Test the hypothesis ₁ = ₂ = ₂ = M₁ = μs at 0.05 level of significance for the data of Table 13.1 on absorption of moisture by various types of cement aggregates. Absorption of Moisture in Concrete Aggregates Aggregate 1 3 4 5 551 563 457 631 522 613 656 679 Total Mean 450 731 499 632 3320 553.33 2 595 580 508 639 615 511 417 449 517 583 438 633 415 517 677 555 3416 569.33 573 648 3663 2791 3664 610.50 465.17 610.67 16854 561.80
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- 9.48) Acid gases must be removed from other refinery gases in chemical production facilities in order to minimize corrosion of the plants. Two methods for removing acid gases produced the corrosion rates (in mm/yr) are listed below in experimental tests: Method A: 0.3, 0.7, 0.5, 0.8, 0.9, 0.7, 0.8 Method B: 0.7, 0.8, 0.7, 0.6, 2.1, 0.6, 1.4, 2.3 Estimate the difference in mean corrosion rates for the two methods, using a confidence coefficient of 0.90. What assumptions must you make for your answer to be valid?Total plasma volume is important in determining the required plasma component in blood replacement therapy for a person undergoing surgery. Plasma volume is influenced by the o health and physical activity of an individual. Suppose that a random sample of 46 male firefighters are tested and that they have a plasma volume sample mean of x = 37.5 ml/kg (milliliters plasma per kilogram body weight). Assume that o = 7.80 ml/kg for the distribution of blood plasma. %3D (a) Find a 99% confidence interval for the population mean blood plasma volumne in male firefighters. What is the margin of error? (Round your answers to two decimal places.) lower limit upper limit margin of error (b) What conditions are necessary for your calculations? (Select all that apply.) the distribution of volumes is normal the distribution of volumes is uniform o is unknown n is large o is known (c) Interpret your results in the context of this problem. O The probability that this interval contains the true average…3.6 Determine the IPR for a well at the time when the average reservoir pressure will be 1,500 psig. The fol- lowing data are obtained from laboratory tests of well fluid samples: Reservoir properties Present Future Average pressure (psig) Productivity index J" (stb/day-psi) Oil viscosity (cp) Oil formation volume factor (rb/stb) Relative permeability to oil 2,200 1.25 3.55 1,500 3.85 1.20 1.15 0.82 0.65
- “Passive and Active Smoke” in Appendix B includes cotinine levels measured in a group of nonsmokers exposed to tobacco smoke (n = 40, Mean = 60.58 ng>mL, s = 138.08 ng>mL) and a group of nonsmokers not exposed to tobacco smoke (n = 40, Mean = 16.35 ng>mL, s = 62.53 ng>mL). Cotinine is a metabolite of nicotine, meaning that when nicotine is absorbed by the body, cotinine is produced. Use a 0.05 significance level to test the claim that nonsmokers exposed to tobacco smoke have a higher mean cotinine level than nonsmokers not exposed to tobacco smoke. Based on your hypothesis tests and confidence intervals, what do you conclude about the effects of second hand smoke? Why?“Passive and Active Smoke” in Appendix B includes cotinine levels measured in a group of nonsmokers exposed to tobacco smoke (n = 40, Mean = 60.58 ng>mL, s = 138.08 ng>mL) and a group of nonsmokers not exposed to tobacco smoke (n = 40, Mean = 16.35 ng>mL, s = 62.53 ng>mL). Cotinine is a metabolite of nicotine, meaning that when nicotine is absorbed by the body, cotinine is produced. Use a 0.05 significance level to test the claim that nonsmokers exposed to tobacco smoke have a higher mean cotinine level than nonsmokers not exposed to tobacco smoke. 1. Construct a confidence interval estimate of the difference betwen the mean continen levels fo the two groups of nonsmokers. What confidence level would be appropriate? 2. Find the margin of error E using the formula. 3. What is the confidence interval? Explain the meaning of the confidence interal and what the limit represents.The table below shows the 1 results from the specific gravity (S.G.) test performed in a soil laboratory including twenty samples of sand. Determine the Coefficient of Quartile Variation. * Specific Gravity 2.30-2.39 Number of Samples 1 2.40-2.49 2 2.50-2.59 3 2.60-2.69 2.70-2.79 7 2.80-2.89
- Radiation in Baby Teeth Listed below are amounts of strontium-90 (in millibecquerels, or mBq) in a simple random sample of baby teeth obtained from Pennsylvania residents born after 1979 (based on data from “An Unexpected Rise in Strontium-90 in U.S. Deciduous Teeth in the 1990s,” by Mangano et. al., Science of the Total Environment).A professor of mathematics teaches at a catholic school and at a public school. He wants to know whether attending a catholic school leads to higher grades. He collected data from 7,430 students from both school's types. Considering this information, please answer the following questions: a) How can you interpret the OLS outcome on whether the child attended a catholic school (cathhs) is important to explain students' grade? Source Model Residual Total cathhs lfaminc SS 143098.058 521610.985 motheduc fatheduc 664709.043 df math12 Coefficient Std. err. 17887.2573 8 7,421 70.2885036 7,429 89.4749015 1.623158 .4106669 1.447696 .1431147 .0613988 .7345401 .832485 . 055494 black -5.226717 hispan -1.221689 asian 3.28341 female _cons MS .3915981 .3328612 .4438741 -1.066021 .1950649 16.72931 1.355485 Number of obs F(8, 7421) Prob > F R-squared Adj R-squared Root MSE t P>|t| 3.95 0.000 0.000 10.12 11.96 0.000 15.00 0.000 -13.35 0.000 -3.67 0.000 7.40 0.000 -5.46 0.000 12.34 0.000 = .8181343…Calculate the average values of the Cp and Cpk capabil-ity indices for the BOD data , assuming that LSL=5 mg/L and USL=35 mg/L. Do these val-ues of the indices indicate that the process performance is satisfactory?
- The depth of wetting of a soil is the depth to which water content will increase owing to extemal factors. The article "Discussion of Method for Evaluation of Depth of Wetting in Residential Areas" (J. Nelson, K. Chao, and D. Overton, Journal of Geotechnical and Geoenvironmental Engineering, 2011:293-296) discusses the relationship between depth of wetting beneath a structure and the age of the structure. The article presents measurements of depth of wetting, in meters, and the ages, in years, of 21 houses, as shown in the following table. Age Depth 7.6 4 4.6 6.1 9.1 3 4.3 7.3 5.2 10.4 15.5 5.8 10.7 4 5.5 6.1 10.7 10.4 4.6 7.0 6.1 14 16.8 10 9.1 8.8 Compute the least-squares line for predicting depth of wetting (y) from age (x). b. Identify a point with an unusually large x-value. Compute the least-squares line that results from deletion of this point. Identify another point which can be classified as an outlier. Compute the least-squares line that results from deletion of the outlier,…Hydrogen content is conjectured to be an important factor in porosity of aluminum alloy castings. An article gives the accompanying data on x = content and y = gas porosity for one particular measurement technique. 0.18 0.20 0.21 0.21 0.21 0.22 0.23 0.23 0.24 0.24 0.25 0.28 0.30 0.37 0.48 0.71 0.42 0.44 0.55 0.44 0.24 0.48 0.22 0.82 0.86 0.72 0.70 0.74 Minitab gives the following output in a response to a Correlation command: Correlation of Hydrcon and Porosity = 0.425 (a) Test at level 0.05 to see whether the population correlation coefficient differs from o. State the appropriate null and alternative hypotheses. O Ho: p = 0 Hip 0 O Ho: p = 0 Hip + 0 O Ho: p+0 Hip = 0 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 Fail to reject H: The data does not suggest that the population correlation coefficient differs significantly…Fifty male subjects drank a measured amount x (in ounces) of a medication and the concentration y (in percent) in their blood of the active ingredient was measured 30 minutes later. The sample data are summarized by the following information: n = 50 Ex = 112.5 Ex? = 356.25 %3D Ey = 4.83 Ey = 0.667 Exy = 15.255 0 < x < 4.5 Or= 0.875 Or= 0.709 Or= -0.846 Or=0.460 Or= 0.965