Two-way tables #5 - Questions and Answers

Question 1 Baseball star David Ortiz-nicknamed "Big Papi"-is known for his ability to deliver hits in high-pressure situations. Here is a two-way table of his hits, walks, and outs in all of his regular-season and post-season plate appearances from 1997 through 2014. Choose a plate appearance at random. Are the events "Hit" and "Post-season" independent? Justify your answer. Answer DEFINITIONS Two events are independent, if the probability that one event occurs in no way affects the probability of the other event occurring. Definition conditional probability: P(A∩B) = P(A and B) P(A/B) = P(A) SOLUTION We note that the table contains P(A) information about 8883 at-bats (given in the bottom right corner of the table). Moreover, 2110 of the 8883 at-bats are Hits, because 2110 is mentioned in the row " Total" and in the column" Hit" of the table. The probability is the number of favorable outcomes divided by the number of possible outcomes: P(Hit) = #of favorable outcomes = 2110 ≈ 0.2375 = 23.75% #of possible outcomes 8883 Next, we note that 352 of the 8883 at-bats are post-season, because 352 is mentioned in the row "Post" and in the column "Total" of the given table. P(Post) = #of favorable outcomes = 352 #of possible outcomes 8883 Next, we note that 87 of the 8883 at-bats are hits in post-season, because 87 is mentioned in the row "Post" and in the column "Hit" of the given table. P(HitandPost) = #of favorable outcomes = 87 #of possible outcomes 8883 Use the definition of conditional probability: P(Hit|Post) = P(Hit and Post) P(Post) = 87/8883 352/8883 = 87 352 ≈ 0.2472 = 24.72% If events A and B are independent, then P(A|B) = P(A) and P(B|A) = P(B) In this case, we note that P(Hit|Post) = 0.2472 is not the same as P(Hit) = 0.2375 and thus the two events are not independent. Question 2 The two-way table shows the eye color of 200 cats participating in a cat show. Green Blue Yellow Total Male 40 24 16 80 Female 30 60 30 120 Total 70 84 46 200 Make a two-way relative frequency table to show the distribution of the data with respect to gender. Round to the nearest tenth of a percent, as needed. Answer We simply have to divide the values in all cells in the table with the total count of given row that has that cell that we need to calculate. 57.2% 28.6% 65.3% 100% 42.8% 71.4% 34.7% 100% 35% 42% 23% 100% Result: 57.2% 28.6% 65.3% 100% 42.8% 71.4% 34.7% 100% 35% 42% 23% 100% Question 3 Researchers carried out a survey of fourth-, fifth-, and sixth-grade students in Michigan. Students were asked whether good grades, athletic ability, or being popular was most important to them. This two-way table summarizes the survey data. | 4th grade | 5th grade | 6th grade | Total Grades 49 50 69 168 Grade Mostimportant Athletic 24 36 38 98 Popular 19 22 28 69 Total 92 108 135 335 Suppose we select one of these students at random. What's the probability that: The student is a sixth grader or a student who rated good grades as important? Answer Definitions: Completed rule P(A') = P(¬A) = 1 – P(A) General addition rule for any two events:P(AorB) = P(A) + P(B) – P(AandB) Solution | 4th | 5th | 6th | Total Grades 49 | 50 | 69 | 168 Grade Mostimportant | Athletic 24 | 36 | 38 | 98 Popular 19 | 22 | 28 | 69 Total 92 | 108 | 135 | 335 S = Sixth grader G = Grades We note that 135 of the 335 people in the table are 6th grades, because 135 is mentioned in the row" Total" and in the column "6th grade" of the given table. The probability is the number of favorable outcomes divided by the number of possible outcomes: P(5th grade) = #of favorable outcomes= 135 We note that 168 of the #of possible outcomes 335 335 people in the table rated good grades as important, because 168 is mentioned in the row" Grades" and in the column "Total" of the given table. P(Athletic and 5th grade) = #of favorable outcomes = 168 We note that 69 of the 335 people #of possible outcomes 335 in the table are 6th graders who rated good grades as important, because 69 is mentioned in the row" Grades" and in the column "6th gradel" of the given table. P(G) = #of favor about comes = 69 #of possible out comes 335 Use the general addition rule: P(S or G) = P(S) + P(G) - P(S and G) 135 + 168 - 69 335 335 335 135+168-69 = 335 234 = 335 ≈ 0.6985 = 69.85% Question 4 A group of children and adults were polled about whether they watch a particular TV show. The survey results, showing the joint relative frequencies and marginal relative frequencies, are shown in the two-way table. | Yes | No | Total Children | 0.3 | 0.4 | 0.7 Adults | 0.25 | x | 0.3 Total | 0.55 | 0.45 | 1 What is the value of x? Answer We have: Yes No Total Children 0.3 0.4 0.7 Adults 0.25 x 0.3 Total 0.55 0.45 1 We observe that x is listed in the column "No" and the row "Adults." Then, by deducting the known value in the column from the column total, we may determine the value of x. (of subtracticg the row total by the known value in the row). Let us consider the column "No". We note that the column total is 0.45 (given at the bottom of the column), while the other known value in the table is 0.4 x=0.45-0.4=0.05 Question 5 Who seems to have more variability in their shoe sizes, men or women? a) Men b) Women c) Neither group show variability d) Flag this Question In general, why use the estimate of n 1 rather than n in the computation of the standard deviation and variance? a) The estimate n-1 is better because it is used for calculating the population variance and standard deviation b) The estimate n-1 is never used to calculate the sample variance and standard deviation c) n 1 provides an unbiased estimate of the population and allows more variability when using a sample and gives a better mathematical estimate of the population d) The estimate n-1 is better because it is use for calculation of both the population and sample variance as well as standard deviation. | Shoe Size (in cm) | Gender (M of F) | | 25.7 | M | | 25.4 | F | | 23.8 | F | | 25.4 | F | | 26.7 | M | | 23.8 | F | | 25.4 | F | | 25.4 | F | | 25.7 | M | | 25.7 | F | | 23.5 | F | | 23.1 | F | | 26 | M | | 23.5 | F | | 26.7 | F | | 26 | M | | 23.1 | F | | 25.1 | F | | 27 | M | | 25.4 | F | | 23.5 | F | | 23.8 | F | | 27 | M | | 25.7 | F | | Shoe Size (in cm) | Gender (M of F) | | 27.6 | M | | 26.9 | F | | 26 | F | | 28.4 | M | | 23.5 | F | | 27 | F | | 25.1 | F | | 28.4 | M | | 23.1 | F | | 23.8 | F | | 26 | F | | 25.4 | M | | 23.8 | F | | 24.8 | M | | 25.1 | F | | 24.8 | F | | 26 | M | | 25.4 | F | | 26 | M | | 27 | M | | 25.7 | F | | 27 | M | | 23.5 | F | | 29 | F | Answer Step 1 We have to show whose shoe size shows maximum variability ic (man of female). We will compute CV (coefficient of variation) for this purpase and accordingly give the result. Firstly we will compute for males Shoe size (in cm) (xi-x̄)² 25.7 0.9604 26.7 0.0004 25.7 0.9604 26 0.4624 26 0.4624 27 0.1024 27 0.1024 27.6 0.8464 28.4 4.6656 28.4 4.6656 25.4 1.6384 24.8 3.5344 26 0.4624 26 0.4624 27 0.1024 27 0.1024 29 5.3824 Total = 435.7 Mean(x̄) = Total / Number of Observation = 453.7 17 = 26.68 Variance(σ²) = 1/n * ∑(xi - x̄)² = 1/17 * 26.7564 = 1.57 ≈ √1.57 = √1.57 CV= √1.57/26.68 *100 = √1.57/26.68 *100 = 4.696 Step 2 Now, we will compute for females Shoe size (in cm) (xi-x̄)² 25.4 0.7569 23.8 6.1009 25.4 0.7569 23.8 6.1009 25.4 0.7569 25.4 0.7569 25.7 0.3249 23.5 7.6729 23.1 10.0489 23.5 7.6729 26.7 0.1849 23.1 10.0489 25.1 1.3689 25.4 0.7569 23.5 7.6729 23.8 6.1009 25.7 0.3249 26.9 0.3969 26 0.0729 23.5 7.6729 27 0.5329 25.1 1.3689 23.1 10.0489 23.8 6.1009 26 0.0729 23.8 6.1009 25.1 1.3689 24.8 2.1609 25.4 0.7569 25.7 0.3249 23.5 7.6729 Total = 814.6 x̄ = 814.6 /31 = 26.27 σ² = 1/n * ∑(xi - x̄)² = 112.0599 /31 = 3.614 σ = √3.614 = 1.90 CV = 1.90/26.27 * 100 = 7.23 Step 3 Answer (1): If we look at CV of the both males and females. The more variability is in female shoe size as compared to male shoe size. So option (B) is correct. Answer (2): (n - 1) provides an umbiased estimate of population and allouss more variability when using a sample and gives a better mathematical estimate of the population. Option (C) is correct.