Mar 31, 2017
- Solve the problem.
- 1. Find the critical value of zα/2 that corresponds to a degree of confidence of 91%.
- 1.34
- 1.70
- 1.75
- 1.645
- 2. The following confidence interval is obtained for a population proportion, p:
0.817 < p < 0.855
Use these confidence interval limit to find the point estimate, p̂.- 0.817
- 0.833
- 0.836
- 0.839
- 1. Find the critical value of zα/2 that corresponds to a degree of confidence of 91%.
- Find the margin of error for the 95% confidence interval used to estimate the population proportion.
- 3. n = 186, x = 103
- 0.0714
- 0.00260
- 0.0643
- 0.125
- 3. n = 186, x = 103
- Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p.
- 4. n = 158, x = 108; 95 percent
- 0.611 < p < 0.756
- 0.610 < p < 0.758
- 0.625 < p < 0.743
- 0.626 < p < 0.742
- 4. n = 158, x = 108; 95 percent
- Find the minimum sample size you should use to assure that your estimate of p̂ will be within the required margin of error around the population p.
- 5. Margin of error: 0.002; confidence level: 93%, p̂ and q̂ are unknown.
- 410
- 405
- 204,757
- 204,750
- 6. Margin of error: 0.07; confidence level: 95%; from a prior study, p̂ is estimated by the decimal equivalent of 92%.
- 4
- 51
- 174
- 58
- 5. Margin of error: 0.002; confidence level: 93%, p̂ and q̂ are unknown.
- Solve the problem.
- 7. 61 randomly selected light bulbs were tested in a laboratory, 50 lasted more than 500 hours. Find a point estimate of the true proportion of all light bulbs that last more than 500 hours.
- 0.803
- 0.180
- 0.450
- 0.820
- 7. 61 randomly selected light bulbs were tested in a laboratory, 50 lasted more than 500 hours. Find a point estimate of the true proportion of all light bulbs that last more than 500 hours.
- Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p.
- 8. When 343 college students are randomly selected and surveyed, it is found that 110 own a car. Find a 99% confidence interval for the true proportion of all college students who own a car.
- 0.256 < p < 0.386
- 0.279 < p < 0.362
- 0.271 < p < 0.370
- 0.262 < p < 0.379
- 8. When 343 college students are randomly selected and surveyed, it is found that 110 own a car. Find a 99% confidence interval for the true proportion of all college students who own a car.
- Solve the problem.
- 9. A researcher is interested in estimation the proportion of voters who favor a tax on e-commerce. Based on a sample of 260 people, she obtains the following 99% confidence interval for the population proportion p:
0.113 < p < 0.171.
Which of the statements below is a valid interpretation of this confidence interval?
A: There is a 99% chance that the true value of p lies between 0.113 and 0.171.
B: If many different samples of size 250 were selected and, based on each sample, a confidence interval were constructed, 99% of the time the true value of p would lie between 0.113 and 0.171.
C: If many different samples of size 250 were selected and, based on each sample, a confidence interval were constructed, in the long run 99% of the confidence intervals would contain the true value of p.
D: If 100 different samples of size 250 were selected and, based on each sample, a confidence interval were constructed, exactly 99 of these confidence intervals would contain the true value of p.- D
- C
- B
- A
- 9. A researcher is interested in estimation the proportion of voters who favor a tax on e-commerce. Based on a sample of 260 people, she obtains the following 99% confidence interval for the population proportion p:
- Determine whether the given conditions justify using the margin of error E = zα/2 σ/ √n when finding a confidence interval estimate of the population mean μ.
- 10. The sample size is n = 9, and σ is not known, and the original population is normally distributed.
- No
- Yes
- 10. The sample size is n = 9, and σ is not known, and the original population is normally distributed.
- Use the confidence level and sample data to find the margin of error E.
- 11. Systolic blood pressures for women aged 18–24: 94% confidence; n = 92, x̅ = 114.9 mm Hg, σ = 13.2 mm Hg.
- 9.6 mm Hg
- 47.6 mm Hg
- 2.3 mm Hg
- 2.6 mm Hg
- 11. Systolic blood pressures for women aged 18–24: 94% confidence; n = 92, x̅ = 114.9 mm Hg, σ = 13.2 mm Hg.
- Use the confidence level and sample data to find a confidence interval for estimation the population μ.
- 12. A group of 52 randomly selected students have a mean sore of 20.2 with a standard deviation of 4.6 on a placement test. What is the 90 percent confidence interval for the mean score, μ, of all students taking the test.
- 18.7 < μ < 21.7
- 19.0 < μ < 21.5
- 18.6 < μ < 21.8
- 19.1 < μ < 21.3
- 12. A group of 52 randomly selected students have a mean sore of 20.2 with a standard deviation of 4.6 on a placement test. What is the 90 percent confidence interval for the mean score, μ, of all students taking the test.
- Use the margin of error, confidence level, and standard deviation σ to find the minimum sample size required to estimate an unknown population mean μ.
- 13. Margin of error: $100, confidence level: 95%, σ = $403.
- 91
- 44
- 108
- 63
- 13. Margin of error: $100, confidence level: 95%, σ = $403.
- Do one of the following, as appropriate:
(a) Find the critical value zα/2;
(b) Find the critical value tα/2;
(c) State that neither the normal nor the t distribution applies.- 14. 99%; n = 17; σ is unknown; population appears to be normally distributed.
- zα/2 = 2.583
- zα/2 = 2.921
- zα/2 = 2.567
- zα/2 = 2.898
- 14. 99%; n = 17; σ is unknown; population appears to be normally distributed.
- Find the margin of error.
- 15. 99% confidence interval; n = 201; x̅ = 175; s = 21.
- 6.0
- 4.6
- 8.2
- 3.9
- 15. 99% confidence interval; n = 201; x̅ = 175; s = 21.
- Use the given degree of confidence and sample data to construct a confidence interval for the population mean μ. Assume that the population has a normal distribution.
- 16. The football coach randomly selected ten players and timed how long each player took to perform a certain drill. The times (in minutes) were:
7.6 10.4 9.7 8.4 11.8
7.0 6.5 11.1 10.4 12.4
Determine a 95 percent confidence interval for the mean time for all players.- 8.00 < μ < 10.98
- 8.13 < μ < 10.93
- 8.03 < μ < 11.03
- 8.06 < μ < 11.00
- 16. The football coach randomly selected ten players and timed how long each player took to perform a certain drill. The times (in minutes) were:
- Solve the problem.
- 17. Find the critical value X2L corresponding to a sample size of 15 and a confidence level of 90 percent.
- 31.319
- 23.685
- 29.141
- 21.064
- 17. Find the critical value X2L corresponding to a sample size of 15 and a confidence level of 90 percent.
- Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation σ. Assume that the population has a normal distribution.
- 18. College students’ annual earnings: 90% confidence; n = 9, x̅ = $3605, s = $800.
- $629 < σ < $1044
- $505 < σ < $1764
- $486 < σ < $1566
- $540 < σ < $1533
- 18. College students’ annual earnings: 90% confidence; n = 9, x̅ = $3605, s = $800.
- Find the appropriate minimum sample size.
- 19. You want to be 95% confident that the sample variance is within 40% of the population variance.
- 14
- 11
- 56
- 224
- 19. You want to be 95% confident that the sample variance is within 40% of the population variance.
- Use the give degree of confidence and sample data to find a confidence interval for the population standard deviation σ. Assume that the population has a normal distribution.
- 20. The football coach randomly selected ten players and timed how long each player took to perform a certain drill. The times (in minutes) were:
7 12 10 14 10
14 9 11 7 14
Find a 95 percent confidence interval for the population standard deviation σ.- (0.7, 2.2)
- (1.8, 4.5)
- (1.9, 4.5)
- (1.9, 4.9)
- 20. The football coach randomly selected ten players and timed how long each player took to perform a certain drill. The times (in minutes) were:
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