BRM Midterm 2 Questions With Verified Answers | LATEST UPDATE

BRM Midterm 2 Questions With Verified Answers | LATEST UPDATE What should be the sample size to provide a 95% confidence interval with a margin of error (E) of 8. Consider the standard deviation as 36. - ANSWER n= (1.96^2 * 36^2) / 8^2 n=77.7924 According to a study, the price of fuel average price of fuel was $3.99. Based on the preliminary analysis, the sample standard deviation is $0.05. Recommend an appropriate sample size to determine a margin of error at 95% confidence level. a. ±0.05$ b. ±0.03$ c. ±0.01$ - ANSWER a. n = [(1.96)2 * 0.052]/0.052 = 4 b. n = [(1.96)2 * 0.052]/0.032 = 11 c. n = [(1.96)2 * 0.052]/0.012 = 97 An Auto company produced as low as 782 and a highest of 1303 vehicles per hour. A 95% confidence interval estimate for vehicle production is desired. What is the planning value for the population standard deviation? - ANSWER Planning Value = Data range / 4 Planning Value = (1303 - 782) / 4 = 130.25 = standard deviation An Auto company produced as low as 782 and a highest of 1303 vehicles per hour. A 95% confidence interval estimate for vehicle production is desired. How large a sample of production plant hours should be taken with the following margin of error? a. 100 cars b. 50 cars c. 30 cars - ANSWER (1303 - 782) / 4 = 130.25 = standard deviation a. n = [(1.96)2 * 1302]/1002 = 7 b. n = [(1.96)2 *1302]/502 = 26 c. n = [(1.96)2 * 1302]/302 = 73 A bank decided to collect information about their customers and assumed a planning value for a standard deviation of $1500 and decided to use $200 as the desired margin of error for the interval estimate of the population mean. Determine the sample size for the following: 1. A 90% confidence interval 2. A 95% confidence interval 3. A 99% confidence interval - ANSWER 1. n = (1.6452*15002)/2002 = 153 2. n = (1.962*15002)/2002 = 217 3. n = (2.5762*15002)/2002 = 374 In a survey of 3121 people, 412 are under-vaccinated. What is the proportion of under-vaccinated people in the local population? - ANSWER p̂ = x /n = 412/3121 = 0.132 Suppose you want to estimate the percentage of the time (with 95% confidence) you're expected to get a red light at a certain intersection. You take a random sample of 100 different trips through this intersection and you find that a red light was hit 53 times. - ANSWER p̂ = 53/100 = 0.53 0.53 - 1.96(√0.53*(1-0.53)/100) = 0.43 0.53 + 1.96(√0.53*(1-0.53)/100) = 0.63 In a recent poll of 200 households, 152 households had at least one computer, estimate the proportion of households in the population that have at least one computer. Construct a 95% interval to estimate population proportion. - ANSWER 0.76 - 1.96(√0.76(1-0.76)÷200) = 0.701 0.76 + 1.96(√0.76(1-0.76)÷200) = 0.819 In words, we are 95% confident that the proportion of households in the population