Question

Please help with part B. I keep redoing the problem and keep getting the wrong answer.

df 818 2. The Q Test. A student doing the titrations in Example II repeated her experiment twice more. She now has obtained five values: 0.555, 0.565, 0.564, 0.567, 0.563 M a. Use the Q test to show that the first value should be rejected (use the tabulated Q for 5 values) b. Recalculate the values for <x> and<d> omitting the value 0.555. SHOW WORK. (The correct value to report is 0.565 ± 0.001 M.)

Answer 1

Since confidence level is missing in the question so I have assumed the confidence level as 95% and all the calculations are based on the same confidence level.

a. 0.555, 0.563,0.564, 0.565, 0.567 Q- gap/range 10.555-0.5631/0.567-0.555 0.667 suppose the confidence level is 95% so for the 5 interval Q value from table is 0.710 which is more than the calculated value from the Q test and hence we can say that at 95% confidence level 0.555 is not a outliner where as at 90% confidence level the Q calculated is more than from the Q-table and hence at 90% confidence 0.555 is an outliner. So at 95% confidence level this value could be excluded. Mean of sample 0.5628 (sum of 0.555, 0.563, 0.564, 0.565, 0.567 and divided by 5) Standard deviation (s) -0.0046 Z* at 95% confidence level :1.96 Value at 95 % confidence level = mean + sz*-0.5628 ± + 000046 X1.96-0.5628± 0,004