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QUANTITATIVE RESEARCH METHODS I, IB&M EXAM Questions and Verified Correct Answers Latest Update 2024 GRADE A+
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a. The probability that both event A and event B occur is at least as large as the probability that event A occurs. b. Once you know the probability that event A occurs and you know the probability that event B occurs, then you can always calculate the conditional probability that event A occurs given that event B occurs. c. The probability that event A or event B occur is at least as large as the probability that event A occurs. CORRECT d. Suppose that event A and event B are independent. Then the conditional probability that event A occurs given that event B occurs is equal to the conditional probability that event B occurs given that event A occurs.
b. The median is not affected by the addition of the two observations. c. The mean can but need not change due to the addition of the two observations. d. The number of outliers increases due to the addition of the two observations. CORRECT
8 2 8 where x refers to a tree’s age and y to the number of apples harvested from a tree. The botanist tests at a significance level of 0.05 whether there is a linear relationship between age and number of apples. What does she conclude? a. The relevant t - test statistic lies inside the rejection region; she consequently does not reject the null hypothesis that there is no linear relationship between age and number of apples. b. The relevant t - test statistic lies inside the rejection region; she consequently does reject the null hypothesis that there is no linear relationship between age and number of apples. CORRECT c. The relevant t - test statistic lies outside the rejection region; she consequently does not reject the null hypothesis that there is no linear relationship between age and number of apples. d. The relevant t - test statistic lies outside the rejection region; she consequently does reject the null hypothesis that there is no linear relationship between age and number of apples.
The scatter diagram must contain the 7 data points. Furthermore, the diagram should be clear. So, the axes should be labelled and have scales on both axes. Secondly, axes should not be truncated (a break in an axis is fine). A perfect scatter diagram yields 5 points. You know the value of the population variance σ^2 , but you are unaware of the fact that the true population mean is 90. This test has a certain power. What would happen with the power of the test if you would double the sample size? a. The power of the test decreases. b. The power of the test remains the same c. The power of the test increases. CORRECT d. You need to know the value of σ^2 in order to be able to say some- thing about possible changes in the power of the test.
Using the table one obtains: Points: 1 for x ¯, 1 for y ¯, 2 for s^2 x^ , 2 for s^2 y , 2 for sxy. The sample coefficient of correlation is therefore: r = sxy sxsy
= 0_._ 729_._ (2 points) The estimate of the slope is b 1 = (^) ss 2 xy^ = (^109) = 0_._ 655. (2 points) The estimate of the intercept is b 0 = y ¯ − b 1 x ¯ = 110 − 0_._ 655 × 100 = 44_._ 47. (2 points) So, the least squares estimate for the linear regression model is: x^166_.^33 y ˆ = 44._ 47 + 0_._ 655 x ( 2 points) NB: 1 point subtracted if the hat on y is missing or if the equation features ǫ. b. Calculate the sample coefficient of correlation. xi ( xi − x ¯)^2 yi ( yi − y ¯)^2 ( xi − x ¯)( yi − y ¯) 86 196 90 400 280 106 36 119 81 54 111 121 127 289 187 84 256 105 25 80 108 64 111 1 8 115 225 111 1 15 90 100 107 9 30 x ¯ = 100 s
x y ¯=^110 s^2 = 134_._ 33 y sxy^ =^109 c. Calculate the least squares estimate for the linear regression model y = β 0 + β 1 x + ǫ, where the IQ of the mother is the independent variable and the IQ of the daughter is the dependent variable.
Bob, a colleague of Anne, has discovered that the population standard deviation regarding Belgian average temperatures is σB = 1_._ 7. He wants to perform the following test using Anne’s (Belgian) data: H 0 : μB = 15 , H 1 : μB > 15 , α = 0_._ 02_._ c. Determine the rejection region of Bob’s test and make a sketch of it. Then calculate the probability that Bob makes a type II error if the true value of the mean Belgian average temperature is μB = 16. The hypotheses are: H 0 : μB − μN = 0 , H 1 : μB − μN > 0_._ (2 points) From part a. we infer that we can assume that σN^2 = σ^2 B^. So, we have to use the pooled variance estimator 2 2 2 2 s 2 = ( nN − 1) s + ( n (^) B − 1) s s + s p N nN + nB − 2
B (^) = 3_._ 125_._ (1 point) Under the null hypothesis the test statistic t = ,^ X ¯ B − X ¯ N p nB nN t - distributed with 15 + 15 − 2 = 28 degrees of freedom. (1 point) You reject H 0 if t > t 0_._ 10;28 = 1_._ 313. (1 point) The test statistic is s^2 ( 1 + 1 ) is Student t = q^
= 2_._ 324_._ (1 point) (^25) ( 1 + 1 ) 8 15 15 Since 2_._ 324 > 1_._ 313, we reject the null hypothesis and we conclude that there is statistical evidence that the mean Belgian average tem- perature exceeds the mean Dutch average temperature. (2 points) Because t 0_._ 025;28 = 2_._ 048 and t 0_._ 010;28 = 2_._ 467, the p - value of the test lies in between 0.010 and 0.025. (2 points)
Making a type I error can be seen as a ‘success’. Since the samples are independent the number of type I errors Bob makes is binomially distributed with 8 trials. (2 points) The probability that a a type I error occurs when executing a test is equal to the level of significance. So, the succes probability is α = 0_._ 02. (2 points) Therefore: P (3 type I errors) =
× 0_._ 023 × 0_._ 985 = 0_._ 0004_._ (2 points) d. Bob investigates the average (yearly) temperatures of 8 Euro- pean countries. For each country he uses a sample of 15 observa- tions (the 8 samples are independent) and a significance level of α = 0_._ 02 to test whether the mean average temperature exceeds some value. What is the probability that Bob makes exactly 3 type I errors when executing these 8 tests? Under the null hypothesis X ¯ B − 15 √ Bob therefore rejects H 0 if x ¯ B^ − 15 √ is standard normally distributed. > 2_._ 05. Equivalently, Bob rejects H 0 if x ¯ B > 15 + 2_._ 05 × (^) √^1_.^7 = 15._ 90. ( 3 points) The sketch should clearly reveal that the graph of the density function of the normal distribution is symmetric around its mean and bell- shaped. (1 point) The rejection region must be indicated in the graph. (1 point) A type II error occurs if the null hypothesis is not rejected even though it is false. (1 point) If the true value of the mean is μB = 16, then the probability that a type II error occurs is β ( X ¯^ < 15_._ 90 | μ = 16 ). ( 1 point) Hence: β = P X ¯ (^) B − μB 15_._ 90 − 16 σB/ nB
(2 points)