculate P (10 ≤ X ≤ 19) , where X follows a normal distribution with mean μ = 15 and standard deviation σ = 4 . Using standardization, we have : P (10 ≤ X ≤ 19) = P 10 - 15 4 ≤ X - 15 4 ≤ 19 - 15 4 = P ( - 1 . 25 ≤ Z ≤ 1) = P ( Z ≤ 1) - P ( Z < - 1 . 25) = 0 . 8413 - 0 . 1056 = 0 . 7357 . The answer is C. 4. We’re looking for information on humidity levels in three cities: Gatineau, Montreal and Quebec City. For each city, we selected a sample of 100 houses and recorded their humidity levels. We then created 3 data sets of 100 observations each, called ”Gatineau”, ”Montreal” and ”Quebec City”. Below are box-and-whisker plots and histograms for these data sets. Which of the following is correct ? (a) Gatineau (b) Qu´ ebec (c) Montr´ eal (d) (e) (f) A) Histogram (d) is for Gatineau, (e) Montr´ eal, and (f) Qu´ ebec. B) Histogram (f) is for Gatineau, (e) Montr´ eal, and (d) Quebec. C) Histogram (e) is for Gatineau, (f) Montr´ eal, and (d) Quebec. D) Histogram(f) is for Gatineau, (d) Montr´ eal, and (e) Quebec. E) Histogram(e) is for Gatineau, (d) Montr´ eal, and (f) Quebec. Solution (Section 7.1): We need to match each histogram with the corresponding boxplot. Note that histogram (f) is approximately symmetrical, histogram (d) is positively asymmetrical and histogram (e) is negatively asymmetrical. To match (f), we need to look at a box whose two halves are of roughly equal length, so (b) matches (f). We should be able to identify a mosaic far from the center for an asymmetrical distribution. Note that the center of the distribution of (a) is the highest of the three distributions, and that the left portion of the box and the left whiskers are longer than on the right side, so (a) corresponds to (e). The center of the distribution of (c) is the lowest of the three distributions, it is a positive asymmetric 3