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1 Mock Exam - 2 hours - use of basic (non-programmable) calculator is allowed - all exercises carry the same marks - exam is strictly individual Question 1. Suppose you want to estimate the percentage of videos on YouTube that are cat videos. It is impossible for you to watch all videos on YouTube so you use a random video picker to select 1000 videos for you. You find that 2% of these videos are cat videos. Determine which of the following is an observation, a variable, a sample statistic, or a population parameter. (a) Percentage of all videos on YouTube that are cat videos. (b) 2%. (c) A video in your sample. (d) Whether or not a video is a cat video. (a) Population parameter (b) Sample statistics (c) Observation (d) Variable Question 2. Suppose you have used a Normal distribution to compute the 95% confidence interval for the mean of a population with known variance, using a sample of size 25 and obtained the interval [ 1; 3]. (a) What is the variance of the population? (b) What is the sample size needed to reduce the length of such interval by half? (a) The computation of a 95% confidence interval is ˆx ± z SE, thus ˆx = 3+( 1) = 1 and 2 z SE = 2. We know that SE = σ n, and from the z-table in order to leave area in each tail, we have z = 1.96: z σ SE = 1.96 = 2 σ = = σ2 = (b) We want z SE = n = 1, therefore n = 100. Question 3. In a given one-sample t-test against null hypothesis H 0 of zero mean, the test statistic is found to be The student who conducted the test concludes that she cannot reject the null hypothesis. Which of the following scenarios does not justify the student s decision (df means degrees of freedom; alpha is the significance level of the test)? (a) df = 10, alpha = 0.01, and a one-sided greater than alternative hypothesis. (b) df = 60, alpha = 0.05, and a two-sided alternative hypothesis. (c) df = 10, alpha = 0.05, and a one-sided greater than alternative hypothesis. (d) df = 500, alpha = 0.01, and two-sided alternative hypothesis. (e) all of these scenarios would justify the failure to reject H 0. (a) T = 1.80, df = 10, one-side greater than as alternative, implies 0.05 < pvalue < 0.1 (from t-table with one-tail). So student is justified. (b) T = 1.80, df = 60, two-sided alternative, implies 0.05 < pvalue < 0.1 (from t-table with

2 two-tails, df = 30 as approximation, or could use Normal table since df is large, would result in pvalue=0.0718). So student is justified. (c) T = 1.80, df = 10, one-side greater than as alternative, implies 0.05 < pvalue < 0.1 (from t-table with one-tail). So student is justified. (d) T = 1.80, df = 500, two-sided alternative, imples 0.05 < pvalue < 0.1 (from t-table with two-tails, df = 30, or could use Normal table since df is large, would result in pvalue=0.0718). So student is justified. (e) This is the correct answer. Question 4. Thirty randomly chosen people were asked in Utrecht if they ever had their bike stolen, and 20 answered yes. Two hundred random people were asked in Amsterdam, and 76 answered yes. (a) Construct a 90% confidence interval for the difference in percentage of people who had a bike stolen in Amsterdam minus Utrecht. (b) With a significance level of 0.05, check if the percentage of people who had a bike stolen in Amsterdam is higher than in Utrecht. Proportion in Utrecht is p 1 = 20/n 1 = , with n 1 = 30. Proportion in Amsterdam is p 2 = 76/n 2 = 0.38, with n 2 = 200. We have at least 10 people in each group, so we will approximate with the Normal distribution. (a) The standard error for a confidence interval of proportions of two groups is: SE = p 1 (1 p 1 ) + p 2(1 p 2 ) = n 1 n 2 The 90% confidence interval is (from z-table we have z = 1.65): (p 2 p 1 ) ± z SE = ± = ( , ). (b) The null hypothesis would be H 0 : percentage of people in Amsterdam less than or equal to in Utrecht (p 2 p 1, or equally ok is p 2 = p 1 ); alternative H A : percentage of people in Amsterdam is higher than in Utrecht (that is, p 2 > p 1 ). There is no reason to run a test, since the difference is already towards more cases in Utrecht, so we are sure that we cannot reject the null hypothesis. Hypothetically, let us assume that the question was check if the number in Amsterdam is lower than in Utrecht. Then H 0 : percentage of people in Amsterdam greater than or equal to in Utrecht (p 2 = p 1, or equally ok is p 2 p 1 ); alternative H A : percentage of people in Amsterdam is lower than in Utrecht (p 2 < p 1 ). Under the null hypothesis, the proportions should be the same, so we can use the pooled estimation to compute the SE: pooledˆp = = SE = pooledˆp (1 pooledˆp) pooledˆp (1 pooledˆp) + = n 1 n = The Z statistic is Z = (p 2 p 1 ) 0 = , so from the Normal table we find pvalue= (1 SE ) = , and we would reject the null hypothesis. Note that the calculation using the SE from the non-pooled version would lead to the same conclusion (very small difference in the numbers). The pooled SE is preferrable for this hypothesis testing, but the difference is very often negligible.

3 Question 5. An independent random sample is selected from an approximately normal population with unknown standard deviation. Find the degrees of freedom and the critical t-value (t ) for the given sample size and confidence level. (a) n = 6, CL = 98% (b) n = 12, CL = 95% (c) n = 29, CL = 99% Results obtained by inspecting the table for two-tails since it is a confidence interval: (a) df = 6 1 = 5, t 5;0.02 = (b) df = 12 1 = 11, t 11;0.05 = (c) df = 29 1 = 28, t 28;0.99 = Question 6. Air quality measurements were collected in a random sample of 40 country capitals in 2013, and then again in the same cities in We would like to use these data to decide if the average air quality has decreased in 2014 with respect to (a) Should we use a one-sided or a two-sided test? Explain your reasoning. (b) Should we use a paired or non-paired test? Explain your reasoning. (c) Should we use a t-test or a z-test? Explain your reasoning. (a) One-sided, we are evaluating whether the air quality has decreased. (b) Paired, data are recorded in the same cities at two different time points. (c) t-test, population standard deviation is unknown. Question 7. Suppose you tossed a coin 10 times independently and obtained heads exactly twice. Using the ideas of hypothesis testing, is there strong evidence to support that the coin is not uniformly distributed? Compute an appropriate p-value and explain your reasoning. The null hypothesis is H 0 : p = 0.5, where p is the chance of heads. The alternative is H A : p 0.5. Pvalue represents the chance of observing something as extreme or more against the null hypothesis, considering that the null hypothesis is true. In this case, we observed 2 in 10, which using the binomial has probability: P r(x = 2) = ( n 2 ) p 2 (1 p) n 2 = ( ) = = 45/ More extreme cases against H 0 would be k = 1 and k = 0: ( ) ( ) n 10 P r(x = 1) = p 1 (1 p) n 1 = = = 10/1024, 1 2 ( ) ( ) n 10 P r(x = 0) = p 0 (1 p) n 0 = = = 1/ Values X=8,9,10 are also as extreme or more than the observed. By symmetry, we have P r(x = 0) = P r(x = 10), P r(x = 1) = P r(x = 9), P r(x = 2) = P r(x = 8). So the pvalue is 2 ( ) = 108/1024 =

4 Question 8. Sally gets a cup of coffee and a muffin every day for breakfast from one of the many coffee shops in her neighbourhood. She picks a coffee shop each morning at random and independently of previous days. The average price of a cup of coffee is 1.40 with a standard deviation of 0.30, the average price of a muffin is 2.50 with a standard deviation of 0.15, and the two prices are independent of each other. (a) What is the mean and standard deviation of the amount she spends on breakfast daily? (b) What is the mean and standard deviation of the amount she spends on breakfast weekly (7 days)? (a) Let X be the coffee price with µ X = 1.4 and σ X = 0.3 and Y the muffin price with µ Y = 2.5 and σ Y = Since they are independent, we have E(X +Y ) = = 3.9, var(x +Y ) = var(x)+var(y ) = = and so the standard deviation is = (b) We can write X 1,..., X 7 and Y 1,..., Y 7 to represent the days of the week. Since they are all independent, the same reasoning holds: E((X 1 + Y 1 ) (X 7 + Y 7 )) = = 27.3, var((x 1 + Y 1 ) (X 7 + Y 7 )) = var(x 1 + Y 1 ) var(x 7 + Y 7 ) = = , and so the standard deviation will be = Question 9. In each part below, there is a value of interest and two scenarios (I and II). For each part, report if the value of interest is larger under scenario I, scenario II, or whether the value is equal under the scenarios. (a) The standard error of sample mean when sample standard deviation is 120 and (I) n=25 or (II) n=125. (b) The margin of error of a confidence interval when the confidence level is (I) 90% or (II) 80%. (c) The p-value for a Z-statistic of 2.5 when (I) n = 500 or (II) n = (a) SE = s n, hence larger under scenario (I). (b) To achieve greater confidence, we need a wider interval, so margin of error is larger under scenario (I). (c) Z = PointEstimate NullValue = 2.5, so n is already taken into account when we compute the Z SE statistic and does not matter afterwards (the distribution is Normal and does not depend on n). Thus, there is no difference in pvalues between the scenarios. Question 10. Rock-paper-scissors is a hand game played by two or more people where players choose to sign either rock, paper, or scissors with their hands. You want to evaluate whether players choose between these three options randomly, or if certain options are favoured above others. You ask two friends to play rock-paper-scissors and count the times each option is played. The data are: Rock 43 times, Paper 21 times, Scissors 35 times. Compute the Chi-Square statistics and its p-value to evaluate whether players choose between these three options randomly, or if certain options are favoured above others.

5 The total number of counts is = 99. The expected counts for each situation are the same: (1/3) * 99 = 33 for each option. We can use a Chi-Square with df = 3 1 = 2. The statistic is χ 2 stat for df=2 = ((43 33) 2 )/33 + ((21 33) 2 )/33 + ((35 33) 2 )/33 = 7.52, and so 0.02 < pvalue < 0.05 (by inspecting the table for 7.52 with df = 2). Since a significance level was not defined, we cannot decide if we can reject the null hypothesis. Nevertheless, the pvalue is reasonably small. For those who want to discuss further the topic: First of all, 99 samples can be considered strange, since there are two players each time. We could think that once one of the players did not show a hand, which happens (and he/she lost that game). Second, one may argue about the independence if the same players keep playing the game. We can assume that you ask two friends to play, then other two friends, and so on (you need many friends though). Question 11. Let X be a random variable that takes the value 0 with probability 1/2 (that is, P r(x = 0) = 1/2), and takes the value 1 with probability 1/2. Let Z be a random variable, independent of X, that takes the value -1 with probability 1/2, and takes the value 1 with probability 1/2. Let U be a random variable constructed as U = X Z. (a) Prove that X and U are not independent (use the definition of independence). (b) Compute the expected value E(U) and the variance var(u). (a) We can show that the rule of independence (that is, P r(u X) = P r(u)) does not hold: P r(u = 0 X = 0) = 1, while P r(u = 0 X = 1) = 0, and therefore P r(u = 0) = 1/2 and P r(u = 0 X = 0) P r(u = 0). (b) VERSION1: First, let us compute E(Z). E(Z) = x Z x P r(z = x) = = 0. Because X and Z are independent, we have cov(x, Z) = 0 and hence E(XZ) = E(X)E(Z), so E(U) = E(XZ) = E(X)E(Z) = 0. As for the variance: var(u) = E(U 2 ) (E(U)) 2 = E(U 2 ) = E(X 2 Z 2 ) = E(X 2 )E(Z 2 ), E(X 2 ) = x X x 2 P r(x = x) = = 1 2, E(Z 2 ) = x 2 P r(z = x) = ( 1) = 1. Therefore var(u) = 1 2. x Z (b) VERSION2: Another possible approach is to write down all possible values of U and their probabilities, and go from there. We know that U = XZ with X independent of Z, hence P r(u = 0) = 1/2, P r(u = 1) = 1/4 and P r(u = 1) = 1/4. Therefore, E(U) = x P r(u = x) = ( 1) = 0, x U E(U 2 ) = x U x 2 P r(u = x) = ( 1) = 1 2, var(u) = E(U 2 ) (E(U)) 2 = E(U 2 ) = 1 2.

6 Formulas (assumptions/conditions to use these formulas are not explicit here but must be considered) Sample x 1,..., x n. Sample mean: x = 1 n n i=1 x i. Sample variance: s 2 = 1 n 1 n i=1 (x i x) 2. ˆx µ T-statistic, Z-statistic:, where ˆx depends on what is being tested. For instance, it can SE be x, or a proportion ˆp, or a difference of means, etc. Degrees of freedom of T are n 1, or min(n 1, n 2 ) 1 for two samples. s Standard error: SE = 2, where the sample variance n s2 depends on what is being estimated (if population variance σ 2 is known, then we use it instead of s 2 ). For instance, for proportions we have s 2 = p (1 p), where p may come from the null hypothesis, or from the sample(s), or pooled from 2 groups. For a binomial distribution over n trials and success chance of p, a normal approximation would have µ = n p and σ 2 = n p (1 p). For difference of estimators from two samples, we use s 2 1 SE = + s2 2, n 1 n 2 where s 1, n 1 come from group 1, and s 2, n 2 from group 2. Margin of error: ME = v SE, where v comes from the appropriate table (Normal, t-table with appropriate degrees of freedom). Confidence interval: (point-estimate M E; point-estimate +M E), where point-estimate depends on what is your target (mean, proportion, a difference). Chisquare statistics: k i=1 (O i E i ) 2 E i, where k is the number of types/categories/possibilities, O i the number of observations of type i, E i the number of expected cases of type i. Degrees of freedom are k 1. Expected value (E) of a discrete variable X: E(X) = x X x P r(x = x), E(f(X)) = x X f(x) P r(x = x). Variance (var) and covariance (cov) of random variables X, Y in terms of their expected values: var(x) = E(X 2 ) (E(X)) 2, var(x + Y ) = var(x) + var(y ) + 2 cov(x, Y ), cov(x, Y ) = E(X Y ) E(X) E(Y ). If X,Y are independent of each other, then var(x + Y ) = var(x) + var(y ). Binomial distribution (n trials, success p): P r(x = k) = ( n k) p k (1 p) n k. Geometric distribution (success p): P r(x = k) = p(1 p) k 1.

7 Normal Probability Table Y positive Z Second decimal place of Z Z For Z 3.50, the probability is greater than

8 t-probability Table One tail One tail Two tails Figure 1. Tails for the t-distribution. one tail two tails df

9 Chi-Square Probability Table Figure 2. Areas in the chi-square table always refer to the right tail. Upper tail df In the case of df > 50, one may use the fact that χ 2 df df 2 df Normal(µ = 0, σ = 1) and so the number from the table of the standard Normal distribution can be employed after such transformation.

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