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# hw7sol - 1 11(a Scatter plot Figure 2 Scatterplot Caliper 1...

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STAT401: Homework 7 Solutions (Spring 2009) 6.9 2. (a) Median = X (11) + X (12) 2 = 28 . 64; 25 th percentile = Q 1 = X (6) = 27 . 98; 75 th per- centile = Q 3 = X (17) = 29 . 53. (b) IQR = Q 3 - Q 1 = 1 . 55. (c) The percentile for 19 th ordered value: 100 19 - 0 . 5 22 = 84 . 09%. (d) 100 i - 0 . 5 22 = 90 i = 20 . 3. So the approximate will be the average of 20 th and 21 th ordered value: 29 . 79+29 . 88 2 = 29 . 835. 4. Let X be the starting salary, then V ar ( X ) = 312 . 3. (a) Let Y = X + 5000 be the second year salary, then V ar ( Y ) = V ar ( X ) = 312 . 3 (b) Let Y = 1 . 05 X be the second year salary, then V ar ( Y ) = 1 . 05 2 V ar ( X ) = 344 . 3108 9. Boxplot: The mean of the concrete strength exposed to Temp 2(15 degrees) is greater Figure 1: Boxplot than the concrete strength mean of Temp 1(-8 degrees). The variance of the concrete strength of Temp 2 is greater than that of Temp 1.

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Unformatted text preview: 1 11. (a) Scatter plot.: Figure 2: Scatterplot Caliper 1 vs Caliper 2 (b) Pearson’s correlation coeﬃcient=0.985 (c) Spearman’s correlation coeﬃcient=0.982 12. (a) Y=194+0.934X (b) Y=194+0.934*300=474.2. (c) Average sediment conductivity increases by 0.934*50=46.7. 13. (a) Y=-2.18+0.66X (b) Let Y = 12 Y and X = 12 X , Y = 12 Y = 12 * (-2 . 18) + 0 . 66(12 X ) = 12 * (-2 . 18) + 0 . 66 X Thus the intercept is 12*(-2.18) and the slope is 0.66. The intercept increases by a factor of 12 and the slope remains the same. (c) Y=-2.18=0.66*14=7.06. (d) Average distance increases by 4*slope=4*0.66=2.64. 2...
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hw7sol - 1 11(a Scatter plot Figure 2 Scatterplot Caliper 1...

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