Two-way tables #1 - Questions and Answers

Question 1 Mark All That Are True. We can use the Normal (Z) table to find probabilities about: You must choose all five correct statements to get credit. A) individuals, if the population is Normal B) individuals, if the population is NOT Normal C) averages based on small n, if the population is Normal D) averages based on small n, if the population is NOT Normal E) averages based on large n, if the population is Normal F) averages based on large n, if the population is NOT Normal G) count of successes out of n independent trials H) sample proportion of successes out of n independent trials, when np and n(1-p) is large enough Answer Step 1 The correct options are below: A) [Individuals, if the population is Normal]. C) [Averages based on small n, if the population is Normal]. E) (Averages based on large n, if the population is Normal]. F) (Averages based on large n, if the population is NOT Normal]. H) [Sample proportion of successes out of n independent trails, when np and n(1 – p) is large enough]. Question 2 A group of 125 truck owners were asked what brand of truck they owned and whether or not the truck has four-wheel drive. The results are summarized in the two-way table below. Suppose we randomly select one of these truck owners. Four-wheel drive No four-wheel drive Ford Chevy 28 32 20 17 18 10 Dodge What is the probability that the person owns a Dodge or has four-wheel drive? (a) 20 (b) 20 80 (c) 125 (d) 125 90 110 (e) 125 Answer Given: Four-wheel drive No four-wheel drive Ford 28 17 Chevy 32 18 Dodge 20 10 Let us first determine the row/column totals, which is the sum of all counts in the row/column. Four-wheel drive No four-wheel drive Ford 28 Chevy 32 Dodge 20 Total 28+32+20 = 80 17 18 10 17+18+10 = 45 Total 28+1745 32+18=50 20+10=30 45+50+30 = 125 The table contains 125 people in total, which is given in the bottom right corner of the table. The people who own a dodge or have four-wheel drive are given in the row "Dodge" and given in the column "Four-wheel drive" of the above table, which corresponds with 28+ 32+20+10= 90 people. Thus 90 of the 125 people own a dodge or have four-wheel drive. The probability is the number of favorable outcomes divided by the number of possible outcomes: P(Dodge or four-wheel drive) of favorable outcomes 90 of possible outcomes 125 Answer: (d) 125 90 Question 3 A table of values for f,g, f', and g' is given. f(x) g(x) f'(x) f(x) a) If h(x)=f(g(x)), find h'(3) b) If H(x)=g(f(x)), find H'(1). Answer 1 Given: To determine: a) h(x) = f(g(x)) To find h'(3) h(x) = f(g(x)) Then,h'(x) = f'(g(x)).g(x) Putting x=3, h'(3) = f'(g(3)). g'(3) From the given table, g(3) = 2 and g'(3) = 9 So, h'(3) = f'(2)-9 From the table, f'(2) = 5 h'(3) = 5.9 1 3 2 4 6 2 1 8 5 7 3 7 2 7 9 h'(3) = 45 Answer 2 b) H(x)=g(f(x)) Then, H'(x) = g(f(x)) · f'(x) Putting x=1 H'(1) = g(f(1)) • f'(1) From the given table, f(1) = 3 and f'(1) = 4 H'(1) = 9.4 H'(1) = 36 Question 4 In a Two-way table, which type of distribution looks at percentages based on one variable only? Answer 20 IN ATWO-WAY TABLE MARGINAL DISTRIBUTION OF EACH VARIABLE LOOKS AT PERCENTAGES BASED ONE VARIABLE ONLY. Question 5 Find the expected count and the contribution to the chi-square statistic for the (Control, Disagree) cell in the two-way table below. Strongly Agree Agree Neutral Disagree Strongly Disagree Control Treatment 38 60 47 2 12 45 9 4 11 2 Round your answer for the excepted count to one decimal place, and your answer for the contribution to the chi-square statistic to three decimal places. Expected count? Contribution to the chi-square statistic ? Answer Step 1 a) The expected frequencies are calculated by using the following formula: Expected frequency = Row total-Column total Overall total Step 2 Then, expected frequencies are calculated as follows: Strongly Agree Agree Neutral Disagree Strongly Disagree Control 46.9 44.0 5.3 7.7 Treatment 51.1 48.0 5.7 8.3 6.2 6.8 The formula for finding the test statistic is obtained as follows: 2² = ∑ (0;-) The chi-square statistic is calculated as follows: E Observed count (0) Expected count (E) (O-E)2 E 38 46.9 1.678 47 44.0 0.205 2 5.3 2.021 12 7.7 2.470 11 6.2 3.679 60 51.1 1.539 45 48.0 0.188 9 5.7 1.853 4 8.3 2.264 2 6.8 3.372 Total: 230 x² = 19.269 Thus, the chi-square statistic is 19.269