Experimental Table 1: Pooled Class Data from Different Indicators Indicator n 10

Question

Experimental Table 1: Pooled Class Data from Different Indicators Indicator n 10096 (Mk Sp M 24 24 24 30 12 0.1031 0.0976 0.1017 0.1015 0.1038 0.0084 0.0054 0.0088 0.0056 0.0142 8.15 5.57 8.70 5.54 13.7 0.0038 0.0026 0.0080 0.0063 0.0137 MR BG MO 10 BB-Bromothymol blue MR-Methyl red BG- Bromocresol green MO - Methyl orange E- Erythrosine n-Number of measurements S- Number of students x-Mean HCI molanty S: -Standard deviation of all n measurements (degrees of freedomn- 1) Sp-Pooled standard deviation for S students (degrees offreedom .S) Table 2: Individual Data from Bromothymol Blue and Methyl Red Indicators

Explanation / Answer

The two most different standard deviations are :

The lowest one : for MR = 0.0026

The highest one : for E = 0.0137

first we will calculate 95% confidence interval for both.

(i) 95% confidence interval for standard deviation of MR

Lower limit = sqrt [(n-1)s2/x20.025,7] = sqrt [ 7 * 0.00262/16.02176] = 0.00172

Upper limit = sqrt [(n-1)s2/x20.975,7] = sqrt [ 7 * 0.00262/1.6898] = 0.0053

so 95% confidenc interval for standard deviation of MR = (0.0017, 0.0053)

95% confidence interval for standard deviation of E

Lower limit = sqrt [(n-1)s2/x20.025,3] = sqrt [ 3 * 0.01372/9.3484] = 0.0078

Upper limit = sqrt [(n-1)s2/x20.975,3] = sqrt [ 7 * 0.00262/0.2158] = 0.0053

so 95% confidenc interval for standard deviation of MR = (0.0078, 0.051)

Here Now F test for standard deviation of MR and E

F = (sE/sMR)2 = (0.0137/0.0026)2 = 27.76

Here dF1 = 7 and dF2 = 3

so,

Fcritical = F(0.05, 7,3) = 8.8867 < F

so here we can say that these two standard deviations are different from eachother.

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