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Prediction and Tolerance Interval Fifteen (15) samples of water were collected from a certain treatment facility in order to gain some insight regarding the amount of coliform in the urban pipeline. The concentration of the coliform is measured in parts per million (ppm), per liter. Suppose that the mean at the treatment facility is not as important as the upper extreme of the distribution of the amount of coliform detected. The concern is whether the concentration of coliform is too large. Readings for the 15 water samples gave a sample mean of 3.84 ppm per liter and a standard deviation of 3.07 ppm per liter. Assume that the readings are a random sample from a normal distribution. Calculate a Prediction Interval (upper 95% prediction limit) and a Tolerance Limit (95% upper tolerance limit that exceeds 95% of the population values). Interpret both; that is, tell what each communicates about the upper extreme of the distribution of coliform at the sampling in the treatment facility. O a. Prediction Upper Limit: Xn+159.424. This means that a new observation will have a chance of 95% to fall below the upper limit. Tolerance Upper Limit: 11.72. Hence, we are 95% confident that a limit of 11.72 will contain 95% of the coliform measurement in the water samples O b. Prediction Upper Limit: Xn+1 59.398. This means that a new observation will have a chance of 95% to fall below the upper limit. Tolerance Upper Limit: 11.86. Hence, we are 95% confident that a limit of 11.86 will contain 95% of the coliform measurement in the water samples O C. Prediction Upper Limit: Xn+15 10.597. This means that a new observation will have a chance of 95% to fall below the upper limit. Tolerance Upper Limit: 11.72. Hence, we are 95% confident that a limit of 11.72 will contain 95% of the coliform measurement in the water samples

Prediction and Tolerance Interval Fifteen (15) samples of water were collected from a certain treatment facility in order to gain some insight regarding the amount of coliform in the urban pipeline. The concentration of the coliform is measured in parts per million (ppm), per liter. Suppose that the mean at the treatment facility is not as important as the upper extreme of the distribution of the amount of coliform detected. The concern is whether the concentration of coliform is too large. Readings for the 15 water samples gave a sample mean of 3.84 ppm per liter and a standard deviation of 3.07 ppm per liter. Assume that the readings are a random sample from a normal distribution. Calculate a Prediction Interval (upper 95% prediction limit) and a Tolerance Limit (95% upper tolerance limit that exceeds 95% of the population values). Interpret both; that is, tell what each communicates about the upper extreme of the distribution of coliform at the sampling in the treatment facility. O a. Prediction Upper Limit: Xn+159.424. This means that a new observation will have a chance of 95% to fall below the upper limit. Tolerance Upper Limit: 11.72. Hence, we are 95% confident that a limit of 11.72 will contain 95% of the coliform measurement in the water samples O b. Prediction Upper Limit: Xn+1 59.398. This means that a new observation will have a chance of 95% to fall below the upper limit. Tolerance Upper Limit: 11.86. Hence, we are 95% confident that a limit of 11.86 will contain 95% of the coliform measurement in the water samples O C. Prediction Upper Limit: Xn+15 10.597. This means that a new observation will have a chance of 95% to fall below the upper limit. Tolerance Upper Limit: 11.72. Hence, we are 95% confident that a limit of 11.72 will contain 95% of the coliform measurement in the water samples

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Prediction and Tolerance Interval Fifteen (15) samples of water were collected from a certain treatment facility in order to gain some insight regarding the amount of coliform in the urban pipeline. The concentration of the coliform is measured in parts per million (ppm), per liter. Suppose that the mean at the treatment facility is not as important as the upper extreme of the distribution of the amount of coliform detected. The concern is whether the concentration of coliform is too large. Readings for the 15 water samples gave a sample mean of 3.84 ppm per liter and a standard deviation of 3.07 ppm per liter. Assume that the readings are a random sample from a normal distribution. Calculate a Prediction Interval (upper 95% prediction limit) and a Tolerance Limit (95% upper tolerance limit that exceeds 95% of the population values). Interpret both; that is, tell what each communicates about the upper extreme of the distribution of coliform at the sampling in the treatment facility. O a. Prediction Upper Limit: Xn+1<9.424. This means that a new observation will have a chance of 95% to fall below the upper limit. Tolerance Upper Limit: 11.72. Hence, we are 95% confident that a limit of 11.72 will contain 95% of the coliform measurement in the water samples O b. Prediction Upper Limit: Xn+1 $9.398. This means that a new observation will have a chance of 95% to fall below the upper limit. Tolerance Upper Limit: 11.86. Hence, we are 95% confident that a limit of 11.86 will contain 95% of the coliform measurement in the water samples O C. Prediction Upper Limit: Xp+1s 10.597. This means that a new observation will have a chance of 95% to fall below the upper limit. Tolerance Upper Limit: 11.72. Hence, we are 95% confident that a limit of 11.72 will contain 95% of the coliform measurement in the water samples O d. Prediction Upper Limit: Xn+15 10.641. This means that a new observation will have a chance of 95% to fall below the upper limit. Tolerance Upper Limit: 12.91. Hence, we are 95% confident that a limit of 12.91 will contain 95% of the coliform measurement in the wate vater samples
Transcribed Image Text:Prediction and Tolerance Interval Fifteen (15) samples of water were collected from a certain treatment facility in order to gain some insight regarding the amount of coliform in the urban pipeline. The concentration of the coliform is measured in parts per million (ppm), per liter. Suppose that the mean at the treatment facility is not as important as the upper extreme of the distribution of the amount of coliform detected. The concern is whether the concentration of coliform is too large. Readings for the 15 water samples gave a sample mean of 3.84 ppm per liter and a standard deviation of 3.07 ppm per liter. Assume that the readings are a random sample from a normal distribution. Calculate a Prediction Interval (upper 95% prediction limit) and a Tolerance Limit (95% upper tolerance limit that exceeds 95% of the population values). Interpret both; that is, tell what each communicates about the upper extreme of the distribution of coliform at the sampling in the treatment facility. O a. Prediction Upper Limit: Xn+1<9.424. This means that a new observation will have a chance of 95% to fall below the upper limit. Tolerance Upper Limit: 11.72. Hence, we are 95% confident that a limit of 11.72 will contain 95% of the coliform measurement in the water samples O b. Prediction Upper Limit: Xn+1 $9.398. This means that a new observation will have a chance of 95% to fall below the upper limit. Tolerance Upper Limit: 11.86. Hence, we are 95% confident that a limit of 11.86 will contain 95% of the coliform measurement in the water samples O C. Prediction Upper Limit: Xp+1s 10.597. This means that a new observation will have a chance of 95% to fall below the upper limit. Tolerance Upper Limit: 11.72. Hence, we are 95% confident that a limit of 11.72 will contain 95% of the coliform measurement in the water samples O d. Prediction Upper Limit: Xn+15 10.641. This means that a new observation will have a chance of 95% to fall below the upper limit. Tolerance Upper Limit: 12.91. Hence, we are 95% confident that a limit of 12.91 will contain 95% of the coliform measurement in the wate vater samples
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