Question

PLEASE DO BOTH

1. + -/12 points By NO sh Touria=ા On the basis of extensive tests, the yield point of a particular type of mild steel-reinforcing bar is known to be normally distributed with a = 100. The composition of the bar has been slightly modified, but the modification is not believed to have affected either the normality or the value of 0. (a) Assuming this to be the case, if a sample of 25 modified bars resulted in a sample average yield point of 8467 lb, compute a 90% CI for the true average yield point of the modified bar: (Round your answers to one decimal place.) 16 (b) How would you modify the interval in part (a) to obtain a confidence level of 96%? (Round your answer to two decimal places.) ---Select-- should be changed to You may need to use the appropriate table in the Appendix of Tables to answer this question. Submit Answer 2. -/12 points My Notes + Ask Your Teacher It was reported that in a survey of 4763 American youngsters aged 6 to 19, 18% were seriously overweight (a body mass index of at least 30; this Index is a measure of welght relative to height). Calculate a confidence interval using a 99% confidence level for the proportion of all American youngsters who are seriously overweight. (Round your answers to three decimal places.) Interpret the resulting interval. We are 99% confident that the true proportion of American youngsters who are seriously overweight is within this interval. We are 99% confident that the true proportion of American youngsters who are seriously overweight is above this interval. We are 99% confident that the true proportion of American youngsters who are seriously overweight is below this interval. We are 99% confident that the true proportion of American youngsters who are serlously overweight is outside this interval.

Answer 1

á = 8467 100 n critical : २७ 2 value at x = 0.10 is 1645. :: 90% confidence interval is it =) 8467 + (1.645)x logo =) 8467 I (1645) * 100 20 =) 8462 I (1645) * 20 ao) confidence interval is 58434.1, 8499-97 Critical 2 value at 2=0.04 is 2.0537 96% confidence interval is Ž I 2012 * = -). 8467 + (2.0537) * 100 8467 I (2.0537) 100 -) 8467 + (2.0537) *20. - $425,926, 8506-014 *), confidence intered wal suas.9, 841]

So 92% confidence intoval for proportion is P*(1-P) V n P= 18% = 0.18. , n = 4763 K = 0.01 , al2 = 0.005 2 critical value at y = 0.005 is 2-5758 Margin of Error ME = 28. Pre (1-P) 0.18*(1-0-18) = (2-5758) * 4763 ME = (2-5158)* 50:18 * 0.82 4763 99% =) M6 = 0.014339. confidence interval for proportion 0.18 + (0.014339). [0.166, 0.194] option A (within interval).