Question

1) Find the minimum number of football fans required in a poll to estimate, within 1.5%, the percentage who believe that LeBron James could play in the NFL. Use 90% confidence, and assume no prior estimate of the percentage in question is available.

Select one:

a. 2037

b. 23

c. 609

d. 3007

2)Your professor wishes to estimate the proportion of high school students enrolled in college-level courses each school year. How large a sample is necessary if she wishes to be 90% confident with a margin of error of 3.5 percent? From an old 1999 study, the percentage of high school students enrolled in college-level courses was estimated to be 18.3%.

Select one:

a. 330

b. 469

c. 810

d. 331

3)In a sample of 56 tax returns filed by local hotel managers in a recent year, the taxable incomes are found to have a mean of $41225 and a population standard deviation of $6112. Construct the 99% confidence interval for the mean taxable income of all local hotel manager that year.

Select one:

a. 39255 < Mu < 41259

b. 39624 < Mu < 42826

c. 39122 < Mu < 43328

d. 39121 < Mu < 43327

4)Noise levels at various area urban hospitals were measured in decibels. The mean of the noise levels in 84 corridors was 61.2 decibels, and the population standard deviation was 7.9. Find the margin of error for a 90% confidence interval for the mean.

Select one:

a. 10.9844

b. 6.1392

c. 1.4179

d. 2.2579

In a study of 10 insurance sales representatives from a certain large city, the average age of the group was 48.6 years with a standard deviation of 4.1 years. Find the 90% confidence interval of the population mean age of all insurance sales representatives in that city.

Select one:

a. 45.63 < Mu < 48.50

b. 46.23 < Mu < 50.99

c. 46.22 < Mu < 50.98

d. 46.47 < Mu < 50.73

5)Your professor wishes to estimate the proportion of ALL high school students enrolled in college-level courses each school year. A sample of 1500 students revealed that 18.3% were enrolled in college-level courses. Find the margin of error for a 99% confidence interval for a proportion.

Select one:

a. .00988

b. .01642

c. .01957

d. .02571

Answer 1

1) At 90% confidence interval the critical value is z_0.05 = 1.645

Margin of error = 0.015

or, z_0.05 * sqrt(p(1 - p)/n) = 0.015

or, 1.645 * sqrt(0.5 * 0.5/n) = 0.015

or, n = (1.645 * sqrt(0.5 * 0.5)/0.015)^2

or, n = 3007

Option - D is the correct answer.

2) Margin of error = 0.035

or, z_0.05 * sqrt(p(1 - p)/n) = 0.035

or, 1.645 * sqrt(0.183 * (1 - 0.183)/n) = 0.035

or, n = (1.645 * sqrt(0.183 * (1 - 0.183))/0.035)^2

or, n = 331

Option - D is the correct answer.

3) At 99% confidence interval the critical value is z_0.005 = 2.58

The 99% confidence interval is

+/- z_0.005 *

= 41225 +/- 2.575 * 6112/sqrt(56)

= 41225 +/- 2103

= 39122, 43328

Option - C is the correct answer.

4)a) Margin of error = z_0.05 *

                             = 1.645 * 7.9/sqrt(84)

                             = 1.4179

Option - C is the correct answer.

b) The 90% confidence interval for population mean is

+/- z_0.05 *

= 48.6 +/- 1.645 * 4.1/sqrt(10)

= 48.6 +/- 2.13

= 46.47, 50.73

Option - d is the correct answer.

5) at 99% confidence interval the critical value is z_0.005 = 2.575

Margin of error = z_0.005 * sqrt(p(1 - p)/n)

                        = 2.575 * sqrt(0.183 * (1 - 0.183)/1500)

                        = 0.02571

Option - d is the correct answer.