York University MATH 1131 (FALL 2005): Introduction to Statistics Mid Term Test Friday, Oct 28, 2005

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1 York University MATH 1131 (FALL 2005): Introduction to Statistics Mid Term Test Friday, Oct 28, 2005 Last Name: Given Names: Student Number: Signature : DO NOT WRITE IN THIS AREA Read the following instructions carefully 1. This exam is closed book and closed notes. 2. You are allowed a calculator. You may NOT borrow a calculator from another student. 3. All work must be done in the free space provided below for each problem or on the back of the preceding page. 4. No other sheets of paper will be accepted or marked. 5. Show all work. Answers without sufficient supporting work will receive *no credit*, even if the answers are correct. 6. Your final grade is 1

2 Question 1 Anthropologists often study soil composition for clues to how the land was used during different periods. The following data on D = soil depth (in cm) and M = percentage of montmorillonite in the soil was taken from an article in a journal. i D i M i You are given D = 70, s D = 21.60, M = 33, sm = 11.68, and r DM = 0.833, where r DM denotes Pearson s sample correlation coefficient between soil depth D and percentage of montmorillonite in the soil M. (a) 7 points Draw a scatterplot of D versus M. D * 90 + * 80 + * 70 + * 60 + * 50 + * 40 + * M (b) 7 points Give the formula for Pearson s sample correlation coefficient r DM. So- 2

3 7 D i D M i M i=1 s lution: r = D s M 7 1. (c) 7 points Find the equation of the least-squares regression line you would use for predicting D = soil depth. Solution: ˆD = M. (d) 7 points The following table gives the residuals e i except for e 5. Compute the residual e 5 and draw the residual plot of e i versus M i. i e i % Plot of e*m. Symbol used is *. % % % % 20 + % % % * %R 10 + * %e * %s %i * %d 0 + %u %a * %l % % * % % * % % % % % % M 3

4 (e) 7 points Use the 1.5(IQR) rule to detect a potential outlier in M. Solution: Q 1 = 27, Q 3 = 34, 1.5(IQR) = 10.5, Q = 16.5, Q = Therefore 58 is clearly an outlier, as can be seen from the residual graph and the scatterplot. (f) 6 points If D is measured in millimeters rather than centimeters, does the value of the sample correlation coefficient change and if so, by how much? Solution: It does not change. Question 2 The heights (in inches) of 20 randomly selected people from a population are listed below You are given the sample mean is and the sample standard deviation is (a) 7 points What percentage of values are within one standard deviation of the mean. Do your results agree with the empirical rule? Explain Solution = 57.29, = There are 14 out of 20, which is 70%. This is pretty close to what the empirical rule predicts which is 68%. (b) 7 points What percentage of values lies within 3 standard deviations of the mean according to Chebyshev s rule? Is this verified by the data? Solution At least 89%.In this data set, the number of values between and is 20 i.e. 100%. So, the prediction of Chebyshev s rule is verified. (c) 7 points Give the five-number summary of the data set. Solution Min. 1st Qu. Median Mean 3rd Qu. Max Question 3 According to the Arizona Chapter of the American Lung Association, 7% of the population has lung disease. Of those having lung disease, 90% are smokers; of those not having lung disease, 25.3% are smokers. 4

5 (a) 8 points Draw a tree diagram for selecting a person from this population (outcomes: having lung disease or not having lung disease) and then recording his or her smoking status (outcomes: smoker or non-smoker). (b) 8 points What is the probability that a person is smoker for a randomly chosen person from this population? Solution: P(S) = P(SD)P(D) + P(S D)P( D) =.07(.90) +.93(.0.253) = (c) 8 points What is the probability that a person has lung disease given that this person is smoker? Solution: P(DS) = P(D)P(SD) = P(SD)P(D)+P(S D)P( D) Question 4 The probability that 0, 1, 2, 3, or 4 people will be placed on hold when they call a radio talk show is shown in the distribution. Let X be the number of people being placed on hold. x p(x) (a) 7 points Compute the mean value µ X. Solution:µ =.34(1) +.23(2) +.21(3) +.04(4) = (b) 7 points Compute the standard deviation σ X. Solution:σ 2 = 1.2, σ = 1.1 5