Question

2. The following arsenic concentrations (in parts per billion, ppb) were reported for 1.8 19 24 250 300 340 580 ground waters of southeastern New Hampshire (Boudette et al, 1985): 2.6 | 2.8 9.5 4.0 4.8 3.5 12 80 14 100 23 120 190

Suppose it was determined that an error was made in recording the last value. if it was 58.0 (rather that 580) what would be the 95% confidence interval for the population mean be

Answer 1

Given Data: 1.3, 1.5, 1.8, 2.6, 2.8, 3.5, 4, 4.8, 8, 9.5, 12, 14, 19, 23, 41, 80, 100, 110, 120, 190, 240, 250, 300, 340, 58 We calculate sample mean and std deviation from given data Sample Mean, - Sample Variance, 2 Sample std dev, s- Vs10881.4396104.314139 Σ(r) = 1936.8 = 77.472 25 Σ(z-2)2 2611545504 24 - 10881.4396 95% CI for μ using t-dist Sample Mean = 77.472 Sample Standard deviations 104.314139 Sample Size -n-25 Significance level-α-1-0.95-0.05 Degrees of freedom for t-distribution, d.f.-n -1 24 Critical Value - ta/2,df to.025.df-24 2.064 (from t-table, two-tails, d.f. - 24 104.314139 Margin of Error = E-ta/2,df × E 2.064 x 20.862828 2.064 × Margin of Error, E43.060877 Limits of 95% confidence interval are given by Lower limit -- E 77.472 -43.06087734.41112334.411 Upper limit--+ E = 77.472 + 43.060877-e 120.532877 120.533 95% confidence interval is: x - (34.411123, 120.532877) E-77.472士43.060877 95% CI using t-dist: 34.411 < μ < 120.533 Note that exact answer using technology is: 77.472-104.314139022474/SQRT (25) *T.INV (1-(1-0.95)/2,25-1, 77.472 104.314139022474/SQRT (25) *T.INV (1-(1-0.95)/2,25-1 (34.4132397028 120.5307602972) @seoThank ,youJ0户