MATLAB: An Introduction with Applications
6th Edition
ISBN: 9781119256830
Author: Amos Gilat
Publisher: John Wiley & Sons Inc
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- a sample size n=99 is drawn from a population whose standard deaviation is 11 (a) find the margin of error for a confidence interval for u (population mean). round the answer to at least three decimal places the margin of error for a 90% confidence for u (population mean) isarrow_forwardFind the margin of error for a 95% confidence interval for estimating the population mean when the sample standard deviation equals 100, with a sample size of (i) 400 and (ii) 1600. What is the effect of the sample size? (i) Find the margin of error for a 95% confidence interval for estimating the population mean when the sample standard deviation equals 100 with a sample size of 400. 9.83 (Round to two decimal places as needed.) (ii) Find the margin of error for a 95% confidence interval for estimating the population mean when the sample standard deviation equals 100 with a sample size of 1600. (Round to two decimal places as needed.)arrow_forwardWhat critical value t∗t∗ from Table C would you use for a confidence interval for the mean of the population in each of the following situations? (a)96% confidence interval based on n = 25 observations.(b) A 95% confidence interval from an SRS of 17 observations.(c) A 90% confidence interval from a sample of size 13.arrow_forward
- Let X equal the length of life of a 60-watt light bulb marketed by a certain manufacturer. We do not know the distribution of X except that Var(X) = 1158. Let u be the mean of X. Suppose a random sample of n = 25 bulbs is tested until they burn out, yielding a sample mean of = 1368 hours. (i) Compute an approximate 95% confidence interval for u. (ii) Compute an approximate 95% one-sided confidence interval for that provides a lower bound for u, i.e. compute & such that ɛ P(X - ≤ μµ)≈ 0.95.arrow_forwardA sample of size n=77 is drawn from a population whose standard deviation is 0= 33. Part 1 of 2 (a) Find the margin of error for a 95% confidence interval for µ. Round the answer to at least three decimal places. The margin of error for a 95% confidence interval for µ is Part 2 of 2 (b) If the confidence level were 90%, would the margin of error be larger or smaller? (Choose one) ▼ because the confidence level is (Choose one) ▼arrow_forwardOn average, 0(zero) is expected for the t statistic when the null hypothesis is true. Why? The value for an unknown population mean is estimated with a confidence interval using the t statistic? Why?arrow_forward
- What critical value of t* should be used for a 99.5% confidence interval for the population mean based upon a random sample of 18 observations? Find the t-table here. t* = 2.878 t* = 2.898 t* = 3.197 t* = 3.222arrow_forwardA random sample of 28 items is drawn from a population whose standard deviation is unknown. The sample mean is = 790 and the sample standard deviation is s= 15. Use Appendix D to find the values of Student's t (a) Construct an interval estimate of u with 99% confidence. (Round your answers to 3 decimal places.) The 99% confidence interval is from to (b) Construct an interval estimate of u with 99% confidence, assuming that s= 30. (Round your answers to 3 decimal places.) The 99% confidence interval is from to (c) Construct an interval estimate of p with 99% confidence, assuming that s= 60. (Round your answers to 3 decimal places.) The 99% confidence interval is from to (d) Describe how the confidence interval changes as s increases. O The interval stays the same as s increases. O The interval gets wider as s increases. O The interval gets narrower as s increases.arrow_forwardWhat price do farmers get for their watermelon crops? In the third week of July, a random sample of 40 farming regions gave a sample mean of = $6.88 per 100 pounds of watermelon. Assume that σ is known to be $1.90 per 100 pounds. (a) Find a 90% confidence interval for the population mean price (per 100 pounds) that farmers in this region get for their watermelon crop (in dollars). What is the margin of error (in dollars)? (For each answer, enter a number. Round your answers to two decimal places.)lower limit $ upper limit $ margin of error $ (b) Find the sample size necessary for a 90% confidence level with maximal error of estimate E = 0.37 for the mean price per 100 pounds of watermelon. (Enter a number. Round up to the nearest whole number.) farming regions (c) A farm brings 15 tons of watermelon to market. Find a 90% confidence interval for the population mean cash value of this crop (in dollars). What is the margin of error (in dollars)? Hint: 1 ton is 2000…arrow_forward
- A simple random sample of 60 items from a population with o = 8 resulted in a sample mean of 32. (Round your answers to two decimal places.) (a) Provide a 90% confidence interval for the population mean. to (b) Provide a 95% confidence interval for the population mean. to (c) Provide a 99% confidence interval for the population mean. toarrow_forwardA random sample of 90 observations produced a mean of ?⎯⎯⎯=26.3 from a population with a normal distribution and a standard deviation ?=2.08.arrow_forwardIn each case, use the information given to compute a confidence interval for the population mean ?. Assume that all necessary conditions are present for using the method described in Section 11.2. (Round your answers to two decimal places.) (a) x = 79, s = 9, n = 25, 95% confidence level to (b) x = 102, s = 9, n = 25, 95% confidence level ___to___ (c) x = 102, s = 19, n = 81, 99% confidence level ___to___ (d) x = 102, s = 19, n = 81, 90% confidence level ___to___arrow_forward
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