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eBook Given are five observations for two variables, and y. T4712 15 19 7 15 7 25 20 Develop the 95% confidence and prediction intervals when z-9. Explain why these two intervals are different. (to 4 decimals) (to 3 decimals) (to 4 decimals) (to 4 decimals) t-value pred Confidence Interval for the Mean Value: (to 2 decimals) (to 2 decimals) Prediction Interval for an Individual Value: (Enter negative values as negative number.) The two intervals are different because there is more variability associated with predicting an select your ans ver value than there is a select your ans er value

Choices are: "mean" or "individual"

math   Statistics-And-Probability   
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Answer #1

From the given data,

X Y X^2 Y^2 XY
4 7 16 49 28
7 15 49 225 105
12 7 144 49 84
15 25 225 625 375
19 20 361 400 380
Total: 57 74 795 1348 972

- 57 χ=-= 11.4, y =-= 14.8 - 74 (Ex)(Zy) = 972-274-1284 (5x) 57 SS - 2-1095.2-- 252.8 SS 128.4 b= 0.8843 S145.2 a=y-bx = 14.8-0.8843 * 1 1.4=4.719 The regression equation of Y on X is Y= 4.719 + 0.8843x Given X-then Y-4.719 + 0.8843(9) = 12.678 y128.4-0.8843 *128.4 5-2 = 2.2253 t-value = 3.18245 (x-x) 1(9-114) 5 145.2 = 2.2253 = 1.0894 1 (x-x) 1 · (9-114) 5 145.2 = 2.225311 +- = 2.4M7 The 95%confidence inter alfor the forecasted values yofxis У ± t(/r-2)d) =[12.678-3.1824 5 * 1 .0 894, 12.678 + 3.1824 5 * 1.0894-[9.21 10, 1 б. 145] The 95%prediction intervalfor the forecastedvalues yofxīs У ±fa/-2wSB. =[12.678-3.1824 5 * 2.4777. 12.678+3. 18 245 * 2.47771-47928, 20.5632]

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