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Question 1: Please solve all parts!
\fOn 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 o'. (a) Assuming this to be the case, if a sample of 36 modified bars resulted in a sample average yield point of 8483 lb, compute a 90% CI for the true average yield paint of the modified bar. (Round your answers to one deCImaI place.) (3,:th (b) How would you modify the interval in part (a) to obtain a confidence level of 94%? (Round your answer to two decimal places.) The value ofz v should be changed to . You may need to use the appropriate table in the Appendix of Tables to answer this question. Need Help? The negative effects of ambient air pollution on children's lung function has been well established, but less research is available about the impact of indoor air pollution. The authors of an article investigated the relationship between indoor air-pollution metrics and lung function growth among children ages 6-13 years liVing in four Chinese cities. For each subject in the study, the authors measured an important lung- capacity index known as FEVI, the forced volume (in ml) of air that is exhaled in 1 second. Higher FEV1 values are assomated With greater lung capaCIty. Among the children in the study, 516 came from households that used coal for cooking or heating or both. Their FEV1 mean was 1427 With a standard deviation of 329. (A complex statistical procedure was used to show that burning coal had a clear negative effect on mean FEV1 levels.) (a) Calculate and interpret a 95% (two-sided) confidence interval for true average FEV1 level in the population of all children from which the sample was selected. (Round your answers to one decimal place.) (13986 _~/) Interpret the resulting interval. Q We are 95% confident that the true population mean lies below this interval. 0 We are 95% confident that this interval does not contain the true population mean. We are 95% confident that this interval contains the true population mean. 0 We are 95% confident that the true population mean lies above this interval. Does it appear that the parameter of interest has been accurately estimated? O This interval is quite narrow relative to the scale of the data values themselves, so it could be argued that the mean has not been accurately estimated. (9 This interval is quite narrow relative to the scale of the data values themselves, so it could be argued that the mean has been accurately estimated. 0 This interval is quite wide relative to the scale of the data values themselves, so it could be argued that the mean has been accurately estimated. 0 This interval is quite Wide relative to the scale of the data values themselves, so it could be argued that the mean has not been accurately estimated. J (b) Suppose the investigators had made a rough guess of 330 for the value of s before collecting data. What sample Size would be necessary to obtain an interval width of 40 ml for a confidence level of 95%" (Round your answer up to the nearest whole number.) 1045.9 x children A sample of ultimate tensile strength observations (ksi) was taken. Use the accompanying descriptive statistics output from Minitab to calculate a 99% lower confidence bound for true average ultimate tensile strength, and Interpret the result. (Round your answer to two deCimal places.) N Mean Median TrMean stDev SE Mean M;n;mum Maximum Q1 Q3 153 135.35 135.36 135.37 4.42 0.36 124.40 148.90 131.75 136.25 We are 99% condent that the true average ultimate tensile strength is greater than x ksi. You may need to use the appropriate table in the Appendix of Tables to answer this question. A survey collects demographic, socioeconomic, dietary, and healthrelated information on an annual basis. Here is a sample of 20 observations on HDL cholesterol level (mg/dl) obtained from the survey (HDL is \"good" cholesterol; the higher its value, the lower the risk for heart disease). 36 48 53 55 54 52 50 46 BS 37 45 34 40 44 40 64 94 36 29 47 (a) Calculate a point estimate (in mg/dl) of the population mean HDL cholesterol level. mg/dl (b) Making no assumptions about the shape of the population distribution, calculate a point estimate (in mg/dl) of the value that separates the largest 50% of HDL levels from the smallest 50%. mg/dl (c) Calculate a point estimate (in mg/dl) of the population standard deviation. (Round your answer to three decimal places.) mg/dl (d) An HDL level of at least 60 is considered desirable as it corresponds to a significantly lower risk of heart disease. Making no assumptions about the shape of the population distribution, estimate the proportion p of the population having an HDL level of at least 60
