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Course: STAT 101, Fall 2009
School: UPenn
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Silvestro Section Joe 003 Statistics 101, Homework 7 Chapter 13 22) Population = 9,000+ readers of the magazine Parameter of Interest = Rating of products within magazine Sampling Frame = Readers of magazine Sampling Size = 225 readers Sampling Design = Simple random sampling of population Sources of Biases or other problems = Sample size represents less than 2.5% of all readers. Since the sample represents such...

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Silvestro Section Joe 003 Statistics 101, Homework 7 Chapter 13 22) Population = 9,000+ readers of the magazine Parameter of Interest = Rating of products within magazine Sampling Frame = Readers of magazine Sampling Size = 225 readers Sampling Design = Simple random sampling of population Sources of Biases or other problems = Sample size represents less than 2.5% of all readers. Since the sample represents such a small portion of the population the results may not be representative of the entire population. 28a) This represents a voluntary response strategy. The voluntary response bias which exists in all voluntary response surveys creates a strong bias in this data, since people with very strong opinions are more likely to respond, as are people with more negative opinions. b) This represents cluster sampling. The bias which may exist in this case is that only the people who choose to attend the meeting have their opinions recorded, leaving the survey biased towards the opinions of those people more likely to attend the meeting. c) This represents cluster sampling. The bias which could exist in this case is that the results of the survey may only reflect the opinions of people who happened to be in a particular geographic area on a given day. As a result, the survey will not fully reflect the beliefs of the entire employee population. d) This represents systematic sampling. The bias which could exist in this case is that the customers could be recorded in a particular pattern, making the each hundredth person biased towards a particular result. 32) The company cannot determine the proportion of all customers at this dealership who believe that their service is extremely good based upon this survey. This is because that 55% of customers felt this way when surveyed represents a sample statistic, rather than a population parameters (which would include the mean and variation of all responses). The results of a survey of a sample cannot be applied to a population at large because reality is not reflected by the surveying of a sample. The 55% sample statistic can only be to used help estimate the proportion of people at this dealership who are extremely satisfied. Nothing can be inferred about other dealerships, but the results at this dealership can be compared to identical surveys at other dealerships to determine if tampering has occurred. Chapter 14 26a) Mean = 1,020 g SD = 8 g 1,000 = 20 / 8 = 2.5 SDs from mean Chance of falling more than 2.5 SDs below mean = 6.21% according to normal tables Type I error rate = 6.21% b) Type II error rate = Chance that no action is taken to correct a problem Probability that this occurs = 1 P(Type I error) = 93.8% 28a) The company could establish a durability test which will measure quality, specifically testing the durability of the ink on the products. Only a sample of photos needs to be tested each time the process is undertaken, so that a mean quality can be established. Once that is done then the process can be halted only when the mean of a sample falls far enough away from the mean quality to warrant a change. b) The company should group the photos into a large batch to increase sample size so that it is more than 10 times the size of the measured skewness and absolute value of the kurtosis, which is a prerequisite for normal distribution to be used. By making n larger the company can help shrink both Type I and Type II error rates. 30) Control Limits for x-bar chart = (mean) (+/-) ((SD) / (square root of n)) a) Mean = 100, SD = 20, n = 25 100 + 20 / (square root of 25) = 104 = Upper control limit 100 20 / (square root of 25) = 96 = Lower control limit Lower control limit = 96 Upper control limit = 104 b) Mean = 2000, SD = 2000, n = 100 2000 + 2000 / (square root of 100) = 2200 = Upper control limit 2000 2000 / (square root of 100) = 1800 = Lower control limit Lower control limit = 1800 Upper control limit = 2200 32a) Control limit = (+/-) 3 Probability that one sample falls outside of 3 = 2.7% Probability of (A u B u C u D) = P(A) + P(B) 2.7% x 10 samples = 27% chance that at least one falls beyond limit There is a 27% chance that at least one sample falls beyond the control limit.
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