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Unformatted text preview: 5E Final practice answers 1. In general, slope parameter b 1 has a mean of 1 with a standard deviation of b 1 = / q ( x i x ) 2 . (page 513 of text) Here we have 1 = . 1 and = . 7. To find q ( x i x ) 2 we can use the sample standard deviation of x , which is given as 1.2. From the formula for sample st dev we get q ( x i x ) 2 = 1 . 2 30 1 = 6 . 462. So b 1 = . 7 / 6 . 462 = . 1083. At this point, we just recognize that that b 1 is normally distributed with mean .1 and st dev .1083. To find the probability that b 1 is greater than 0, we find the z score of 0 and look it up in the table. z = (0 ( . 1)) /. 1083 = . 9232. The area to the right of that is .179. Note: We know to use z here, not t, since we know the pop st dev, b 1 , based on . If we were estimating this using s b 1 , then it would be a t score. If we have 100 points, then the only difference is in the calculation of ( x i x ) 2 . It is now 1 . 2 100 1 = 11 . 94. So b 1 = . 7 / 11 . 94 = . 058. Now z = (0 ( . 1)) /. 058 = 1 . 72 giving a probability of .043. If = . 4, then the modified calculations lead to z = 1 . 62 (for n = 30), so a probability of .053. 2. To find the conf int for b 1 we need the standard error s b 1 . Getting that takes a couple steps. Its convenient to first use the sample st dev of x to get q ( x i x ) 2 = s x n 1 = 12 . 3 29 = 66 . 24. Also, s = q SSE/ ( n 2) = 3 . 51, and so s b 1 = 3 . 51 / 66 . 24 = . 053. To find the conf int for053....
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 Fall '08
 LANDRIGAN
 Standard Deviation

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