(2) A CHEM 215 student performed 5 measurements of the wt% of SO4²- of a solid power sample. The results are 8.75%, 8.85%, 8.89%, 8.71%, 8.64%. The standard deviation is 0.10%. What is the confidence interval for this student's experiment at 95% confidence level. The t values for 95% confidence for degree of freedom from 1 to 5 are 12.71, 4.303, 3.182, 2.776, 2.571, respectively. Show your work!
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- A clinical chemist obtained the following data for the alcohol content of a sample of blood: % C2H5OH: 0.084, 0.089, and 0.079. Calculate the 95% confidence interval for the mean assuming that (a) the three results obtained are the only indication of the precision of the method and that (b), from previous experience on hundreds of samples, we know that the standard deviation of the method s 5 0.005% C2H5OH and is a good estimate of s.Does the data set of 155.8, 158.4, 194.9, 164.8, 154.5, 184.3, 114.8 and 157.8 ppm contain outlier(s)? (Given that Qcrit.90 % confidence interval (n = 8) D 0.468) %3D %3D O All data should be retained O It is difficult to judge O Both 114.8 and 194.9 ppm are outliers O 114.8 ppm is the outlier O 194.9 ppm is the outlierAnalysis of several plant-food preparations for potassium ion yielded the following data: Sample Percent K+ 1 3.90, 3.96, 4.16, 3.96 2 4.48, 4.65, 4.68, 4.42 Find the mean ; standard deviation variance and RSD for each sample.
- 2 (1) When performing chemical measurements, replicate experiments are often needed. This is to mitigate (Systematic or Random) error. (2) A CHEM 215 student performed 5 measurements of the wt% of SO4²- of a solid power sample. The results are 8.75%, 8.85%, 8.89%, 8.71%, 8.64%. The standard deviation is 0.10%. What is the confidence interval for this student's experiment at 95% confidence level. The t values for 95% confidence for degree of freedom from 1 to 5 are 12.71, 4.303, 3.182, 2.776, 2.571, respectively. Show your work!3. Seven analyses for the phosphorus content of a fertilizer resulted in 16.2, 17.5, 15.4, 15.9, 16.8, 16.3, and 17.1%. Find (a) the standard deviation, s, , and (b) the 95% confidence interval for the true value, H. 4. The following five values were obtained for the wt % of an organic acid in a sample: 30.3, 31.1, 32.6, 36.7, 28.9. Determine if the value of 36.7 may be rejected at the 90% (two-tailed) confidence level. 5. The concentration of a NaOH solution is determined in a single experiment by titrating a weighed sample of pure, dried KHP (MW = 204.229) with the NaOH. The data are: Weight of KHP = (11.6723 + 0.0001) - (10.8364 ±0.0001) g Volume NaOH = (32.68 + 0.02) - (1.24 + 0.02) mL Where the standard deviations are o, values. (a) What is the molarity of the NaOH? (b) Based on the propagation-of-error formula, what is the standard deviation of the molarity,oM ?Q5. Two methods were used to measure fluorescence lifetime of a dye. The average result for Method 1 is = 1.382 ns with a standard deviation of s1 = 0.039 and n1= 7. The average result for Method 2 is = 1.346 ns with a standard deviation of s2 = 0.015 and n2= 5. Is there a difference between the two methods at the 95% confidence level using F test?
- he variance of chemical experimentdata is square of standard deviation. tru or falseThe following molarities were calculated from replicate standardization of a solution: 0.5025, 0.5012, 0.5012, 0.5028, 0.5023, 0.5015, and 0.5039 M. Assuming no determinate errors, what is the upper limit of the 95% confidence interval?The ages of a group of 122 randomly selected adult females have a standard deviation of 18.5 years Assume that the ages of female statistics students have less variation than ages of females in the general population, so let o = 18 5 years for the sample size caloulation How many female statistics student ages must be obtained in order to estimate the mean age of all female statistics students? Assume that we want 95% confidence that the sample mean is within one-half year of the population mean. Does it seem reasonable to assume that the ages of female statistics students have less variation than ages of females in the general population? The required sample size is (Round up to the nearest whole number as needed)
- Calculate the 95% confidence interval for each set of data in Problem 2 if s is a good estimate of σ and has a value of *set A, 0.015; set B, 0.30 ; *set C, 0.070 ; set D, 0.20 ; *set E, 0.0090; and set F, 0.15.(b) The random error, r', on the result, R, is given by F = √()*² + ( )² R where v', and v, are the random errors on V', and Vt, respectively (1) Let V=26.00 mL, V = 24.00 mL; v' = ±0.01 (3) mL, vt = ±0.01 (2) mL. What is the percentage random error, r'/R x 100%, on R? (2) What is r'/Rx 100 if the conditions of each titration vary (e.g., different portions of the burette stem are used) so that V₁= 26.00 mL, V₁ = 24.00 mL; v' = ±0.05(2) mL, vt = ±0.04(8) mL?In the technique of high performance liquid chromatography (HPLC) the peak area response for a particular analyte is directly proportional to concentration Analgesic tablets typically consist of aspirin and caffeine as the active ingredients, the remainder made up of inert binding material. The concentrations of aspirin and caffeine may be routinely monitored using HPLC, whereby the peak areas obtained from a solution of the tablet are compared with those obtained from standard solutions of the analytes. In an analysis to determine the concentration of caffeine in such a tablet, the following data were recorded: Solutions Analysed: Tablet Sample: Mass of tablet (dissolved in 250 cm methano): 0.523 g Caffeine Standard: Mass of caffeine (diessolved in 500 cm methanol) 0.105 g HPLC results: Solution Peak Area (catfeine) 1. Tablet Sample 325 2. Caffeine Std. 865 The concentration of caffeine in the tablet (expressed as a mass (w/wl percentage) was Oa 1.89 Ob 13.4 O 3.77 Od. 267 Oe 7.54 You…