Past Exam 1

QUEEN’S UNIVERSITY FINAL EXAMINATION FACULTY OF ARTS AND SCIENCE BIOL243/GPHY247/KNPE251/NURS323/PSYC 202 Dr. Yakimowski and Dr. Flanagan December 11 th , 2022 INSTRUCTIONS TO STUDENTS • This examination is 3 HOURS in length. There are 2 sections to this exam. • Answer all multiple choice questions on the provided crowdmark bubble sheet at the back of the exam. Answer all short answer questions directly on the exam page in the space provided. • Write your first name, last name and student ID in the box above. • The following aids are allowed: Casio FX-991 calculator • Formulas and tables provided at the back of the exam Proctors are unable to respond to queries about interpretation of exam questions. Do your best to answer exam questions as written. GOOD LUCK! This material is copyrighted and is for the sole use of students registered in BIOL 243 / GPHY 247 / KNPE 251 / NURS 323 / PSYC 202 and writing this exam. This material shall not be distributed or disseminated. Failure to abide by these conditions is a breach of copyright and may also constitute a breach of academic integrity under the University Senate's Academic Integrity Policy Statement. HAND IN Answers recorded on exam paper

2 Multiple Choice Questions (22 marks total) You are interested in whether participating in daily exercise changes the amount of uninterrupted sleep people get at night. You randomly select 60 adult participants and conduct an initial survey of the number of days per week each adult exercises. Five people already exercise daily and are not included in the study, and so you continue with 55 participants. The 55 participants record their maximum hours of uninterrupted sleep per night for 2 weeks. Then participants begin a regime of exercising at least 30 minutes per day, for a minimum of 5 days per week. One week into the adoption of this exercise regime, participants continue the exercise regime and begin recording their maximum hours of uninterrupted sleep per night for another 2 weeks. The following two questions are based on this study (questions 1 and 2). Q1. Select the null and alternative hypotheses most appropriate for the question of whether the exercise regime increased hours of uninterrupted sleep. B=before exercise regime; A=after adoption of exercise regime [1 mark] a) H 0 : μ A = μ B H A : μ A > μ B b) H 0 : μ A ≥ μ B H A : μ A < μ B c) H 0 : μ A ≤ μ B H A : μ A > μ B d) H 0 : μ A = μ B H A : μ A ≠ μ B e) H 0 : μ ≤ 0 H A : μ > 0 Q2. Your study (above) finds that before adopting a regular exercise regime participants were sleeping an average of 6.2 uninterrupted hours per night. While exercising regularly participants were sleeping an average of 8.1 hours per night. Thus, overall on average people slept 1.9 hours longer (s.d. 1.5) without interruption following the adoption of a regular exercise regime. What is the observed t value? [1 mark] a) 9.81 b) 1.26 c) 10.5 d) 22.7 e) 9.39

4 Q5. Select the plot that is most appropriate for visualizing these data: [1 mark] a) Scatter plot b) Grouped bar chart c) Bar chart d) Grouped box plot e) Box plot Q6. What is the best statistical test to evaluate whether dietary habits depend on exercise habits? [1 mark] a) Single-factor ANOVA b) One-sample t -test c) Two-factor ANOVA d) Chi-square test e) Two-sample t -test Q7. As a government scientist you have data on crop damage due to insects for 100 agricultural sites across Canada. You are interested in whether variation in insect damage is related to climate, specifically drought conditions. Insect damage is quantified as proportion of leaves damaged. You estimate an index of drought by downloading precipitation data for your sites from the past 20 years and examining the difference between the 20 year annual average and the current year for which you have insect damage data. Select the plot that is most appropriate for visualizing these data: [1 mark] a) Scatter plot b) Grouped bar chart c) Bar chart d) Grouped box plot e) Box plot Q8. You work at a hospital and observe an increasing number of patients coming in over the summer with a unique rash that is persistent and associated with fever. You are interested in whether this rash might be associated with exposure to a novel biological organism due to time spent in outdoor environments (hiking, camping, fishing etc.) or other lifestyle factors (diet, level of substance abuse etc.). By the end of the summer you have 61 patients who have presented with the rash and are otherwise healthy. They all agree to participate in a study and fill out a survey to characterize their behaviours over the course of the summer. You also ask

5 ~60 healthy people from the region to participate in the survey. Which of the following best describes the study design? [1 mark] a) Cohort survey design b) Cluster survey design c) Experimental study design d) Case-control survey design e) Simple random survey Q9. You are interested in whether video game players have better (ie: quicker) reaction times than non-gamers. You find 100 gamers and 100 non-gamers between the ages of 20-40 (mean age 30) and estimate their reaction times for pressing a stop button. Select which of the following null and alternative hypotheses are most appropriate for the two-sample t-test that answers the following question: “Is the mean reaction time for video game players (V) better than for non-gamers (N)?” [1 mark] a) H o : μ ≤ 0 H A : μ > 0 b) H o : μ V ≥ μ N H A : μ V < μ N c) H o : μ N ≥ μ V H A : μ N < μ V d) H o : μ N = μ V H A : μ N ≠ μ V e) H o : μ N = μ V H A : μ N < μ V A researcher was interested in whether early exposure to CBN (cannabinol), one of the 3 primary cannabinoids in cannabis, affects embryonic development. Research was conducted on the model animal system, zebrafish. The researcher exposed embryos during the gastrulation stage, where development from one-dimensional cells to multi-dimensional cells occurs. Embryos were randomly assigned to the control (no CBN), low (0.01 mg L −1 CBN), medium (1 mg L −1 CBN) or high (4 mg L −1 CBN). The activity level, estimated as amount of time swimming, was quantified for each embryo in the experiment. (questions 10 and 11) Q10. Select the plot that is most appropriate for visualizing these data: [1 mark] a) Scatter plot b) Grouped bar chart c) Bar chart d) Grouped box plot e) Box plot

7 standard normal figure given above to calculate the probability of seeing a mean of 45 or greater based on last years’ data. [1 mark] Question options: a) 0 ≤ ANSWER < 0.01 b) 0.01 ≤ ANSWER < 0.2 c) 0.2 ≤ ANSWER < 0.4 d) 0.4 ≤ ANSWER < 0.8 e) 0.8 ≤ ANSWER < 1.0 Q14. You are interested in the number of flowers produced by the native plant Aquilegia canadensis because it is an important metric of reproductive potential. You sample 9 Aquilegia plants in late spring when total flower production can be observed, and find that the average number of flowers produced is 14.3 with a standard deviation of 3.4. What is the 95% confidence interval for average Aquilegia flower production based on this sample? [1 mark] a) 6.5 to 22.1 b) 8.0 to 20.6 c) 11.7 to 16.9 d) 12.2 to 16.4 e) 13.6 to 18.8 Q15. The F-test is used for analysis of variance. Which of the following is statements about what the F-ratio represents is true? [1 mark] a) It represents the effect size presented as a ratio b) A large residual (MS) variance relative to the group (MS) variance results in a large F-ratio, making it unlikely the null hypothesis will be rejected. c) A large residual (MS) variance relative to the group (MS) variance results in a large F-ratio. d) A large group (MS) variance relative to residual (MS) variance results in a large F- ratio, making it likely the null hypothesis will be rejected. e) A large group (MS) variance relative to residual (MS) variance results in a large F- ratio, making it unlikely the null hypothesis will be rejected.

8 Q16. Which of the following statements about statistical errors is TRUE? [1 mark] a) Type I and Type II errors are independent of one another. b) Type II error rate is a threshold set by the researcher c) Type I and Type II error are positively related: the larger your type I error the larger the type II error d) Type I error is set by the researcher and is the probability of rejecting the null hypothesis when the null hypothesis is actually true. e) Type I error is based on the alternative hypothesis and is the probability of rejecting the null hypothesis when the null hypothesis is actually true. Q17. You are interested in whether the diversity of wetland organisms (ie: number of species) differs between wetland communities located within urban regions compared to wetlands located in non-urban regions. You randomly select urban and non-urban wetlands across Ontario for which you survey vertebrate, invertebrate and plant diversity. You conduct a t -test test to compare your estimates of diversity between urban and non-urban wetlands. Which of the following null t -distributions (alpha shaded in gray) is most appropriate for determining whether the mean diversity differs between urban and non-urban wetlands? [1 mark]

10 Tukey multiple comparisons of means 95% family-wise confidence level Fit: aov(formula = x) $`file$passenger_class` diff lwr upr p adj 2nd-1st -11.367459 -14.345803 -8.389115 0.0000000 3rd-1st -15.148115 -18.192710 -12.103521 0.0000000 3rd-2nd -3.780656 -6.871463 -0.689849 0.0116695 a) We fail to reject the null hypothesis as there is no evidence for significant differences in age among passenger classes. b) We fail to reject the null hypothesis as there is evidence for significant differences in age among passenger classes c) We reject the null hypothesis as there is evidence for significant differences in age among all passenger classes. d) We reject the null hypothesis as there is evidence for a significant differences in age among 1 st and 2 nd , 1 st and 3 rd (but not 2 nd and 3 rd ) passenger classes. e) We reject the null hypothesis as there is no evidence for significant differences among passenger classes. Q20. As a wildlife biologist you are interested in how morphological traits of animals scale with one another. Bats can use their tongues to prey on a variety of insects, and some bats also use their tongues to sip nectar from flowers. You are interested in the variation in tongue length for bats, and whether tongue length is related to palate size (as estimate of the bat’s mouth size). You collect data from the published scientific literature for 11 bat species on palate size and tongue length. You conduct the follow correlation analysis. Which of the following is an appropriate statistical conclusion? [1 mark] Pearson's product-moment correlation data: file2$palate_length and file2$tongue_length t = 0.70211, df = 9, p-value = 0.5004 alternative hypothesis: true correlation is not equal to 0 95 percent confidence interval: -0.4308992 0.7282088 sample estimates: cor 0.2278793

11 a) There is a weak positive correlation between palate size and tongue length for bats (r=22.8, df=9, P=0.50) b) There is a strong correlation between palate size and tongue length for bats (r=22.8, df=9, P=0.50) c) There is a strong correlation between palate size and tongue length for bats (r=70.2, df=9, P=0.50) d) There is a weak correlation between palate size and tongue length for bats (r=70.2, df=9, P=0.23) e) There is no significant correlation between palate size and tongue length for bats (r=0.228, df=9, P=0.50). Q21. The Public Health Agency of Canada is worried about the increase in Lyme disease, which is caused by a bacterium transmitted by ticks. One proposed solution is to vaccinate people against the bacterium. To evaluate the potential effectiveness of the vaccine, researchers conducted a study by vaccinating laboratory rabbits with different concentrations of the vaccine and then exposing them to the Lyme bacterium. Effectiveness was quantified by measuring the number of bacteria in the blood of each rabbit. The researchers were interested in predicting how sick the rabbits would get based on how much of the vaccine they were given, and specifically wanted to answer whether increasing the concentration of the vaccine would predict decreased bacterial load in the rabbits. [1 mark] > reg=lm(num_bact~vacc_conc) > summary(reg) Call: lm(formula = num_bact~vacc_conc) Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) -3378.352 3626.416 -0.932 0.404 Vacc_conc 1.686 1.801 0.936 0.402 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 7.534 on 4 degrees of freedom Multiple R-squared: 0.1797, Adjusted R-squared: -0.02543 F-statistic: 0.876 on 1 and 4 DF, p-value: 0.4023

13 i) Explain which statistical test is appropriate for answering the question (above) and why. [1 mark] ii) Write the null and alternative hypotheses. [1 mark] iii) Calculate the expected counts. [2 marks] Female Male 1 st 2 nd 3 rd iv) Fill in the (O-E) 2 /E values (to 2 decimal places) for all cells in the table. [2 marks] Female Male 1 st 2 nd 3 rd v) Calculate your observed test score to two decimal places. [1 mark]

14 vi) Find and report the c 2 critical score for a Type I error rate of 5%. [1 mark] vii) Write both your statistical conclusion (with relevant statistical parameters in brackets) and your scientific conclusion. [2 marks] Q24. Previous studies have suggested that the navigation and memory required to be a taxi driver in large cities with complex street (non-grid) streets such as London, UK can result in enlarging of the posterior hippocampus region of the brain where this intensive work occurs. A group of scientists was interested in this potential association between navigation and brain structure. They estimated the adjusted volume of the posterior hippocampus for taxi drivers that varied in the length of time they had been driving a taxi (range: 54 to 349 months) in a large, complex city. This is the result of a linear regression they performed. Call: lm(formula = file3$posterior_hippocampus_volume_adjusted ~ file3$months_as_taxi_driver) Residuals: Min 1Q Median 3Q Max -3.5530 -1.3811 0.3619 1.3812 3.0255 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) -3.539934 1.055333 -3.354 0.00518 ** file3$months_as_taxi_driver 0.014408 0.005166 2.789 0.01535 * --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 2.125 on 13 degrees of freedom Multiple R-squared: 0.3744, Adjusted R-squared: 0.3263 F-statistic: 7.779 on 1 and 13 DF, p-value: 0.01535

16 STUDY 2 Q26. You are interested in whether participation in martial arts increases or decreases blood cortisol levels. You recruit 50 random participants; they all agree to participate in a martial arts class. You take a blood sample from each participant 20 minutes before their 1 hour class, and another blood sample from each participant in the class within 20 minutes of the end of their class. All classes in the study take place at a standard time (6:00 pm eastern) to account for daily temporal cortisol fluctuations. Cortisol levels (nmol/L) are estimated from each blood sample collected. [2 marks] i) Identify the most appropriate statistical test and explain why it is appropriate. ii) State the null and alternative hypothesis. STUDY 3 Q27. You have been working for an animal rescue and started tracking how long animals stay at the shelter depending on whether they are small (<4kg), medium (4-10kg) or large (10kg+) in size. You are interested in evaluating whether the length of stay (days) depends on the size of the animal. [2 marks] i) Identify the most appropriate statistical test and explain why it is appropriate. ii) State the null and alternative hypothesis.

17 Q28. The Loblolly pine is one of the most common trees in the southern United States, and can reach heights upwards of 100 feet. Understanding the relationship between age and size of trees is critical to managing forests. Below is a linear regression of age (in yrs) and height (in feet) for Loblolly pine. Call: lm(formula = Loblolly$height ~ Loblolly$age) Residuals: Min 1Q Median 3Q Max -7.0207 -2.1672 -0.4391 2.0539 6.8545 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) -1.31240 0.62183 -2.111 0.0379 * Loblolly$age 2.59052 0.04094 63.272 <2e-16 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 2.947 on 82 degrees of freedom Multiple R-squared: 0.9799, Adjusted R-squared: 0.9797 F-statistic: 4003 on 1 and 82 DF, p-value: < 2.2e-16 i) Explain the meaning of the P -value associated with the slope of the regression. [1 mark] ii) Write your statistical and scientific conclusions based on the above linear regression. [2 marks]

19 Formulae Sheet ࠵? = 1 ࠵? % ࠵? ! " !#$ ࠵? % = 1 ࠵? − 1 %(࠵? ! − ࠵?̅) % " !#$ ࠵? = , ࠵? % Where: c i = column marginal proportion CI = confidence interval d = mean difference df = degrees of freedom E i = expected frequency IQR = inter quartile range n = sample size O i = observed frequency Q1 = first quartile Q3 = third quartile r i = row marginal proportion s = sample standard deviation s 2 = sample variance SE = standard error t crit = critical t -statistic t obs = observed t- statistic x = observed value x ̅ = sample mean z = z -statistic μ = population mean or reference mean χ 2 = observed Chi-squared statistic ࠵? = ࠵? − ࠵? ࠵? ࠵?࠵? = ࠵? √࠵? ࠵?࠵? = 1 ࠵?࠵? $ ࠵? $ % + ࠵?࠵? % ࠵? % % ࠵?࠵? $ + ࠵?࠵? % 5 1 ࠵? $ + 1 ࠵? % 6 ࠵? &’( = ࠵? − ࠵? ࠵?࠵? ࠵? &’( = ࠵? ࠵?࠵? ࠵? &’( = ࠵? $ − ࠵? % ࠵?࠵? ࠵? % = % (࠵? ! − ࠵? ! ) % ࠵? ! " !#$ ࠵? ! = ࠵? ! ࠵? ! ࠵? ࠵?࠵? = ࠵?̅ ± ࠵? )*!+ × ࠵?࠵? ࠵?࠵?࠵? = ࠵?3 − ࠵?1

20 0.025 0.05 0.95 0.975 1 qt(0.025, 1) = -12.706 qt(0.05, 1) = -6.314 qt(0.95, 1) = 6.314 qt(0.975, 1) = 12.706 2 qt(0.025, 2) = -4.303 qt(0.05, 2) = -2.92 qt(0.95, 2) = 2.92 qt(0.975, 2) = 4.303 3 qt(0.025, 3) = -3.182 qt(0.05, 3) = -2.353 qt(0.95, 3) = 2.353 qt(0.975, 3) = 3.182 4 qt(0.025, 4) = -2.776 qt(0.05, 4) = -2.132 qt(0.95, 4) = 2.132 qt(0.975, 4) = 2.776 5 qt(0.025, 5) = -2.571 qt(0.05, 5) = -2.015 qt(0.95, 5) = 2.015 qt(0.975, 5) = 2.571 6 qt(0.025, 6) = -2.447 qt(0.05, 6) = -1.943 qt(0.95, 6) = 1.943 qt(0.975, 6) = 2.447 7 qt(0.025, 7) = -2.365 qt(0.05, 7) = -1.895 qt(0.95, 7) = 1.895 qt(0.975, 7) = 2.365 8 qt(0.025, 8) = -2.306 qt(0.05, 8) = -1.86 qt(0.95, 8) = 1.86 qt(0.975, 8) = 2.306 9 qt(0.025, 9) = -2.262 qt(0.05, 9) = -1.833 qt(0.95, 9) = 1.833 qt(0.975, 9) = 2.262 10 qt(0.025, 10) = -2.228 qt(0.05, 10) = -1.812 qt(0.95, 10) = 1.812 qt(0.975, 10) = 2.228 11 qt(0.025, 11) = -2.201 qt(0.05, 11) = -1.796 qt(0.95, 11) = 1.796 qt(0.975, 11) = 2.201 12 qt(0.025, 12) = -2.179 qt(0.05, 12) = -1.782 qt(0.95, 12) = 1.782 qt(0.975, 12) = 2.179 13 qt(0.025, 13) = -2.16 qt(0.05, 13) = -1.771 qt(0.95, 13) = 1.771 qt(0.975, 13) = 2.16 14 qt(0.025, 14) = -2.145 qt(0.05, 14) = -1.761 qt(0.95, 14) = 1.761 qt(0.975, 14) = 2.145 15 qt(0.025, 15) = -2.131 qt(0.05, 15) = -1.753 qt(0.95, 15) = 1.753 qt(0.975, 15) = 2.131 16 qt(0.025, 16) = -2.12 qt(0.05, 16) = -1.746 qt(0.95, 16) = 1.746 qt(0.975, 16) = 2.12 17 qt(0.025, 17) = -2.11 qt(0.05, 17) = -1.74 qt(0.95, 17) = 1.74 qt(0.975, 17) = 2.11 18 qt(0.025, 18) = -2.101 qt(0.05, 18) = -1.734 qt(0.95, 18) = 1.734 qt(0.975, 18) = 2.101 19 qt(0.025, 19) = -2.093 qt(0.05, 19) = -1.729 qt(0.95, 19) = 1.729 qt(0.975, 19) = 2.093 20 qt(0.025, 20) = -2.086 qt(0.05, 20) = -1.725 qt(0.95, 20) = 1.725 qt(0.975, 20) = 2.086 21 qt(0.025, 21) = -2.08 qt(0.05, 21) = -1.721 qt(0.95, 21) = 1.721 qt(0.975, 21) = 2.08 22 qt(0.025, 22) = -2.074 qt(0.05, 22) = -1.717 qt(0.95, 22) = 1.717 qt(0.975, 22) = 2.074 23 qt(0.025, 23) = -2.069 qt(0.05, 23) = -1.714 qt(0.95, 23) = 1.714 qt(0.975, 23) = 2.069 24 qt(0.025, 24) = -2.064 qt(0.05, 24) = -1.711 qt(0.95, 24) = 1.711 qt(0.975, 24) = 2.064 25 qt(0.025, 25) = -2.06 qt(0.05, 25) = -1.708 qt(0.95, 25) = 1.708 qt(0.975, 25) = 2.06 26 qt(0.025, 26) = -2.056 qt(0.05, 26) = -1.706 qt(0.95, 26) = 1.706 qt(0.975, 26) = 2.056 27 qt(0.025, 27) = -2.052 qt(0.05, 27) = -1.703 qt(0.95, 27) = 1.703 qt(0.975, 27) = 2.052 28 qt(0.025, 28) = -2.048 qt(0.05, 28) = -1.701 qt(0.95, 28) = 1.701 qt(0.975, 28) = 2.048 29 qt(0.025, 29) = -2.045 qt(0.05, 29) = -1.699 qt(0.95, 29) = 1.699 qt(0.975, 29) = 2.045 30 qt(0.025, 30) = -2.042 qt(0.05, 30) = -1.697 qt(0.95, 30) = 1.697 qt(0.975, 30) = 2.042 40 qt(0.025, 40) = -2.021 qt(0.05, 40) = -1.684 qt(0.95, 40) = 1.684 qt(0.975, 40) = 2.021 50 qt(0.025, 50) = -2.009 qt(0.05, 50) = -1.676 qt(0.95, 50) = 1.676 qt(0.975, 50) = 2.009 60 qt(0.025, 60) = -2 qt(0.05, 60) = -1.671 qt(0.95, 60) = 1.671 qt(0.975, 60) = 2 70 qt(0.025, 70) = -1.994 qt(0.05, 70) = -1.667 qt(0.95, 70) = 1.667 qt(0.975, 70) = 1.994 80 qt(0.025, 80) = -1.99 qt(0.05, 80) = -1.664 qt(0.95, 80) = 1.664 qt(0.975, 80) = 1.99 90 qt(0.025, 90) = -1.987 qt(0.05, 90) = -1.662 qt(0.95, 90) = 1.662 qt(0.975, 90) = 1.987 100 qt(0.025, 100) = -1.984 qt(0.05, 100) = -1.66 qt(0.95, 100) = 1.66 qt(0.975, 100) = 1.984 Degrees of freedom Critical t-values