I Math 119 Exam 3 version 4 (page 2/2) The workers at XYZ. Inc. took a random sa

Question

I Math 119 Exam 3 version 4 (page 2/2) The workers at XYZ. Inc. took a random sample of 800 man- hole covers and found that 40 of them were defective. What is the 95 percent Conf. Interval for p, the true proportion of de tetive manhole envers, based on thi·sampler p, the true proportion of deA 37 28 D C015, 085) 9. Decreasing the confidence level, while holding the sample size BI make it smaller the same, will do what to the length of your confidenoe inter- val? cannot be determined from the given information D make it bige E none of these A it will stay the same 10. Suppose we are interested in finding a 95 percent confidence What is the 95 percent confidence interval for the population interval for the mean SAT Verbal score of students at a certain high school. Five students are sampled, and their SAT Verbal n scores are 560, 500, 470, 660, and 640 mean? (4863. 6457) T (4023, 6e97) (462.3, 600.7) C (4605 ( 662.1) D 492.8, 639 2 11. A survey of 49 people revealed that the mean number of min-

Explanation / Answer

8. n = 800, p = 40/800 = 4/80 = 1/20 = 0.05, q = 1 - p = 0.95

For a 95% CI, z = +/- 1.96 and the CI = z*sqrt(p*q/n) + p = +/- 1.96*sqrt(0.05*0.95/800) + 0.05 = (0.0349,0.0651)

C is correct

9. As is evident from the CI expression CI = z*sqrt(p*q/n) + p,: Decreasing the confidence level decreases the confidence interval

B is correct

10.560,500,470,660,640
mean = 566
sd = 83.55
n = 5

CI = z*sd/sqrt(n) + mean = +/- 1.96*83.55/sqrt(5) + 566 = (492.8,639,23)

D is correct

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