# Statistics for Environmental Engineers

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The hypothesis that the population variances are equal can be tested using the F statistic. The upper 5% level of the F statistic for (4, 4) degrees of freedom is F4,4 = 6.39. A ratio of two variances as large as this is expected to occur by chance one in twenty times. The ratio of the two variances in this problem is Fexp = sB SA 2.369/0.641 = 3.596, which is less than F44 = 6.596. The conclusion is that a ratio of 3.596 is not exceptional. It is accepted that the variances for Methods A and B are estimating the same population variance and they are pooled to give:

s2 — [4( 0.000641) + 4( 0.002369)]/8 = 0.001505 The variance of each coefficient is:

Var (b) — 0.001505/16 = 0.0000941 and the standard error of the true value of each coefficient is:

SE(b) — VVar (b) — 70.0000941 = 0.0097 The half-width of the 95% confidence interval for each coefficient is:

SE(b) x r8>0.025 = 0.0097(2.306) = 0.0224

Judging the magnitude of each estimated coefficient against the width of the confidence interval, we conclude:

b0 = 2.2278 ± 0.0224 b1 = 1.1348 ± 0.0224 b2 = 0.0478 ± 0.0224 b3 = 0.0183 ± 0.0224

Average — significant Nitrate level — significant Method — significant PMA — not significant

Coefficients b0 and b1 were expected to be significant. There is nothing additional to note about them.

The interactions are not significant:

b12 = 0.0192 ± 0.0224    b13 = -0.0109 ± 0.0224

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