Question

X 7.3.10-T Question Listed below are the amounts of mercury (in parts per million, or ppm) found in tuna sushi sampled at different stores. The sample mean is 0.971 ppm and the sample standa deviation is 0315 ppm. Use technology to construct a 90% confidence interval estimate of the mean amount of mercury in the population 067 062 0.74 096 1.15 1 19 What is the confidence interval estimate of the mean amount of mercury in the population? ppm<< ppm (Round to three decimal places as needed) Enter your answer in the edit fields and then click Check Answer

Answer 1

7.3.10-T The R command to construct the 90% confidence interval estimate is as shown below: > X<-c(1.47,0.67,0.62,0.74,0.96,1.15,1.19) > t.test(x,conf.level=0.90) One Sample t-test data: x t = 8.1528, df = 6, p-value = 0.0001832 alternative hypothesis: true mean is not equal to o 90 percent confidence interval: 0.7398941 1.2029630 sample estimates: mean of x 0.9714286 Hence the confidence interval is 0.740 ppmu<1.203 ppm

7.3.9-T The R command to construct the 95% confidence interval estimate is as shown below: > library (BSDA) > tsum.test (mean.x=4, s.x=5.5, n.x=61, conf.level=0.95) One-sample t-Test data: Summarized x t = 5.6802, df = 60, p-value = 4.174e-07 alternative hypothesis: true mean is not equal to o 95 percent confidence interval: 2.591385 5.408615 sample estimates: mean of x Warning message: In tsum. test (mean.x = 4, s.x = 5.5, n.x = 61, conf.level = 0.95): argument 'var.equal' ignored for one-sample test. Hence the confidence interval is 2.591 kg < u<5.409 kg

7.3.10-T The R command to construct the 90% confidence interval estimate is as shown below: > X<-c(1.12,1.17,0.57,0.86,0.62,0.92,0.78) > t.test(x, conf.level=0.90) One Sample t-test data: X t = 9.958, df = 6, p-value = 5.933e-05 alternative hypothesis: true mean is not equal to 0 90 percent confidence interval: 0.6944812 1.0312330 sample estimates: mean of x 0.8628571 Hence the confidence interval is 0.694 ppmu <1.031 ppm

7.3.20 The R command to construct the 95% confidence interval estimate is as shown below: > X<-c(4045,3884, 3849, 4023, 4323, 4806, 4648, 4035, 5010, 4821, 4343, 4325) > t.test(x,conf.level=0.95) One Sample t-test data: x t = 38.018, df = ll, p-value = 5.04e-13 alternative hypothesis: true mean is not equal to O 95 percent confidence interval: 4091.257 4594.076 sample estimates: mean of x 4342.667 Hence the confidence interval is 4091.3< u<4594.1 7.3.15 a. The best point estimate is 3.8 mg/dL If you have more questions you can send them to us and we will try our best to help you out. Good luck with your studies!!!