Math 1530 Chapter 6 Review Questions
1. Assume that a computer was used to generate the given confidence interval of
(136, 150) for the population mean, µ. Find the margin of error and the point estimate of the mean.
2. Find the critical value z
α/2
that corresponds to a 98% degree of confidence.
2b. Find the critical t
α/2
that corresponds to a 98% degree of confidence. The sample size is 25.
3. Find the margin of error (E) one would use in estimating the population mean µ for weights of eggs at the 95%
confidence level; where n = 49, x-bar = 1.71, and s = 0.39.
4. Write the confidence interval for #3.
5. A random sample of 94 light bulbs had a mean life of x-bar = 548 hrs with s = 29 hrs. Construct a 90%
confidence level for the mean life, µ, of all light bulbs of this type.
6. The drug Lipitor is meant to lower cholesterol levels. In a clinical trial of 863 patients who received 10-mg doses
of Lipitor daily, 47 reported a headache as a side effect. Obtain a point estimate for the population proportion of
Lipitor users who will experience a headache as a side effect.
7. Construct a 90% confidence interval for the population proportion of Lipitor users who will report a headache as a
side effect.
8. A 2004 Harris Poll revealed that of 2114 adults who follow professional football, 381 were Green Bay Packers
fans. Write a 90% confidence interval for the proportion of pro football fans who say their favorite team is the Green
Bay Packers. Express your answer as percentages.
9. A population is normal with a variance of 17.[ σ = sqrt(17)] Suppose you wish to estimate the population mean,
µ. Find the sample size needed to assure with 68.2% confidence that the sample mean will not differ from the
population mean by more than 3 units.
10. Write a 99% confidence interval for the information given in #8. What is the effect of increasing the level of
confidence on the width of the interval?
11. Of 221 adults selected randomly from one town, 48 of them smoke. Construct a 95% confidence interval for the
true proportion of all adults in the town that smoke.
Ch 5 Answers: (Scores calculated using the TI-83 may differ slightly from those using Table A-2due to rounding errors. Some of
the answers have been derived from Table A-2; some from the calculator program. You may use either.)
1a. normcdf(180, 225, 200, 20) = .736 1b. normcdf(-1000, 215, 200, 20) = .773 1c. normcdf(240, 1000, 200, 20) = 0.0228
1d. invNorm(.7, 200, 20) = 210* 1e. normcdf(202, 212, 200, 10/3) = .274 1f. normcdf(198, 1000, 200, 2) = .841
2a. P(.67 ≤ z≤1.67) = .9525 - .7486 = .2039 2b. P(z > 1.67) = 1 - .9525 = .0475 2c. P(z < 0.67) = .7486
2d. P(-0.4 < z < 0.4) = .6554 - .3446 = .3108 2e. x = 75 + (-.44)(15) = 68 2f. x = 75 + (.67)(15) = 85
2g. P(2.33 < z < 9.33) = .9999 - .9901 = .0098 2h. .P(z < -1.8) = .0359 3. normcdf(-1000, 46, 40, 12) = .691
4. 1- normcdf(5.48, 5.82, 5.67, .07) = 1.94% 5. normcdf(85, 120, 100, 15) * 9000 = 6,751 6. normcdf(0,76,,73, 7.8/√(24)) = .970
Ch. 6 Answers: 1. E = 7; p.e. = 143 2. 2.33 2b. 2.492
3. 0.1092 or 0.11 4. use “T Interval”: (1.60, 1.82) 5. use “T Interval”: (543, 553)
6. 47/863 = 0.054 7. (0.042, 0.067) 8. (16.6%, 19.4%)
9. 2 10. (15.9%, 20.2%); it gets wider 11. (0.163, 0.272)