ENGR.ECONOMIC ANALYSIS
14th Edition
ISBN: 9780190931919
Author: NEWNAN
Publisher: Oxford University Press
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- A manufacturer of cell phone batteries wants to estimate the useful life of it'sthe battery (in thousand of hours.) The estimate is the be within 0.10 (100) hours.Assume a 95 percent level of confidence and that the standard deviation of theuseful life of the battery is 0.90 (900 hours). Determine the required sample size.arrow_forwardConsider the estimated equation from Example 4.3, which can be used to study the effects of skipping class on college GPA: 1.39 + .412 hsGPA + .015 ACT – 083 skipped (.33) (.094) n = 141, R = .234. colGPA (.026) (.011) Using the standard normal approximation, find the 95% confidence interval for BraGPA: |(ii) Can you reject the hypothesis H,: (iii) Can you reject the hypothesis Hg: BisGPA = 1 against the two-sided alternative at the 5% level? (i) 4 against the two-sided alternative at the 5% level?arrow_forwardThe average price of 350 cellphones is Php 20,000 with a sample standard deviation of Php 2500. Find the 99% confidence interval of the true average price of the cellphones.arrow_forward
- Find the confidence interval for estimating the population proportion for parts (a) through (c). a. 90% confidence level; n = 500; p=0.30 b. 95% confidence level; n = 220; p=0.04 c. α = 0.07; n = 345; p=0.50 Click the icon to view the standard normal table of the cumulative distribution function. a. The confidence interval isarrow_forwardConsider the estimated equation from Example 4.3, which can be used to study the effects of skipping class on college GPA: colGPA = 1.39.412 hsGPA+.015 ACT- .083 skipped (.33) (.094) (.011) (.026) n = 141, R² = .234. (ii) (i) Using the standard normal approximation, find the 95% confidence interval for BhsGPA- Can you reject the hypothesis Ho: ßhsGPA .4 against the two-sided alternative at the 5% level? Can you reject the hypothesis Ho: PhsGPA = 1 against the two-sided alternative at the 5% level? - (iii)arrow_forwardFind the lower bound of the confidence interval, rounding the result to four decimal places. Р(1 — р) p - E = P – Z«j2\ %3D 0.57(1 – 0.57) = 0.57 1.960 1,007 Find the upper bound of the confidence interval, rounding the result to four decimal places. P(1 — р) p + E : p + z 0.57(1 – 0.57) = 0.57 + 1.960 1,007 Therefore, a 95% confidence interval for the population proportion of adults who lack confidence they will be able to afford health insurance in the future is from a lower bound of X to an upper bound ofarrow_forwardQuestion of the day. The mean weekly sales of the chocolate bar in candy stores was 146-3 bars per store. After an advertising campaign the mean weekly sales in 22 stores for a typical week increased to 153-7 and showed a standard deviation of 17-2. Was the advertising campaign successful?arrow_forwardA sample of 21 minivan electrical warranty repairs for “loose, not attached” wires (one of several electrical failure categories the dealership mechanic can select) showed a mean repair cost of $45.66 with a standard deviation of $27.79. (a) Construct a 95 percent confidence interval for the true mean repair cost. (b) How could the confidence interval be made narrower? (Data are from a project by MBA student Tim Polulak.)arrow_forward7. It is well known that car buyers often add accessories to their new cars. A sample of 179 Ottawa purchasers yielded a sample mean of $5000 worth of accessories added to the purchase above the base sticker price. Suppose the cost of accessories purchased for all Ottawa has a standard deviation of $1500. (a) Calculate a 95% confidence interval for the average cost of accessories on Ottawa. - Section Break - (b) Determine the margin of error in estimating the average cost of accessories on Ottawa. (c) What sample size would be required to reduce the margin of error by 50%?arrow_forwardA magazine conducts an annual survey in which readers rate their favorite cruise ship. All ships are rated on a 100-point scale, with higher values indicating better service. A sample of 33 ships that carry fewer than 500 passengers resulted in an average rating of 85.59, and a sample of 44 ships that carry 500 or more passengers provided an average rating of 81.40. Assume that the population standard deviation is 4.55 for ships that carry fewer than 500 passengers and 3.97 for ships that carry 500 or more passengers. A) What is the point estimate of the difference between the population mean rating for ships that carry fewer than 500 passengers and the population mean rating for ships that carry 500 or more passengers? (Use smaller cruise ships − larger cruise ships.) (b) At 95% confidence, what is the margin of error? (Round your answer to two decimal places.) (c) What is a 95% confidence interval estimate of the difference between the population mean ratings for the two sizes of…arrow_forwardTom wants to estimate the percentage of students who live more than 3 miles from the school.He wants to create a 95% confidence interval which has a margin of error of at most 3% .How many students must be polled to create the confidence interval?arrow_forwardSuppose that the average student buys 12 books a year. You have a theory that by offering a discount you can increase that number. You run a pilot project by offering the discount to a random sample of 25 students and you find that x = 14 and s = 4.32 State the null and alternative hypotheses Give the first symbol in the expression for Hoarrow_forwardarrow_back_iosSEE MORE QUESTIONSarrow_forward_ios
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