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Unformatted text preview: 5E Final practice – answers 1. In general, slope parameter b 1 has a mean of β 1 with a standard deviation of σ b 1 = σ/ q ∑ ( x i- ¯ x ) 2 . (page 513 of text) Here we have β 1 =- . 1 and σ = . 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 σ 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, σ b 1 , based on σ . 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 σ b 1 = . 7 / 11 . 94 = . 058. Now z = (0- (- . 1)) /. 058 = 1 . 72 giving a probability of .043. If σ = . 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. It’s 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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