Six bottles of wine were analyzed for residual sugar, with the results shown bel
ID: 818931 • Letter: S
Question
Six bottles of wine were analyzed for residual sugar, with the results shown below: Evaluate the standard deviation s for each set of data. Pool the data to establish an absolute standard deviation for the method. In an integrated circuit and packaging company, lead-tin solder material was deposited onto silicon substrates to make solder bumps. A problem with the performance of the solder bumps led to analysis of the material by energy dispersive X-ray spectroscopy (EDS). The results from 5 different solder bumps on the same substrate showed the following percentages of tin: 9.25% 9.55% 9.21% 9.17% 9.24% The above data set has a possible outlying result. Apply the Q test to see if rejection (95% confidence) is warranted. Apply the Tn test to the data set here to determine whether rejection (95% confidence level) of the outlying result is statistically justified. A prosecuting attorney in a criminal case presented as principal evidence small fragments of glass found imbedded in the coat of the accused. The attorney claimed that the fragments were identical in composition to a rare Belgian stained glass window broken during the crime. The averages of triplicate analyses for five elements in the glass are shown below. On the basis of these data, does the defendant have grounds for claiming reasonable doubt as to guilt? Employ the 99% confidence level as a criterion for doubt.Explanation / Answer
Solution 1a) std deviation of each set
s1 = sqrt(((0.05)^2 + (0.10)^2 + (0.08)^2) / 2) = sqrt(0.0189 /2 ) =0.097
s2 = sqrt(((0.06)^2 + (0.05)^2 + (0.09)^2 + (0.06)^2) / 3) = 0.077
s3 = sqrt(((0.05)^2 + (0.12)^2 + (0.07)^2 + (0.08^2) / 4) = 0.0839 = 0.084
s4 = sqrt(((0.05)^2 + (0.10)^2 + (0.06)^2 + (0.09)^2) / 3) = 0.0898 = 0.090
s5 = sqrt(((0.07)^2 + (0.09)^2 + (0.1)^2) / 2) = 0.107
s6 = sqrt(((0.06)^2 + (0.12)^2 + (0.04)^2 + (0.03^2) / 3) = 0.08266 = 0.083