Question

Assume an approximate normal distrihution. IS. A random sample of 12 eniployees of the Wlergate Carpet Company wed an average contribution of $8.00 to the American Cancer Socicty, with a landard deviation of $1.25, Construct u 90% wnfidence interva: for the average contribution by all employces of the Watergate Carpet Company to the American Cancer Society. Assume the contribuions to hc?pproxi- matcly normally distributed. Suction 8.5 16. (a) A random sample of 200 voters from Salem, Virginia, is selected and 120 are found to support an annexation suit. Find the 96%, conti- denee interval for the proportion of the voling population favoring the sust. ize or our (b) What can we assen with 96% confidence ahout the possibl errur if we estimate the proportion of voters favoring the innexation uit to be 0.6? 17. (a) A random sample of 400 cigarette smokers is selected and 80 are found to have a preference for a brand called Choke. Find the 90%, contidence proportion of the population of cigarette ss prefer Choke cigarettes (b) What can we assert with 90% confidence about the possible size of our igarette smokers who prefer Chuke error if we estimate the proportion of c to be 0217 18. In a random sample of 1000 families in Dodge City, it is found thal 228 own a horse. Find the 98 confdence interval for the proportior of families in Dodge City that own a harse 19. A random sampte of 64 students at Royal Oak College is selected and 48 s on campus. Use a 95% confidence interval to estimate n of students at Royal Oak College who have cars on campus. 20. A new rocket-launching system is being considered for deployment of smalt sample of 40 experimental launches is made with the are found to have car the pro partio short-range launches. The existing system has p 0.8 as the probability of a successful launch. A new system and 34 are successful (a) Construct a 95% confidence interval for p. (b) Would you conclude that the new system is better? 21. How large a sample is needed in Exercise 16 if we wish to be 96% confident that our sample proportion will be within 0.02 of the true proportion of the voting population? 22, How large a sample is needed in Exercise 18 if we wish to be 98% confident that our sample proportion will be within 0.05 of the true proportion of families in Dodge City that own a horse? 23. A study is to be made to estimate the percentage of citizens in Stoney Creek who favor having their water fluoridated. How large a sample is needed if

Answer 1

Problem 15)

Given,

n = 12

Mean(µ) = 8 $

Std Deviation(?) = 1.75

Alpha = 0.1

Z_critical = 1.645

Confidence Interval = µ +/- Z_critical * ? / n ^½ = 8 +/- 1.645 * 1.75 / 12^1/2 = (7.17,8.83)

Problem 16)

n = 200

Mean(µ) = 120/200 = 0.6

Std Deviation(?) = (0.6 * (1-0.6)/200)^1/2 = 0.035

Alpha = 0.04

Z_critical = 2.05

a)

96% Confidence Interval of people favouring the suit = µ +/- Z_critical * ? = 0.6 +/- 2.05 * 0.035 = (0.53,0.67)

b)

Margin of Error (ME) = Z_critical * (p*(1-p)/n)^1/2

Assuming ME = 5%

Hence, n = Z_critical ² * p * (1-p) / ME ² = 2.05² * 0.6 * 0.4 / 0.05² ~ 403.4

Problem 17)

n = 400

Mean(µ or p) = 80/400 = 0.2

Std Deviation(?) = (0.2 * (1-0.2)/400)^1/2 = 0.02

Alpha = 0.1

Z_critical = 1.645

a)

90% Confidence Interval of people favouring the Choke cigarette = µ +/- Z_critical * ? = 0.2 +/- 1.645 * 0.02 = (0.17,0.23)

b)

Margin of Error (ME) = Z_critical * (p*(1-p)/n)^1/2

Assuming ME = 0.2

Hence, n = Z_critical ² * p * (1-p) / ME ² = 1.645² * 0.2 * 0.8 / 0.2² ~ 10.8

Problem 18)

Given,

n = 1000

Mean(µ or p) = 228/1000 = 0.228

Std Deviation(?) = (0.228 * (1-0.228)/1000)^1/2 = 0.013

Alpha = 0.02

Z_critical = 2.33

98% Confidence Interval of House owners= µ +/- Z_critical * ? = 0.228 +/- 2.33 * 0.013 = (0.20,0.26)