Concept explainers
If a process variable is
a. With LSL and USL denoting the lower and upper specification limits, one commonly used process capability index is
b. The Cp index described in (a) does not take into account process location. A capability measure that does involve the process
Calculate the value of Cpk for each of the corkproduction processes described in the previous exercise, and comment. [Note: In practice, m and s have to be estimated from process data; we show how to do this in Section 16.2]
c. How do Cp and Cpk compare, and when are they equal?
Trending nowThis is a popular solution!
Chapter 16 Solutions
Probability and Statistics for Engineering and the Sciences
- Respiratory Rate Researchers have found that the 95 th percentile the value at which 95% of the data are at or below for respiratory rates in breath per minute during the first 3 years of infancy are given by y=101.82411-0.0125995x+0.00013401x2 for awake infants and y=101.72858-0.0139928x+0.00017646x2 for sleeping infants, where x is the age in months. Source: Pediatrics. a. What is the domain for each function? b. For each respiratory rate, is the rate decreasing or increasing over the first 3 years of life? Hint: Is the graph of the quadratic in the exponent opening upward or downward? Where is the vertex? c. Verify your answer to part b using a graphing calculator. d. For a 1- year-old infant in the 95 th percentile, how much higher is the walking respiratory rate then the sleeping respiratory rate? e. f.arrow_forwardIf the observed t value is 3.32 and the critical t value is 1.812, then that means.arrow_forwardFind the mean and standard deviation for each uniform continuous model. (Round mean answers to 1 decimal place and standard deviation answers to 4 decimal places)a. U(2,15) Mean= Standard deviation=b. U(110,190) Mean=Standard Deviation=c. U(3,97)Mean= Standard Deviation=arrow_forward
- = 3.84, the Consider the demand for Fresh Detergent in a future sales period when Enterprise Industries' price for Fresh will be x1 average price of competitors' similar detergents will be x2 = 3.93, and Enterprise Industries' advertising expenditure for Fresh will be x3 = .00. A 95 percent prediction interval for this demand is given on the following JMP output: 31 Predicted Demand 8.5665895648 95% PI[ Level of inventory needed = Lower dollar amount = Lower 95% Mean Demand 8.0099247189 Upper 95% Mean Demand 9.1232544107 99% PI[ StdErr Indiv Demand .2708133658 (a) Find and report the 95 percent prediction interval on the output. If Enterprise Industries plans to have in inventory the number of bottles implied by the upper limit of this interval, it can be very confident that it will have enough bottles to meet demand for Fresh in the future sales period. How many bottles is this? If we multiply the number of bottles implied by the lower limit of the prediction interval by the price of…arrow_forwardThe total number of thousands of tons of coal produced per year over a 10-year period for a certain region is provided in the accompanying dataset. Use double exponential smoothing to determine which pairs of values for a and ß minimize MAD for this dataset. α = 0.1, p=0.9; a=0.5, ß= 0.2; x = 1, ß= 0.6 Click the icon to view the coal production data. First find the MAD for each pair of values, a and ß. (Type integers or decimals rounded to two decimal places as needed.) X В MAD 0.9 0.2 0.6 0.1 0.5 1 123 1 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 A Year 1 2 3 4 5 6 7 8 9 10 B Coal Production (thousands of tons) C 434333 434332.9 420420 420419.9 439047 439046.9 477192 477191.9 504182 504181.9 526959 526958.9 546824 546823.9 564878 564877.9 556704 556703.9 570973 570972.9 Darrow_forwardJapan'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. 125.4 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 77.4 77.2 79.9 80.4 159.4 145.7 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; = 2817.4 Fy, = 1575.5, 5x? = 415,815.84, Fxy, = 222,698.08, Fy? = 124,149.53. Also, x = 140.870, y = 78.78, Sy = 18,928.7020, Sy = 757.395, and SSE = 9.222. The estimated standard deviation is o = 0.716 and the equation of the least squares line is…arrow_forward
- 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"+ 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 125.1 98.3 201.5 147.1 145.7 124.6 112.2 120.0 161.2 179.0 77.7 76.7 81.5 79.8 y 78.1 78.3 77.4 77.0 80.3 80.2 X 159.7 145.9 75.0 151.3 144.4 125.1 198.6 132.5 159.8 110.7 79.9 78.9 76.8 78.2 79.6 77.9 81.3 76.9 78.8 78.8 Relevant summary quantities are Σx, = 2817.7 Σy, = 1574.1, Σx2 = 415,920.19, Σxy = 222,534.06, Σy? = 123,930.31. Also, X = 140.885, y = 78.71, Sxx = 18,948.5255, Sxy= 766.982, and SSE 10.091. The estimated standard deviation is a = 0.749 and the equation of the least squares is…arrow_forwardA number of studies have shown lichens (certain plants composed of an alga and a fungus) to be excellent bio-indicators of air pollution. An article gives the following data (n = 13) on x = NO3- wet deposition (g N/m²) and y = lichen (% dry weight): x 0.05 0.10 0.11 0.12 0.31 0.37 0.42 0.58 0.68 0.68 0.73 0.85 0.92 y 0.53 0.58 0.45 0.54 0.57 0.53 1.03 0.89 0.90 1.02 0.91 1.00 1.66 Predictor Constant no3 depo Estimate 0.39329 0.9286 0.1767 S = 0.1866 R-sq = 71.5% Standard Error 0.09566 t Stat 4.11 5.26 p-value 0.002 0.000 R-sq (adj) = 68.9% (a) What are the least square estimates of B and ẞ₁? (Round your answers to three decimal places.) bo = = (b) What is the margin of error for a 95% CI for ẞ₁? (Round your answer to four decimal places.) Margin of Error = (c) Predict lichen N for an NO3 - deposition value of 0.7. (Round your answer to three decimal places.) % dry weight (d) What is the estimate of σ? (Round your answer to four decimal places.) S = % dry weightarrow_forwardFind critical value z0.10arrow_forward
- The accompnaying data were read from a graph that appeared in a recent study. The independent variable is SO2 deposition rate (mg/m2/day) and the dependent variable is steel weight loss (g/m²). 14 18 40 43 45 112 280 350 470 500 560 1200 Calculate the residual for when SO2 deposition rate is 18 mg/m²/day. -44.516 44.516 -54.516 54.516arrow_forwardThe parameter “∆uv” helps to determine the colour balance of LiquidCrystal Display (LCD) panels on a manufacturing line. The targetvalue, upper and lower specification limits for the product are ∆uv =0.0000, ∆uv = 0.0200 and ∆uv = –0.0200 respectively, where ∆uv is aunitless parameter.Prior knowledge of the process provides evidence that ∆uv variesrandomly. Recent measurements at final test indicate that the meanvalue of ∆uv is –0.0027 and the standard deviation is 0.0057. 100,000 units of that LCD panel are produced, Calculate the number of panels that falls below, and the number of panels that falls above, the specification limitsarrow_forwardQ1: The data points below are related to a chemi-thermo-mechanical pulp from mixed density hardwoods. They relate Y (specific surface area of the fibres in cm/g) to the % NaOH (sodium hydroxide) used as a pretreatment chemical and the treatment time (min) for different batches of pulp. The variables are present at three different levels. In this case, it is preferred (for reasons that we will discuss later in the course) to code the levels as shown in the last two columns of the table below, designated by Xı and X2. Y SODIUM ΤΙME Xi X2 HYDROXIDE 5.95 3 30 -1 5.60 3 60 -1 5.44 3 90 -1 1 6.22 9. 30 -1 5.85 9 60 5.61 9. 90 1 8.36 15 30 1 -1 7.30 15 60 1 6.43 15 90 1 1 a. Using the variables Y, X1 and X2 (not actual time and sodium hydroxide! You will see why later!), fit the following multiple linear regression model to the data: (Model A) Y = (b0) + (b1) X1 + (b2) X2; subsequently, estimate the parameters and examine the residual plot (residuals vs Y hat). What does this residual plot…arrow_forward
- Algebra & Trigonometry with Analytic GeometryAlgebraISBN:9781133382119Author:SwokowskiPublisher:CengageCalculus For The Life SciencesCalculusISBN:9780321964038Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.Publisher:Pearson Addison Wesley,