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88 34825 356 10000 12127806 1266 123933 096 001 220

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Unformatted text preview: tput voltage responds to pH. Otherwise there is no change in voltage when pH changes. When the labmate questioned about the small slope, there is concern about whether the slope is so small that it might be zero. 1) We can examine the probability that the slope could reach zero: b − b0 98.4867 − 0 t= = = 19.6470 n 2 30.9902 38.22 s ∑ pH i − pH i =1 ( ) With 6 degrees of freedom, the probability (1.127E ­6) is close to zero. 2) We can also use the results from part b. It shows that the 99% confidence interval of the slope is from 79.9021 to 117.0714, and 0 is not within this interval. The probability that the slope could research 0 is less than 1%. d) What is the value of R ­squared for the relationship you developed? n ˆ SSE = ∑ Vi − Vi i =1 ( ) 2 = 5762.35 n 2 2 SST = ( n − 1) SV = ∑ (Vi − V ) = 376479.50 i =1 The coefficient of determination: R 2 = 1 − SSE 5762.35 = 1− = 0.9847 SST 376479.50 15/19 CEE 202 – Engineering Risk & Uncertainty Spring 2011 (Bond) Homework 11 Diff. between Means; Least ­Squares Alternative solution: treat output voltage as independent variable and pH as dependent variable. 8 7 6 pH 5 4 3 2 1 0 0 200 400 600 800 Output Voltage (mv) The scattering plot between output voltage and pH shows that these two variables are linearly related. a) Determine a best estimate of the coefficients in a linear relationship between voltage and pH. Assume the linear relationship between Voltage (V) and pH is: pH = a + bV 1n V = ∑ Vi = 448.25 n i =1 1n pH = ∑ pH i = 4.44 n i =1 Vi pH i Vi − V pHi − pH Vi 2 2 (V − V ) ( pH i i − pH 2 2 ˆ ˆ ) (V − V ) ( pH − pH ) pH = a + bV ( pH − pH ) i i i i i 100 0.88 -348.25 -3.56 10000 121278.06 12.66 1239.33 0.96 0.01 220 2.12 -228.25 -2.32 48400 52098.06 5.38 529.25 2.16 0.00 295 3.22 -153.25 -1.22 87025 23485.56 1.49 186.77 2.91 0.10 422 3.78 -26.25 -0.66 17808...
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