For this question, we will consider one of the world's most popular video games, Minecraft adapted from Mills & Mills, 2016). In this game, the character (you) can attempt to tame a wolf with a bone. Assume that you get 15 tries to tame the wolf, i.e. there are 15 bones. Each time you tame the wolf, the wolf becomes a dog and a new wolf appears which you can then attempt to also tame. This continues until you have used up all 15 bones, and the goal is to end up with as many dogs as possible at the end of the game. The probability of taming the wolf with each attempt is fixed at 0.44. Let X denote the number of dogs at the end of the game so that X ~ BIN(15, 0.44). Assume you are about to play the game. Based on this information, answer the following questions, rounding to FOUR decimal places where relevant. When answering the questions, some of the following results from R will be useful: dbinom ( 2, 15, 0.44) = 0.0108 . dbinom ( 3, 15, 0.44) = 0.0369 . dbinom( 4, 15, 0.44) = 0.0869 . dbinom( 5, 15, 0.44) = 0.1502 . dbinom( 6, 15, 0.44) = 0.1967 . dbinom( 7, 15, 0.44) = 0.1987 . dbinom ( 8, 15, 0.44) = 0.1561 dbinom( 9, 15, 0.44) = 0.0954 dbinom ( 10, 15, 0.44) = 0.045 . dbinom( 11, 15, 0.44) = 0.0161 . dbinom( 12, 15, 0.44) = 0.0042 . dbinom( 13, 15, 0.44) = 0.0008 . dbinom( 14, 15, 0.44) = 0.0001 . dbinom( 15, 15, 0.44) = 0 dbinom ( 16, 15, 0.44) = 0 . dbinom( 17, 15, 0.44) = 0 . dbinom ( 18, 15, 0.44) = 0 . dbinom( 19, 15, 0.44) = 0 . dbinom ( 20, 15, 0.44) = 0 . dbinom( 21, 15, 0.44) = 0 . pbinom( 2, 15, 0.44) = 0.013 . pbinom( 3, 15, 0.44) = 0.0498 . pbinom( 4, 15, 0.44) = 0.1367 . pbinom( 5, 15, 0.44) = 0.2869 . pbinom( 6, 15, 0.44) = 0.4836 . pbinom( 7, 15, 0.44) = 0.6824 . pbinom( 8, 15, 0.44) = 0.8385 . pbinom( 9, 15, 0.44) = 0.9339 . pbinom( 10, 15, 0.44) = 0.9789 . pbinom( 11, 15, 0.44) = 0.9949 . pbinom( 12, 15, 0.44) = 0.9991 . pbinom( 13, 15, 0.44) = 0.9999 . pbinom( 14, 15, 0.44) = 1 . pbinom( 15, 15, 0.44) = 1 . pbinom( 16, 15, 0.44) = 1 . pbinom ( 17, 15, 0.44) = 1 . pbinom ( 18, 15, 0.44) = 1 . pbinom( 19, 15, 0.44) = 1 . pbinom( 20, 15, 0.44) = 1 . pbinom( 21, 15, 0.44) = 1 1. What is the probability you will tame exactly 3 wolves? (2 marks) 2. What is the probability you will tame less than 4 wolves? (2 marks) 3. What is the probability you will tame more than 4 wolves? (2 marks) 4. If you tame more than half of the wolves, you win. Otherwise, you lose. What is the probability you will win the game? (2 marks) 5. The probability that you will tame some number a or less wolves is 0.9949. What must a be (round to the nearest whole number)? (2 marks) References Mills, P., & Mills, T. 2016. Minecraft and Mathematics. Viniculum, 53(1).Suppose X is normally distributed with a mean of A = 5 and a standard deviation of o = 2. Further suppose that a random sample of n = 50 has been taken from this population. Which of the following statements is/are true about the distribution of the associated sample mean, X? Select one or more: a. The distribution of X will have a variance of of = 4 50 O b. The distribution of X will have a variance of o2 2 V50 C. The distribution of X will have a standard deviation of o = 2 V50 O d. The distribution of X will be unknown O e. The distribution of X will be Normal Of. The distribution of X will have a standard deviation of o = 4 50 9. The distribution of X will have a mean of / > 5 Oh. The distribution of X will have a variance of o IN . The distribution of X will have a standard deviation of o = 4 V50 Suppose a sample of university students were asked the question, "In hours, what was your phone screen time yesterday?' Some researchers wish to know if the average daily phone screen time of university students is significantly different from 4 hours. That is, they wish to test the hypotheses Ho : H = 4 versus H1 : M # 4, where u denotes the mean average daily phone screen time of university students in hours. Using the data collected, a one-sample t-test was carried out with the following results: One Sample t-test data: screentime t = -1.3293, df = 56, p-value = 0. 1891 alternative hypothesis: true mean is not equal to 4 95 percent confidence interval: 3. 577816 4. 085378 sample estimates : mean of x 3. 831597 Answer the following questions, rounding your answers to three decimal places where relevant. a. What is the test statistic? b. What is the degrees of freedom? c. What must the sample size be? d. What is the p-value? e. What is the sample mean? f. What is the LOWER bound of the confidence interval (i.e., the lower number in the confidence interval range)? g. What is the UPPER bound of the confidence interval (i.e., the higher number in the confidence interval range)?Now suppose a different sample of university students were asked the question. \"in hours, what was your phone screen time yesterday?" Some researchers wish to know if the average daily phone screen time of university students is signicantly different from 4 hours. That is. they wish to test the hypotheses H0 :,u, =4versusH1:p3 4, where ,u denotes the mean average daily phone screen time of university students in hours. Using the data collected. a one-sample ttest was carried out with the following results: One Sample ttest data: screentime t = 1.3293. df = 56, p-value = 0.1891 alternative hypothesis: true mean is not equal to 4 95 percent confidence interval: 3.577816 4.085378 sample estimates: mean of x 3.831597 Which ofthe following statement(s) is/are true? [assume a significance level of or = 0.05. also assume that any reported pvalues have been rounded to three decimal places). Select one or more: [I a. Since we have a pvalue that is greater than the significance level, we cannot reject the null hypothesis Cl b. Since we have a pvalue that is less than the significance level, we can reject the null hypothesis [I c. If it is true that the average daily phone screen time of university students is equal to 4 hours, then the probability we would have seen a test statistic at least this extreme is 0.189 El d. If it is true that the average daily phone screen time of university students is equal to 4 hours, then the probability we would have seen a test statistic at least this extreme is 0.05 H e. Since we have a pvalue that is greater than the significance level, we can reject the null hypothesis Cl f. The pvalue suggests that, on average, at least 18.9% of students have a daily phone screen time of less than 4 hours i i g. Since we have a pvaluethat is less than the significance level, we cannot reject the null hypothesis Suppose you are carrying out a onersample trtest for a sample of data with sample size n : 25: and you Wish to check for normality Further suppose you have carried out the ShapiroWilk test with the following results: Shapiroewilk normality test data: x W = 0.93625. p-value = 0412i3 Using the results ofthe Shapiro-Wilktest alone, can normality be assumed? Select one: 0 a Ves,normality can be assumed O b. No, normality cannot be assumed Suppose you are carrying out a onersample trtest for a sample of data with sample size n : 37: and you wish to check for normality Further suppose you have ascertained that the underlying distribution is not normal Based on this information, has the normality assumption been violated? Select one: 0 a: Yes, the normality assumption has been violated O b. No, the normality assumption has not been violated Suppose a group of 50 university students participated in a study which aimed to compare alertness levels before and after coffee consumption. in this study. the students were provided wi h a chart containing nine symbols, each coded with a numberfrom 'l to 9. Then. the students were asked to perform the following coding task: one of the nine symbols, selected at random, appeared to the student. and they needed to correctly code this symbol to its assOCIated number from i to 9. The time taken to complete this task was then recorded S udents completed this coding task both before and after coffee consumption. We wish to est the following hypotheses: Ho :pp = 0 versus H1 :,u,D y 0, where: - #0 deno es the true mean difference in time taken to perform the task before coffee consumption compared with after coffee consumption Using the data collected it was found that the average time taken to perform the task before coffee was .5184 seconds. while the average time taken to perform the task after coffee was i .5008 seconds so that the estimated mean difference was i .5008 , i.5184 : 70.0176 seconds. A paired tetest was carried out with the following results: Paired tetest data: df$reaction.time by df$Group t : 70.12961, df : 49. pevalue : 0,8974 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -O.2913288 0.2560255 sample estimates: mean of the differences 70.0l765166 Answer the following questions, rounding your answers to three decimal places where relevan : (Note that a rounding tolerance has been built into the question: so that small discrepancies in mean differences due to rounding have been allowed for.) a. W at is the test statistic? b. W at is the degrees of freedom? c. W at is the pvalue? d. W at is the sample mean of the difference? e. W at is the LOWER bound of the confidence interval fie, the lower number in the confidence interval range)? f. W at is the UPPER bound ofthe confidence interval (i e, the higher number in the confidence interval range)