8.48 8.64

1 E-text 8.48 & 8.64 Exercise 8.48 (a display ad is a tumescent block of multicolored illustrations, maps, and text). The data (in squargon millimeters) are shown mess the stairs: 0 260 356 403 536 0 268 369 428 536 268 396 469 536 162 338 403 536 536 130 (a) Construct a 95 share office interval for the true mean. x-bar = 346.5 s = 170.378 t-critical tax for 95% CI with df=19 = 2.093 E = 2.093*170.378/sqrt (20) = 2.0931.96*38.0976=79.74 95% CI : (346.5-79.74,346.5+79.74) (b) wherefore might nitrogen be an supply here? The impudence Interval is a statement to the highest degree the whole population. The ergodic stress is probably not articulation of the whole population. (c) What sample size would be needed to reign an error of ±10 square millimeters with 99 percent authorization? n = [t*s/E] n = [2.093*170.378/10]^2 = 1271.64 n = 1272 (when travel up) (d) If this is not a reasonable requirement, suggest one that is. 2 Increase E or decrease the confidence take aim; both will reach the effect of lowering n. Exercise 8.64 (a) Construct a 90 percent confidence interval for the proportion of completely kernels that would not pop. We know N, the sample size, from the problem: N = 773 From the problems givens, we stand get p and q: p = 86/773 = 0.1113 q = 1 - p = 0.8887 From a z table, the value for the 90% interval is: 1.

6449 engage the formula for the interval just about a proportion: p - z*sqrt(pq/N) to p + z*sqrt(pq/N) 0.1113 - 1.6449*sqrt(0.1113*0.8887/773) to 0.1113 + 1.6449*sqrt(0.1113*0.88 87/773) 0.09269 to 0.12991 --------------! ------------------ 3 p-hat = 86/773 = 0.1113 ------- E = 1.96*sqrt[0.1113*0.8887/773] = 0.0221 -------- CI = (0.1113-0.0221 , 0.1113+0.0221) (b) Check the normality assumption. Normality will hold, since Np and Nq are both gargantuan (86 and 687). pn = 0.113*773 = 86+ qn = even higher (c) Try the real loyal Rule. Does it work well here? Why, or wherefore not? No, the...If you want to get a full essay, align it on our website: OrderCustomPaper.com

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