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lapan's high population density has resulted in a multitude of resource-usage problems. One especially serious difficulty concerns waste removal. The article "Innovative Sludge Handling Through Pelletization Thickening"t reported the development of a new compression machine for processing sewage sludge. An important part of the investigation involved relating the moisture content of compressed pellets (y, in %) to the machine's filtration rate (x, in kg-DS/m/hr) Consider the following data. x 125.2 98.2 201.4 147.4 145.7 124.5 112.1 120.0 161.3 178.9 77.8 76.8 81.6 79.9 78.1 78.3 77.3 76.9 80.2 80.2 x 159.5 145.7 75.0 151.4 144.2 125.1 198.6 132.5 159.4 110.7 y 79.7 79.2 76.7 78.3 79.6 77.9 81.5 76.8 79.2 78.8 Relevant summary quantities are x, = 2816.8 y, = 1574.8, x? = 415,645.46, xy, = 222,590.57, y? = 124,042.78. Also, x = 140.840, y = 78.74, S = 18,927.3480, S = 795.738, and SSE = 9.607. The estimated standard deviation is a = 0.731 and the equation of the least squares line is y = 72.825 + 0.042x. Consider the filtration rate-moisture content data introduced above. (a) Compute a 90% CI for B, + 1256,, true average moisture content when the filtration rate is 125. (Round your answers to three decimal places.) (b) Predict the value of moisture content for a single experimental run in which the filtration rate is 125 using a 90% prediction level. (Round your answers to three decimal places.)

lapan's high population density has resulted in a multitude of resource-usage problems. One especially serious difficulty concerns waste removal. The article "Innovative Sludge Handling Through Pelletization Thickening"t reported the development of a new compression machine for processing sewage sludge. An important part of the investigation involved relating the moisture content of compressed pellets (y, in %) to the machine's filtration rate (x, in kg-DS/m/hr) Consider the following data. x 125.2 98.2 201.4 147.4 145.7 124.5 112.1 120.0 161.3 178.9 77.8 76.8 81.6 79.9 78.1 78.3 77.3 76.9 80.2 80.2 x 159.5 145.7 75.0 151.4 144.2 125.1 198.6 132.5 159.4 110.7 y 79.7 79.2 76.7 78.3 79.6 77.9 81.5 76.8 79.2 78.8 Relevant summary quantities are x, = 2816.8 y, = 1574.8, x? = 415,645.46, xy, = 222,590.57, y? = 124,042.78. Also, x = 140.840, y = 78.74, S = 18,927.3480, S = 795.738, and SSE = 9.607. The estimated standard deviation is a = 0.731 and the equation of the least squares line is y = 72.825 + 0.042x. Consider the filtration rate-moisture content data introduced above. (a) Compute a 90% CI for B, + 1256,, true average moisture content when the filtration rate is 125. (Round your answers to three decimal places.) (b) Predict the value of moisture content for a single experimental run in which the filtration rate is 125 using a 90% prediction level. (Round your answers to three decimal places.)

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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(d) Interpret the hypotheses H,: Bo + 1256, = 80 and H,: Bo + 1256, < 80. Assuming the filtration rate is 125 kg-DS/m/h, we would test to see if the average moisture content of the compressed pellets is ---Select-- 80%. Carry out a hypothesis test at significance level 0.01. Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to three decimal places.) t = P-value = State the conclusion in the problem context. O Reject H. There is significant evidence to prove that the true average moisture content for a filtration rate of 125 kg-DS/m/hr is less than 80%. O Reject Ho. There is not significant evidence to prove that the true average moisture content for a filtration rate of 125 kg-DS/m/hr is less than 80%. O Fail to reject Hn. There is not significant evidence prove that the true average moisture content for a filtration rate of 125 kg-DS/m/hr is less than 80%. O Fail to reject H. There is significant evidence to prove that the true average moisture content for a filtration rate of 125 kg-DS/m/hr is less than 80%. You may need to use the appropriate table in the Appendix of Tables to answer this question.
Transcribed Image Text:(d) Interpret the hypotheses H,: Bo + 1256, = 80 and H,: Bo + 1256, < 80. Assuming the filtration rate is 125 kg-DS/m/h, we would test to see if the average moisture content of the compressed pellets is ---Select-- 80%. Carry out a hypothesis test at significance level 0.01. Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to three decimal places.) t = P-value = State the conclusion in the problem context. O Reject H. There is significant evidence to prove that the true average moisture content for a filtration rate of 125 kg-DS/m/hr is less than 80%. O Reject Ho. There is not significant evidence to prove that the true average moisture content for a filtration rate of 125 kg-DS/m/hr is less than 80%. O Fail to reject Hn. There is not significant evidence prove that the true average moisture content for a filtration rate of 125 kg-DS/m/hr is less than 80%. O Fail to reject H. There is significant evidence to prove that the true average moisture content for a filtration rate of 125 kg-DS/m/hr is less than 80%. You may need to use the appropriate table in the Appendix of Tables to answer this question.
Japan's high population density has resulted in a multitude of resource-usage problems. One especially serious difficulty concerns waste removal. The article "Innovative Sludge Handling Through Pelletization Thickening"t reported the development of a new compression machine for processing sewage sludge. An important part of the investigation involved relating the moisture content of compressed pellets (y, in %) to the machine's filtration rate (x, in kg-DS/m/hr). Consider the following data. 125.2 98.2 201.4 147.4 145.7 124.5 112.1 120.0 161.3 178.9 y 77.8 76.8 81.6 79.9 78.1 78.3 77.3 76.9 80.2 80.2 159.5 145.7 75.0 151.4 144.2 125.1 198.6 132.5 159.4 110.7 y 79.7 79.2 76.7 78.3 79.6 77.9 81.5 76.8 79.2 78.8 Relevant summary quantities are = 2816.8 y, = 1574.8, x, = 415,645.46, xy; = 222,590.57, y = 124,042.78. Also, x = 140.840, y = 78.74, S = 18,927.3480, S = 795.738, and SSE = 9.607. The estimated standard deviation is o = 0.731 and the equation of the least squares line is y = 72.825 + 0.042x. Consider the filtration rate-moisture content data introduced above. (a) Compute a 90% CI for B, + 1256,, true average moisture content when the filtration rate is 125. (Round your answers to three decimal places.) (b) Predict the value of moisture content for a single experimental run in which the filtration rate is 125 usin a 90% prediction level. (Round your answers to three decimal places.) How does this interval compare to the interval of part (a)? Why is this the case? The width of the confidence interval in part (a) is -Select--- v the width of the prediction interval in part (b) since the ---Select-- v interval must account for both the uncertainty in knowing the value of the population mean in addition to the data scatter. (c) How would the intervals of parts (a) and (b) compare to a CI and PI when filtration rate is 115? Answer without actually calculating these new intervals. Because the value of 115, denoted by x*, is -Select-- v x than 125, the term (x* – x)? will be Select--- v, making the standard error --Select--- v, and thus the width of the interval is -Select--- (d) Interpret the hypotheses Ho: Bo + 1256, = 80 and H: Bo + 1256, < 80. Assuming the filtration rate is 125 kg-DS/m/h, we would test to see if the average moisture content of the compressed pellets is ---Select- 80%.
Transcribed Image Text:Japan's high population density has resulted in a multitude of resource-usage problems. One especially serious difficulty concerns waste removal. The article "Innovative Sludge Handling Through Pelletization Thickening"t reported the development of a new compression machine for processing sewage sludge. An important part of the investigation involved relating the moisture content of compressed pellets (y, in %) to the machine's filtration rate (x, in kg-DS/m/hr). Consider the following data. 125.2 98.2 201.4 147.4 145.7 124.5 112.1 120.0 161.3 178.9 y 77.8 76.8 81.6 79.9 78.1 78.3 77.3 76.9 80.2 80.2 159.5 145.7 75.0 151.4 144.2 125.1 198.6 132.5 159.4 110.7 y 79.7 79.2 76.7 78.3 79.6 77.9 81.5 76.8 79.2 78.8 Relevant summary quantities are = 2816.8 y, = 1574.8, x, = 415,645.46, xy; = 222,590.57, y = 124,042.78. Also, x = 140.840, y = 78.74, S = 18,927.3480, S = 795.738, and SSE = 9.607. The estimated standard deviation is o = 0.731 and the equation of the least squares line is y = 72.825 + 0.042x. Consider the filtration rate-moisture content data introduced above. (a) Compute a 90% CI for B, + 1256,, true average moisture content when the filtration rate is 125. (Round your answers to three decimal places.) (b) Predict the value of moisture content for a single experimental run in which the filtration rate is 125 usin a 90% prediction level. (Round your answers to three decimal places.) How does this interval compare to the interval of part (a)? Why is this the case? The width of the confidence interval in part (a) is -Select--- v the width of the prediction interval in part (b) since the ---Select-- v interval must account for both the uncertainty in knowing the value of the population mean in addition to the data scatter. (c) How would the intervals of parts (a) and (b) compare to a CI and PI when filtration rate is 115? Answer without actually calculating these new intervals. Because the value of 115, denoted by x*, is -Select-- v x than 125, the term (x* – x)? will be Select--- v, making the standard error --Select--- v, and thus the width of the interval is -Select--- (d) Interpret the hypotheses Ho: Bo + 1256, = 80 and H: Bo + 1256, < 80. Assuming the filtration rate is 125 kg-DS/m/h, we would test to see if the average moisture content of the compressed pellets is ---Select- 80%.
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