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1. For each scenario below, determine whether the sample is a good sample or not by answering Yes or No along with an explanation of

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1. For each scenario below, determine whether the sample is a good sample or not by answering \"Yes" or \"No" along with an explanation of why you think so. When determining whether the sample is good or not, consider the type of sample (e.g., representative, random, convenient), the sample size, and the consent or response rate. a. NORC (National Opinion Research Center) at the University of Chicago describes on its website how it selects participants for the General Social Survey (GSS)see below and reports that from 1975 to 2012, the G88 had a response rate of 70% or higher. 62,400 Total 888 Participants ' \"Your address was selected in order to represent a cross-section of the country. You are the voice for 50,000 households like you! ' The random selection of households from across the United States ensures that the results of the survey are scientifically valid. ' All households from across the country had an equal chance of being selected for this survey. ' We will randomly select an adult member of your household to complete the interview. ' Your opinions cannot be replaced by anyone else!\" b. Dr. Juarez is a pediatrician. She wants to test the effectiveness of a new asthma medication for children. She is in a group practice with ve other pediatricians. She has the ofce manager look up the names of all the parents whose children have asthma, and she invites half of them to have their child participate in the drug trial. 0. Dr. Yin is an English professor at Swarthmore College. She is concerned about her students' writing ability and wants to know how much writing experience they get. She decides to nd out on average how many papers a Swarthmore student writes in a given semester. She has the registrar generate a random list of 50 Swarthmore sophomores. She emails each of those students and asks them to report how many papers they have been assigned this semester. Ten students responded to the request. 2. In rural China, children on average spend 3 hours per day unsupervised. Suppose that the standard deviation is 1.5 hours, and the amount of time spent alone is normally distributed. We randomly surveyed one Chinese child and found that this child spends 4.25 hours per day unsupervised. a. How far away from the mean (indicate if the child is above or below the mean) is this child in terms of standard deviations? b. What is the probability that a child spends more than 4.25 hours per day unsupervised? c. What is the probability that a child spends less than 1 hour per day unsupervised? d. How many hours per day are children unsupervised if they are in the 70% percentile? (Hint: You will need to convert the z-score into an x-score.) 3. The diameter of Ping-Pong balls manufactured at a large company is normally distributed with a population mean of u = 1.3 inches and a standard deviation of o = .04 inches. The quality control department selects several random samples of N =36 Ping-Pong balls. a. What is the expected value of the mean? What is the standard error of the mean (keep 4 decimals for your answer and for calculations b-d)? b. What percentage of the sample means will be less than 1.287 inches? 0. What percentage of the sample means will be between 1.2868 and 1.3132 inches? d. What two sample means form the boundaries for the middle 60% of the distribution? (Hint: You will need to convert the z-values into sample means.) 4. For a population with a u = 72 and o = 20. What is the probability of obtaining a sample mean greater than 75 a. for a random sample of N = 100 scores? b. for a random sample of N = 225 scores? 5. Based on the solved confidence intervals below, explain what affects the width of a confidence interval (the difference between the upper and lower limits) and how. M=20 0=10 /=30 M=20 0=10 /=500 95% CI=M+(1.96*OM) 95% CI=M+(1.96*OM) =20+(1.96*1.8257) =20+(1.96*0.4472) 95% CI [16.42, 23.58] 95% CI [19.12, 20.88] M=20 0=0.5 /=30 M=20 0=.5 /=500 95% CI=M+(1.96*OM) 95% CI=M+(1.96*OM) =20+(1.96*0.0913) =20+(1.96*0.0224) 95% CI [19.82, 20.18] 95% CI [19.96, 20.04]

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