1) The life in hours of a 75-watt light bulb is known to be normally distributed
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Question
1) The life in hours of a 75-watt light bulb is known to be normally distributed with = 23 hours. A random sample of 20 bulbs has a mean life of Overscript bar EndScripts = 1014 hours.
Suppose that we wanted the total width of the two-sided confidence interval on mean life to be six hours at 95% confidence. What sample size should be used? Round up the answer to the nearest integer.
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2)Past experience has indicated that the breaking strength of yarn used in manufacturing drapery material is normally distributed and that = 2 psi. A random sample of 8 specimens is tested, and the average breaking strength is found to be 97 psi. Find a 95% two-sided confidence interval on the true mean breaking strength. Round the answers to 1 decimal place.
............. less-than-or-equal-to mu less-than-or-equal-to.................?
3)Use Table V in Appendix A to determine the t-percentile that is required to construct each of the following two-sided confidence intervals. Round the answers to 3 decimal places.
Confidence level = 99%, degrees of freedom = 16
= .......
4)The Bureau of Meteorology of the Australian Government provided the mean annual rainfall (in millimeters) in Australia 1983–2002 as follows (http://www.bom.gov.au/ climate/change/rain03.txt)
499.2, 555.2, 398.8, 391.9, 453.4, 459.8, 483.7, 417.6, 469.2, 452.4, 499.3, 340.6, 522.8, 469.9, 527.2, 565.5, 584.1, 727.3, 558.6, 338.6
Construct a 90% two-sided confidence interval for the mean annual rainfall. Assume population is approximately normally distributed.
5)An Izod impact test was performed on 20 specimens of PVC pipe. The sample mean is x Overscript bar EndScripts equals 1.21 and the sample standard deviation is s = 0.29. Find a 99% lower confidence bound on the true Izod impact strength. Assume the data are normally distributed.
Round your answer to 3 decimal places.
i need the answers in detailed please ... as soon as posibile ....
Explanation / Answer
(1)
E = 3, = 23, C = 95%
z- score for 95% confidence is z = 1.96
N = (z * /E)^2 = (1.96 * 23/3)^2 = 226
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