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1. (A) Classify the following as an example of nominal, ordinal, interval, or ratio level of measurement, and state why it represents this level: rankings of the top ten best-selling authors Ordinal level of measurement, because it assigns and orders the sellers according to their rankings (B) Determine if this data is qualitative or quantitative: Five violent crimes per week Quantitative (C) In your own line of work, give one example of a discrete and one example of a continuous random variable, and describe why each is continuous or discrete. Discrete variable: Number of family members, because it can take only integral values. Continuous variable: Height, because it can take all values within a range. 2. Two popular over-the-counter antacids are compared to determine which brand is preferred. Three hundred subjects are randomly selected then divided into two groups of 150 each, and given one of the two antacids. The researchers in the study were unaware of which antacid was administered at any time during the study. I. What is the population? All users of the antacids II. What is the sample? 300 selected subjects III. Is the study observational or experimental? Justify your answer. Experimental, because subjects were controlled by the researcher’s treatments IV. What are the variables? Variables are brand of antacid preferred and number of subjects that prefer it. V. For each of those variables, what level of measurement (nominal, ordinal, interval, or ratio) was used to obtain data from these variables? Brank of antacid preferred is nominal level while number of subject preferring is ratio level. 3. Construct both an ungrouped and a grouped frequency distribution for the data given below: 68 74 72 68 61 60 72 73 60 66 73 73 71 65 68 60 71 66 72 61 Ungrouped frequency distribution Data Frequency 60 3 61 2 65 1 66 2 68 3 71 2 72 3 73 3 74 1 Grouped frequency distribution Class 60-62 63-65 66-68 69-71 72-74 Frequency 5 1 5 2 7 4. Given the following frequency distribution, find the mean, variance, and standard deviation. Please show all of your work. Class Frequency 61-63 21 64-66 25 67-69 20 70-72 14 73-75 21 N = 21+25+20+14+21 = 101 Mean = (21*62+25*65+20*68+14*71+21*74)/101 = 67.67 Variance = (21*(62-67.67)^2+25*(65-67.67)^2+20*(68-67.67)^2+14*(71-67.67)^2+21*(7467.67)^2)/100 = 18.522 Standard deviation = sqrt(18.522) = 4.3 5. The following data lists the average monthly snowfall for January in 15 cities around the US: 12 40 24 18 38 47 37 13 14 1 29 11 38 30 10 Find the mean, variance, and standard deviation. Please show all of your work. Mean = (12+40+24+18+38+47+37+13+14+1+29+11+38+30+10)/15 = 24.13 Variance = ((12-24.13)^2+(40-24.13)^2+(24-24.13)^2+(18-24.13)^2+(38-24.13)^2+(4724.13)^2+(37-24.13)^2+(13-24.13)^2+(14-24.13)^2+(1-24.13)^2+(29-24.13)^2+(1124.13)^2+(38-24.13)^2+(30-24.13)^2+(10-24.13)^2)/(15-1) = 192.98 Standard deviation = sqrt(192.98) = 13.89 6. Rank the following data in increasing order and find the positions and values of both the 32nd percentile and 74th percentile. Please show all of your work. 171547730405 Order:0 0 1 1 3 4 4 5 5 7 7 7 Rank: 1 2 3 4 5 6 7 8 9 10 11 12 For 32nd percentile R = (12+1)*32/100 = 4.16. So, 32nd percentile = 1+0.16*(3-1) = 1.32 For 74th percentile, R = (12+1)*74/100 = 9.62. So, 74th percentile = 5+0.62*(7-5) = 6.24 7. For the table that follows, answer the following questions: Xy 1 -8 2 -11 3 -14 4 - Would the correlation between x and y in the table above be positive or negative? Negative - Find the missing value of y in the table. Missing value = -17 - How would the values of this table be interpreted in terms of linear regression? There is strong inverse linear relationship between x and y - If a "line of best fit" is placed among these points plotted on a coordinate system, would the slope of this line be positive or negative? Negative 8. Determine whether each of the distributions given below represents a probability distribution. Justify your answer. (A) x 1 2 3 4 P(x) 1/4 1/12 1/3 1/6 Not a probability distribution, because Σ P(x) ≠ 1 (B) x 3 6 8 P(x) 0.2 2/5 0.3 Not a probability distribution, because Σ P(x) ≠ 1 (C) x 20 30 40 50 P(x) 3/10 -0.1 0.5 0.3 Not a probability distribution because P(30) is negative 9. A set of 50 data values has a mean of 45 and a variance of 9. I. Find the standard score (z) for a data value = 52. Z = (52-45)/sqrt(9) = 2.3333 II. Find the probability of a data value < 52. P(x<52) = P(z< 2.3333) = 0.9902 III. Find the probability of a data value > 52. P(x>52) = P(z>2.3333) = 0.0098 10. Answer the following: (A) Find the binomial probability P(x = 4), where n = 12 and p = 0.70. P(x=4) = (12!/(4!*(12-4)!))*0.7^4*(1-0.7)^8 = 0.00779 (B) Set up, without solving, the binomial probability P(x is at most 4) using probability notation. P(x≤4) = P(0)+P(1)+P(2)+P(3)+P(4) (C) How would you find the normal approximation to the binomial probability P(x = 4) in part A? Please show how you would calculate µ and σ in the formula for the normal approximation to the binomial, and show the final formula you would use without going through all the calculations. µ = np = 12*0.7 = 8.4 σ = sqrt(n*p*(1-p)) = sqrt(12*0.7*0.3) = 1.5875 P(x=4) = e^(-(4-8.4)^2/(2*1.5875^2))/sqrt(2*pi*1.5875^2) 11. Describe what a type I and type II error would be for each of the following null hypotheses: : There is no good plan for the Iraq war. Type I error: Conclude that there is a good plan when actually there is no good plan. Type II error: Conclude that there is no good plan when actually there is a good plan 12. A researcher claims that the average age of people who buy theatre tickets is 49. A sample of 30 is selected and their ages are recorded as shown below. The standard deviation is 7. At a = 0.05 is there enough evidence to reject the researcher's claim? Show all work. 50 46 54 48 52 49 46 44 48 53 44 49 57 60 58 51 56 50 55 53 45 52 45 60 52 59 54 59 51 59 Ho: Average age is 49 H1: Average age is different from 49 Sample mean = 51.967 z-statistic = (51.967-49)/(7/sqrt(30)) = 2.3216 critical value = ±1.96 critical region: z<-1.96 or z>1.96 Since z-statistic lies in critical region, null hypothesis is rejected. There is enough evidence to reject researcher’s claim. 13. Write a correct null and alternative hypothesis that tests the claim that the mean distance a student commutes to campus is no less than 9.7 miles? Null hypothesis: Mean distance a student commutes to campus is less than 9.7 miles. Alternative hypothesis: Mean distance a student commutes to campus is greater than or equal to 9.7 miles.