Chapter6.3 lecture notes

Chapter6.3 lecture notes - v> t α = P(t v< t α =1-α...

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Handout (Chapter 6.3. BRING Tables 3 and 4. 6.3. The Sampling Distribution of the sample mean ( σ unknown) 1. In samples of size n, we often estimate both the mean and the standard deviation using X and S. 2. Standardized sample mean based on sample estimate of standard deviation, S. T = n S X / μ - 3. If the X i are normally distributed then T has “t distribution with ν =n-1 degrees of freedom (df). 4.Using the tables t α for d.f.=1,2…..29, : Table 4 t α ,= Value of T that is exceeded with probability α .. For v 30 t α , 2245 z α . 5 The t v distribution is symmetric about 0. So P(t v < -t α ) = P(t v > t α
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Unformatted text preview: v > - t α ) = P(t v < t α ) =1-α . EXAMPLES of Application of t distribution 1. P( t 2 > 2.92) = 2. P(t 2 > -2.92) 3. P(t 2 > 9)= 4. P(t 2 > 10)= 5. Find t 2, 0.01 6. Find c such that P(-c < t 2 <c )= 0.95. 7. P( X > + 3s) for n=4 and 16. 8. A random sample is taken of the diameters of 4 bolts coming off an assembly line. The sample mean equals 13 mm and the sample estimate of the standard deviation is calculated at 2mm. What is the probability of a sample mean with an error this large when in fact the process is in control and =10?...
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