EMA+3800_6808+Presentation 2 C+3

# EMA+3800_6808+Presentation 2 C+3 - Student's t In practice...

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Student’s t In practice, we usually do not know σ 2 Estimate σ 2 by computing s 2 Thus, we must use a different distribution known as Student’s t instead of Z distribution There is a family of t distributions. The t distributions have one parameter v , the number of degrees of freedom in the estimate of the variance. t ν = y av - μ (1/n) s 2

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Figure 2.7 (p. 33) Several t -distributions.
t v = linear statistic – E(statistic) (V(statistic), estimated ) = Σ a i y i – E( Σ a i y i ) [( Σ a i 2 ) s 2 ] for the 95% confidence limits , statistic + t v, 0.025 (V(statistic), estimated ) where t v , 0.025 is that value of t v leaving 0.025 in the upper tail of the distribution. (Show Fig.2-7 and Table 2-4, then t distribution tables)

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Table 2.4 (p. 48) Tests on Means of Normal Distributions, Variance Unknown
* Copied from S. Hunter’s modules (1968)

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* Copied from S. Hunter’s modules (1968)
EXAMPLE 1 : A continuous vacuum crystallizer is fed with a 35% solution of MgSO 4 at a constant temperature for a fixed time period. In n = 8 repeated trials, the proportion of the magma that is crystallized after vaporization equals: 0.54, 0.48, 0.59, 0.52, 0.46, 0.53, 0.45, 0.43 (To save the student time : Σ y i = 4.00; Σ y i 2 = 2.0204) Estimate the true proportion crystallized. Construct a 95% interval statement for the true proportion crystallized. 1 Adapted from Design of Experiments modules by S. Hunter, published by U. of Kentucky, 1968

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ANSWER y av = 4.00/8 = 0.50 is the best estimate of μ The standard error of a statistic is simply the square root of the variance of the statistic. Thus, the estimated standard error of y av is s 2 /n Since we don’t know σ 2 , we estimate it. Thus: SS df Crude Sum of Squares Σ y 2 = 2.0204 8 Correction Factor ( Σ y) 2 /n = 2.0000 1 --------------------------------------------------------------- Corrected Sum of Squares Sy 2 = 0.0204 7 s 2 = Sy 2 / v = 0.0204 / 7 = 0.0029 with v= 7 df.
The standard deviation is s =0.054. The standard error of y av is S.E.(y av ) = s 2 /n = 0.0029/8 = 0.0190. 95% interval statement for μ are given by y av  +  t v,0.025   s 2 /n. Substituting t 7,0.025 = 2.36. And s 2 /n = 0.0029/8 0.50 + 0.045, or Prob(0.46 < μ < 0.54) = 0.95.

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Example from the text: Portland Cement Formulation (Table 2-1, pp. 22) Observation (sample), j Modified Mortar (Formulation 1) Unmodified Mortar (Formulation 2) 1 16.85 17.50 2 16.40 17.63 3 17.21 18.25 4 16.35 18.00 5 16.52 17.86 6 17.04 17.75 7 16.96 18.22 8 17.15 17.90 9 16.59 17.96 10 16.57 18.15 1 j y 2 j y
Graphical View of the Data; Dot Diagram Form 1 Form 2 16.3 17.3 18.3 Dotplots of Form 1 and Form 2 (means are indicated by lines)

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Box Plots Form 1 Form 2 16.5 17.5 18.5 Boxplots of Form 1 and Form 2 (means are indicated by solid circles)
The Hypothesis Testing Sampling from a normal distribution Statistical hypotheses: 0 1 2 1 1 2 : : H H μ =

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EMA+3800_6808+Presentation 2 C+3 - Student's t In practice...

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