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. x 125.4 98.3 201.6 147.2 146.0 124.5 112.3 120.2 161.1 178.8 y 77.9 76.9 81.3 80.0 78.1 78.5 79.9 80.4 77.4 77.2 x 159.4 145.7 74.9 151.5 144.2 124.8 198.6 132.6 159.6 110.7 79.7 79.2 76.9 78.1 79.6 78.3 81.7 76.8 79.1 78.5 Relevant summary quantities are x, = 2817.4r, = 1575.5, x? = 415,815.84, xy, = 222,698.08, Cy,? = 124,149.53. Also, & = 140.870, y = 78.78, S„ = 18,928.7020, S, = 757.395, and SSE = 9.222. The estimated standard deviation is a = 0.716 and the equation of the least squares line is y = 73.140 + 0.040x. Consider the filtration rate-moisture content data introduced above. (a) Compute a 90% CI for B, + 1268, true average moisture content when the filtration rate is 126. (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 126 using 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-- vx than 126, the term (x - x)2 will be --Select-- , making the standard error -Select-- v , and thus the width of the interval is Select- - (d) Interpret the hypotheses H: B, + 1268, = 80 and H,: Bo + 1268, < 80. Assuming the filtration rate is 126 kg-DS/m/h, we would test to see if the average moisture content of the compressed pellets is -Select-- v 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.) P-value = State the conclusion in the problem context. O Reject Hg. There is significant evidence to prove that the true average moisture content for a filtration rate of 126 kg-DS/m/hr is less than 80%. O Fail to reject Ha. There is significant evidence to prove that the true average moisture content for a filtration rate of 126 kg-DS/m/hr is less than 80%. O Reject Hg. There is not significant evidence to prove that the true average moisture content for a filtration rate of 126 kg-DS/m/hr is less than 80%. Fail to reject Hg. There is not significant evidence to prove that the true average moisture content for a filtration rate of 126 kg-DS/m/hr is less than 80%.

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. x 125.4 98.3 201.6 147.2 146.0 124.5 112.3 120.2 161.1 178.8 y 77.9 76.9 81.3 80.0 78.1 78.5 79.9 80.4 77.4 77.2 x 159.4 145.7 74.9 151.5 144.2 124.8 198.6 132.6 159.6 110.7 79.7 79.2 76.9 78.1 79.6 78.3 81.7 76.8 79.1 78.5 Relevant summary quantities are x, = 2817.4r, = 1575.5, x? = 415,815.84, xy, = 222,698.08, Cy,? = 124,149.53. Also, & = 140.870, y = 78.78, S„ = 18,928.7020, S, = 757.395, and SSE = 9.222. The estimated standard deviation is a = 0.716 and the equation of the least squares line is y = 73.140 + 0.040x. Consider the filtration rate-moisture content data introduced above. (a) Compute a 90% CI for B, + 1268, true average moisture content when the filtration rate is 126. (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 126 using 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-- vx than 126, the term (x - x)2 will be --Select-- , making the standard error -Select-- v , and thus the width of the interval is Select- - (d) Interpret the hypotheses H: B, + 1268, = 80 and H,: Bo + 1268, < 80. Assuming the filtration rate is 126 kg-DS/m/h, we would test to see if the average moisture content of the compressed pellets is -Select-- v 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.) P-value = State the conclusion in the problem context. O Reject Hg. There is significant evidence to prove that the true average moisture content for a filtration rate of 126 kg-DS/m/hr is less than 80%. O Fail to reject Ha. There is significant evidence to prove that the true average moisture content for a filtration rate of 126 kg-DS/m/hr is less than 80%. O Reject Hg. There is not significant evidence to prove that the true average moisture content for a filtration rate of 126 kg-DS/m/hr is less than 80%. Fail to reject Hg. There is not significant evidence to prove that the true average moisture content for a filtration rate of 126 kg-DS/m/hr is less than 80%.

College Algebra

1st Edition

ISBN:9781938168383

Author:Jay Abramson

Publisher:Jay Abramson

Chapter6: Exponential And Logarithmic Functions

Section6.8: Fitting Exponential Models To Data

Problem 3TI: Table 6 shows the population, in thousands, of harbor seals in the Wadden Sea over the years 1997 to...

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