lecture 4 - Lecture 4 Notes #8 from previous homework...

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Lecture 4 Notes #8 from previous homework assignment: A buyer of vegetables wanted the assurance that the average amount of spoilage per carton was less than X points w/ a probability of 95%. The supplier took a sample of 100 cartons and found = . x 5 lbs per cartoon The population Standard Deviation ( σ) was known to be 2 lbs. Solve for X. Solution: This is a one tail lower confidence interval for the population mean with sigma known. < =. Pµ x 95 > Px - . =. μ 1 645σx 95 from central limit theorem Px + . > =. 1 645σx µ 95 rearranged terms P < μ + . x 1 645 =. σx 95 Answer: P < . . + = - μ 1 645 2 5 1 α µ < 5.329 New Topic: Chi-Square Distribution: The Chi-Square Distribution can always be shown to be the sum of the squares of a set of z values, (variables that are normally distributed with a mean of zero, and a variance or standard deviation of 1). In the chi-square distribution, the minimum value is 0, and the graph is skewed to the right. = = = - χ2 i 1nz2 x μσ2 = [( - ) ] = ( - ) Ez2 E x µ σ 2 E x µx 2σx2 = = σx2σx2 1 *Thus, the expected value of calculating z2 once is 1. Therefore, the expected value of calculating z2 n times is n (or more specifically, the degrees of freedom). 1
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The following statistic can be shown to be the sum of z 2 with n-1 degrees of freedom. The following algebraic manipulation is designed to convince you of that. - = - = ( - ) - ( - ) = - = = - χn 12 n 1s2σ2 n 1 xi x2 n 1 σ2 xi x2σ2 i 1nxi xσ2 *Remember that the Chi-Square distribution is not symmetrical. For a two tailed procedure, you will need χ2 associated with alpha over 2, that is ( χ ) α22 and χ2 associated with 1 minus alpha over 2 , that is, ( - ) χ1 α22 . P[
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lecture 4 - Lecture 4 Notes #8 from previous homework...

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