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 follow up 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 protect 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 geld here? The impudence Interval is a statement almost the whole population. The ergodic stress is probably non instance of the whole population. (c) What example size would be needed to receive an error of ±10 square millimeters with 99 percent government agency? n = [t*s/E] n = [2.093*170.378/10]^2 = 1271.64 n = 1272 (when round up) (d) If this is not a reasonable requirement, suggest one that is. 2 Increase E or decrease the confidence level; 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 some 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*0.8887/773) 0.0926 9 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 tumid (86 and 687). pn = 0.113*773 = 86+ qn = even higher (c) Try the real firm 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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