Question

For the final project of an analytical chemistry laboratory course, the students were asked to quantify the lead (Pb2+) content in a drinking water sample. To receive a passing grade for the project, the students must produce a result (Pb2 concentration) that agrees with the result obtained by the course professor, who was using the same method, at the 95% confidence level. The mean (x), standard deviation (Sy) and number of replicate measurements (n) are shown below for the data obtained by the professor (p) and one student (s) chosen at random x,-3.970 ppb s,-0.239 ppb Student:*, Professor: x,-4.620 ppb sp0.150ppb It Calculate the F value. Are the standard deviations from the professor's and student's data sets significantly different from each other at the 95% confidence level (Yes or No)? The value of Fable can be found in the table of critical F value Number Yes O No calc Calculate the value of tcalc to compare the data sets from the student and the professor at the 95% confidence level Number calc Yes Did the student receive a passing grade for the final project? A list of t values can be found in the Student's ttable O No

Answer 1

F_calc= $\frac{_{S_{s}^{2}}}{S_{p}^{2}}$ = (0.239)²/(0.150)² = 2.54

Now, Total degrees of freedon = n-1 = 6-1 = 5

From table, F = 5.05 for n =5

So, F_calc < F_table . So, NO

Now

t_calc =

nsn pooled)

Where S_pooled= $\sqrt{\frac{S_{s}^{2}(n_{s}-1)+S_{p}^{2}(n_{p}-1)}{n_{s}+n_{p}-2}}$

S_pooled = (((0.239)²*5 - (0.150)²*5) / 6+6-2)^1/2 = ((0.2856 - 0.1125)/ 10)^1/2 = 0.132

So, t_calc = | (4.620 - 3.970) | / 0.132 ) * ((6*6) / (6+6))^1/2 = 8.56

t_table at 95% confidence, = 2.228 < t_calc.

So, No

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nsn pooled)

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