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 ? , the proportion that malfunction in the population. Solution: The value of p is 12/40 = 0. 30. The estimated value of ? p 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. 6)(0. 072) = . 16 Upper limit = . 30 + (1. 96)(0. 072) = . 44. The confidence interval is: 0. 16 ? ? ? .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 ?. Solution: For large n, a 90% confidence interval for ? is given by vX±z0. 05? S/ n Using z0. 05 = 1. 645, n = 40, S = 1. 2minutes, and x = 3. minutes, the 90% confidence interval for ? (true mean processing time) is given by 3. 8 ± 1. 645 ? 1. 2/ v40 = 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 ? , amount of PCBs in the soil. Solution: For large n, a 95% confidence interval for ? is given by X ±z0. 025 ? S/vn Using z0. 025 = 1. 96, n = 40, S = . ppm, and x = 3. 56ppm, the 95% confidence interval for ? (true mean amount of PCBs in the soil) is given by ± 1. 96 ? .5/v 40 = 3. 56 ± 0. 155 = (3. 405, 3. 715). Q4. Radiation of microwave ovens has normal distribution with standard deviation ? =0. 6. A sample of 25 microwave ovens produced X = 0. 11. Determine a 95% confidence interval for the mean radiation. Solution : The population is normal, and the observed value X = 0. 11. (0. 11 ? 1. 96 ? 0. 6/v25, 0. 11 + 1. 96 ? 0. 6/v25) = (?. 1252, 3452) is a 95% confidence interval for ?. Q5.
A manufacturer of pharmaceutical products analyzes a specimen from each batch of a product to verify the concentration of the active ingredient. The chemical analysis is not perfectly precise. Repeated measurements on the same specimen give slightly different results. The results of repeated measurements follow a normal distribution quite closely. The analysis procedure has no bias, so the mean ? of the population of all measurements is the true concentration in the specimen. The standard deviation of this distribution is known to be ? = . 0068 grams per liter.
The laboratory analyzes each specimen three times and reports the mean result. Three analyses of one specimen give concentrations 0. 84030. 83630. 8447. We want a 99% confidence interval for the true concentration ?. The sample mean of the three readings is .8403 + . 8363 + . 8447 x = —————————– = . 8404. 3 Solution: For 99% confidence, we see that z? = 2. 576. A 99% confidence interval for ? is therefore x ± z? ?/vn = . 8404± 2. 576 . 0068/v3 = . 8404 ± . 0101 = (. 8303, . 8505). We are 99% confident that the true concentration lies between 0. 8303 and 0. 8505 grams per liter.
