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Chapter Seven Statistical Intervals Based on a Single Sample
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Confidence Intervals For the Population Mean : 100(1- )% Confidence Interval (L,U) = x z /2 n
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Common Z Values 99% CI = (.01)/2 = 2.58 95% CI = (.05)/2 = 1.96 90% CI = (.10)/2 = 1.645
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Confidence Interval ( known) When fission occurs, many of the nuclear fragments formed have too many neutrons for stability. Some of these neutrons are expelled almost instantaneously. These observations are obtained on RV X, the number of neutrons released during fission of plutonium-239: n = 40 Average = 2.80 32222333 33334323 32333331 33333333 33233333
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CI Example Continued Is RV X Normally distributed? Assume = 0.5. What is the 99% Confidence Interval on ? The reported value of is 3.0. Do these data refute this interval?
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Sample Size For Interval Width w: n = (2 z /2 / w) 2 For half- width interval B from u; replace 2/w with 1/B.
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Sample Size ( known) The helium porosity of coal samples taken from any particular coal layer (seam) is Normally distributed with true standard deviation equal to 0.75%. What is the 95% CI for the true average porosity of a certain coal layer if the average porosity for 20 specimens from the seam was 4.85%. What sample size is necessary to estimate true average porosity to within 0.2% with 99% confidence?
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Large Sample Confidence Intervals For the Population mean : 100(1- )% Confidence Interval (L,U) = x z /2 s n With unknown .
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One Sided Confidence Bound For u: (L) = x - z s n (U) = x + z s n
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Confidence Interval (Large n) A random sample of 64 customers at the UB bookstore found that the average shopping time was 33 minutes with a sample variance of 256. Estimate the true average shopping time per customer with a confidence level of 90%.
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Confidence Interval Example (Large n) Using color infrared photography in identification of normal Douglas fir trees, a sample of 69 healthy trees showed a sample mean dye-layer density of 1.028 with a sample standard deviation of 0.163. What is the 95% (two-sided) CI for the true average dye-layer density for all healthy trees? Suppose the investigators made a rough guess of 0.16 for the value of s before collecting data. What sample size would be necessary to obtain an interval width of 0.05 for a confidence level 0f 95%?
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Student’s t Distribution f(t) = (v+1)/2 1+t 2 –(v+1)/2 (v/2) v v - < t <+ With v degrees of freedom. Values listed on Table A.5
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Student’s t Distribution For Random Sample X from a Normal pdf: T = X – u s/ n RV T has t distribution with n – 1 degrees of freedom.
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Small Sample Confidence Intervals For the Population mean : 100(1- )% Confidence Interval (L,U) = x t /2,v s n With unknown .
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Confidence Interval (Small n) A triathlon consisting of swimming, cycling, and running is a very strenuous amateur sport. Research on 9 male triathletes during a swimming performance showed a sample mean maximum heart rate (beats/minute) of 188 with a sample standard deviation of 7.2. Assuming that the heart rate distribution is approximately Normal, find a 98% CI for the true mean heart rate of triathletes while swimming.
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Small Sample CI Example A manufacture of gunpowder has developed a new powder which was tested in 8 shells. The resulting muzzle velocities, in feet per second, were as follows: 3005 2925 2935 2965 2995 3005 2935 2905 What is the 95% Confidence Interval for the true mean average velocity for shells using this new type of powder? (Evidence indicates the velocities to be Normal.) Sample mean = 2959 Sample s = 39.4
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Confidence Intervals for 2 For the Population Variance 2 : 100(1- )% Confidence Interval L = (n-1) s 2 2 /2, U = (n-1) s 2 2 1- /2,
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Variance Confidence Interval Example An experimenter wanted to check the variability of equipment designed to measure the volume of an audio source. Three independent measurements recorded by this equipment for the same sound were 4.1, 5.2, & 10.2. What is an estimate for the population variance with a confidence level of 90%. Evidence indicates that the measurements recorded by this equipment is Normally distributed.
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Interval Estimation on Variability X-ray microanalysis has become an invaluable method of analysis. With the electron microprobe, both quantitative and qualitative measures can be taken and analyzed statistically. One method for analyzing crystals is called the two-voltage technique. These measurements are obtained on the percentage of potassium present in a commercial product which theoretically contains 26.6% K by weight: 21.923.422.122.124.724.6 24.024.124.226.523.825.3 24.824.824.527.824.9 27.225.125.523.726.5 n = 27 22.026.725.223.125.4 What is the 99% Confidence Interval for 2 ? What is the 99% Confidence Interval for ? Suggest a way to improve the interval estimate for based on these data.
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