Question.3194 - 8.3 Why is it not possible in Example 8.1 on page 256 to have 100% confidence? Explain. 8.19 The file Sedans contains the overall miles per gallon (MPG) of 2009 sedans priced under $20.000. 27 31 30 28 27 24 29 32 32 27 26 26 25 26 25 24 Source: Data extracted from “Vehicle Ratings,” Consumer Reports, April 2009, p. 27 a) Construct a 95% confidence interval estimate for the population mean MPG of 2009 sedans (4 cylinder) priced under $20.000, assuming a normal distribution. b) Interpret the interval constructed in (a). c) Compare the results in (a) to those in Problem 8.20 (a) 8.22 One of the major measures of the quality of service provided by any organization is the speed with which t responds to customer complaints. A large family-held department store selling furniture and flooring, including carpet, had undergone a major expansion in the past several years. In particular, the flooring department had expanded from 2 installation crews. The store had the business objective of improving its response to complaints. The variable of interest was defined as the number of days between when the complaint was made and when it was resolved. Data were collected from 50 complaints that were made in the last year. The data were stored in Furniture and are as follows: 54 5 35 137 31 27 152 2 123 81 74 27 11 19 126 110 110 29 61 35 94 31 26 5 12 4 165 32 29 28 29 26 25 1 14 13 13 10 5 27 4 52 30 22 36 26 20 23 33 68 a) Construct a 95% confidence interval estimate for the population mean number of days between the receipt of a complaint and the resolution of the complaint. b) What assumption must you make about the population distrubiton in order to construct the confidence interval estimate in (a)? c) Do you think that the assumption needed in order to construct the confidence interval estimate in (a) is valid? Explain d) What effect might your conclusion in (c) have on the validity of the results in (a)? 8.23 In New York State, savings banks are permitted to sell a form of life insurance called savings bank life insurance (SBLI). The approval process consists of underwriting, which includes a review of the application a medical information bureau check, possible requests for additional medical information and medical exams, and a policy compilation stage in which the policy pages are generated and sent tot the bank for delivery. The ability to deliver approved policies to customers in a timely manner is critical to the profitability of this service to the bank. During a period of one month, a random sample of 27 approved policies was selected, and the total processing time, in days, was a s shown below and stored in Insurance: 73 19 16 64 28 28 31 90 60 56 31 56 22 18 45 48 17 17 17 91 92 63 50 51 69 16 17 a) Construct a 95% confidence interval estimate for the population mean processing time. b) What assumption must you make about the population distribution order to construct the confidence interval estimate in (a)? c) Do you think that the assumption needed in order to construct the confidence interval estimate in (a) is valid? Explain.
Answer Below:
8.3) It is not possible to have 100% confidence because to calculate 100% confidence, one must run the trials infinite number of times and not just for 100 sheets of paper as used in the example 8.1.The entire sheets of paper has to be selected. The confidence interval that we would get from using 100% confidence will be minus infinity to plus infinity, which is not very useful in estimating or predicting any real statistics. The interval will always include 11 (the required size for paper sheets) and that will always indicate that the process is working properly. We use results of only one sample to determine the quality, the manufacturer can never know with 100% confidence that the result of a sample includes the true mean of population. 8.19 a) Data = {27, 31, 30, 28, 27, 24, 29, 32, 32, 27, 26, 26, 25, 26, 25, 24} n = 16 Degrees of freedom = 15 Mean, X? = 27.4375 For 95% confidence, Z α/2 = 1.96 Standard Deviation, σ = Confidence Interval, µ = X? ± Z α/2 = 27.4375 ± 1.96 = 27.4375 ± 1.2608 =26.1767 ≤ µ ≤ 28.6983 b) One can be 95% confident that the mean miles per gallon of 2009 family sedans priced under $20,000 is somewhere between 26.0214 and 28.8536 c) Since the 95% confident interval for population mean miles per gallon of 2007 small SUVs does not overlap with that for the population mean miles per gallon of 2007 family sedans, we are 95% confident that the population mean miles per gallon of 2007 small SUVs is lower than that of 2007 family sedans. 8.22 a) Data = { 54 5 35 137 31 27 152 2 123 81 74 27 11 19 126 110 11 0 29 61 35 94 31 26 5 12 4 165 32 29 28 29 26 25 1 14 13 13 10 5 27 4 52 30 22 36 26 20 23 33 68 } n = 50 Mean, X? = 43.04 Degrees of freedom = 49 For 95% confidence, Z α/2 = 2.0096 Standard Deviation, σ = 41.9261 Confidence Interval, µ = X? ± Z α/2 = 43.04 ± 2.0096 = 43.04 ± 11.92 =31.12≤ µ ≤ 54.96 (b) The population distribution needs to be normally distribution. (c) Normal Probability Plot 0 Ê Ê Ê Ê r rð rà rÏ r¾ -3-2-10123 Z Value Box-and-whisker Plot Days 050100150 From the normal probability plot and the box-and-whisker plot it can be seen that the distribution is skewed to the right. (d) Even though the population distribution is not normally distributed, with a sample of 50, the t distribution can still be used due to the Central Limit Theorem. 8.23 a) Data = { 73 19 16 64 28 28 31 90 60 56 31 56 22 18 45 48 17 17 17 91 92 63 50 51 69 16 17 } n = 27 Degrees of freedom = 26 Mean, X? = 43.889 For 95% confidence, Z α/2 = 2.0555 Standard Deviation, σ = 25.2835 Confidence Interval, µ = X? ± Z α/2 = 43.889 ± 2.0555 = 43.889 ± 10.002 =33.887≤ µ ≤ 53.891 b) The population distribution needs to be normally distributed. c) Normal Probability Plot 0 10 20 30 40 50 60 70 80 90 100 -2-1.5-1-0.500.511.52 Z Value Box-and-whisker Plot Time 1030507090 From the normal probability plot and the box-and-whisker plot it can be seen that the distribution is skewed to the right.More Articles From Statistics
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