Question

9. The amount of time that a certain cell phone will keep a charge is known to be normally distributed with a standard deviation o= 15 hours. A sample of 45 cell phones has a mean time of 145 hours. Let u represent the population mean time that a cell phone will keep a charge. a. What is the point estimate of u? b. What is the standard error of the point estimate? Round to three decimal places. c. Suppose that a 90% confidence interval is to be constructed for the mean time. Find the margin of error for this confidence interval (use three decimal places). e. Construct the 90% confidence interval and interpret your result: f. What sample size would be necessary so that a 98% confidence interval will have a margin of error of 2 hours?

Answer 1

Solution ! Given - mean (8) = 145 hrs 15 hrs 45 sample size a) point estimate ñ 145 b) standard error (ox) = 0 Jn 2.236 J45 c) At gol C 22/2 =1-645 from standard normal tuble 15 margin of error (E)= 22/2 xo Jn 20 = 1.645X 15 545 3.678 e) At gol. , confidence interveel for population mean is, 25. x + Е 145 I 3.678 (141.322 148.678) 1 Interpretation - me are 90 % confident that interval from 141.322 to 148-678 actually does contain the trie value of meun:

f) Given ! margin of error = 2 IS A+ 5 level 12' is 98 - confidence 2 - 1 - 98-) 2= 1-0.98 0.02 2/2 = 0.02/2 0.01 212 2 10 22/2 = 2.326 Sample size (n) = 2 15 (2/285 (2.326 X 15 ) 2 1) 20 Sample size (n) 305