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Carolyn Porterfield
E-text 8.48 & 8.64

Exercise 8.48

(a display ad is a large block of multicolored illustrations, maps, and text). The data (in square millimeters) are shown below:
0 260 356 403 536 0 268 369 428 536
268 396 469 536 162 338 403 536 536 130 (a) Construct a 95 percent confidence interval for the true mean. x-bar = 346.5 s = 170.378 t-critical value 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) Why might normality be an issue here?
The Confidence Interval is a statement about the whole population. The random sample is probably not representative of the whole population. (c) What sample size would be needed to obtain an error of ±10 square millimeters with 99 percent confidence? n = [t*s/E] n = [2.093*170.378/10]^2 = 1271.64 n = 1272 (when rounded up) (d) If this is not a reasonable requirement, suggest one that is.
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Increase E or decrease the confidence level; both will have the effect of lowering "n".

Exercise 8.64 (a) Construct a 90 percent confidence interval for the proportion of all kernels that would not pop.
We know N, the sample size, from the problem: N = 773
From the problem's givens, we can 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
Use the formula for the interval around 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)
0.09269 to 0.12991
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p-hat = 86/773 = 0.1113
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E = 1.96*sqrt[0.1113*0.8887/773] = 0.0221
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CI = (0.1113-0.0221 , 0.1113+0.0221) (b) Check the normality assumption. Normality will hold, since Np and Nq are both large (86 and 687). pn = 0.113*773