Confidence Interval
Q1.
A researcher wishing to estimate the proportion of X-ray machines that malfunction and produce excess radiation. A random sample of 40 machines is taken and 12 of the machines malfunction. The problem is to compute the 95% confidence interval on π, the proportion that malfunction in the population.
Solution:
The value of p is 12/40 = 0.30. The estimated value of σp is
= 0.072.
A z table can be used to determine that the z for a 95% confidence interval is 1.96. The limits of the confidence interval are therefore:
Lower limit = .30 - (1.96)(0.072) = .16
Upper limit = .30 + (1.96)(0.072) = .44.
The confidence interval is: 0.16 ≤ π ≤ .44.
Q2.
A manager at a power company monitored the employee time required to process high-efficiency lamp bulb rebates. A random sample of 40 applications gave a sample mean time of 3.8 minutes and a standard deviation of 1.2 minutes. Construct a 90% confidence interval for the mean time to process μ.
Solution: For large n, a 90% confidence interval for μ is given by
√X±z0.05×S/ n
Using z0.05 = 1.645, n = 40, S = 1.2minutes, and x = 3.8minutes, the 90% confidence interval for μ (true mean processing time) is given by
3.8 ± 1.645 × 1.2/ √40 = 3.8 ± 0.31 = (3.49, 4.11) minutes.
Q.3
The amount of PCBs (polychlorinated biphenyls) was measured in 40 samples of soil that were treated with contaminated sludge. The following summary statistics were obtained. x = 3.56, s = .5ppm Obtain a 95% confidence interval for the population mean μ, amount of PCBs in the soil.
Solution:
For large n, a 95% confidence interval for μ is given by X ±z0.025 ×S/√n
Using z0.025 = 1.96, n = 40, S = .5ppm, and x = 3.56ppm, the 95% confidence interval for μ (true mean amount of PCBs in the soil) is given by
± 1.96 × .5/√ 40 = 3.56 ± 0.155 = (3.405, 3.715).
Q4.
Radiation of microwave ovens has normal distribution with standard deviation σ=0.6. A sample of 25 microwave ovens produced X = 0.11. Determine a 95% confidence
A researcher wishing to estimate the proportion of X-ray machines that malfunction and produce excess radiation. A random sample of 40 machines is taken and 12 of the machines malfunction. The problem is to compute the 95% confidence interval on π, the proportion that malfunction in the population.
Solution:
The value of p is 12/40 = 0.30. The estimated value of σp is
= 0.072.
A z table can be used to determine that the z for a 95% confidence interval is 1.96. The limits of the confidence interval are therefore:
Lower limit = .30 - (1.96)(0.072) = .16
Upper limit = .30 + (1.96)(0.072) = .44.
The confidence interval is: 0.16 ≤ π ≤ .44.
Q2.
A manager at a power company monitored the employee time required to process high-efficiency lamp bulb rebates. A random sample of 40 applications gave a sample mean time of 3.8 minutes and a standard deviation of 1.2 minutes. Construct a 90% confidence interval for the mean time to process μ.
Solution: For large n, a 90% confidence interval for μ is given by
√X±z0.05×S/ n
Using z0.05 = 1.645, n = 40, S = 1.2minutes, and x = 3.8minutes, the 90% confidence interval for μ (true mean processing time) is given by
3.8 ± 1.645 × 1.2/ √40 = 3.8 ± 0.31 = (3.49, 4.11) minutes.
Q.3
The amount of PCBs (polychlorinated biphenyls) was measured in 40 samples of soil that were treated with contaminated sludge. The following summary statistics were obtained. x = 3.56, s = .5ppm Obtain a 95% confidence interval for the population mean μ, amount of PCBs in the soil.
Solution:
For large n, a 95% confidence interval for μ is given by X ±z0.025 ×S/√n
Using z0.025 = 1.96, n = 40, S = .5ppm, and x = 3.56ppm, the 95% confidence interval for μ (true mean amount of PCBs in the soil) is given by
± 1.96 × .5/√ 40 = 3.56 ± 0.155 = (3.405, 3.715).
Q4.
Radiation of microwave ovens has normal distribution with standard deviation σ=0.6. A sample of 25 microwave ovens produced X = 0.11. Determine a 95% confidence