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11 2.3 please help in part 3
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- 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.Wrinkle recovery angle and tensile strength are the two most important characteristics for evaluating the performance of crosslinked cotton fabric. An increase in the degree of crosslinking, as determined by ester carboxyl band absorbance, improves the wrinkle resistance of the fabric (at the expense of reducing mechanical strength). The accompanying data on x = absorbance and y = wrinkle resistance angle was read from a graph in the paper "Predicting the Performance of Durable Press Finished Cotton Fabric with Infrared Spectroscopy".† x 0.115 0.126 0.183 0.246 0.282 0.344 0.355 0.452 0.491 0.554 0.651 y 334 342 355 363 365 372 381 392 400 412 420 Here is regression output from Minitab: Predictor Constant absorb S = 3.60498 Coef 321.878 156.711 SOURCE Regression Residual Error Total SE Coef 2.483 6.464 R-Sq = 98.5% DF 1 9 10 SS 7639.0 117.0 7756.0 T 129.64 24.24 0.000 0.000 R-Sq (adj) = 98.3% MS 7639.0 13.0 F P 587.81 (a) Does the simple linear regression model appear to be…Wrinkle recovery angle and tensile strength are the two most important characteristics for evaluating the performance of crosslinked cotton fabric. An increase in the degree of crosslinking, as determined by ester carboxyl band absorbance, improves the wrinkle resistance of the fabric (at the expense of reducing mechanical strength). The accompanying data on x = absorbance and y = wrinkle resistance angle was read from a graph in the paper "Predicting the Performance of Durable Press Finished Cotton Fabric with Infrared Spectroscopy".t 半 0.115 0.126 0.183 0.246 0.282 0.344 0.355 0.452 0.491 0.554 0.651 334 342 355 363 365 372 381 392 400 412 420 Here is regression output from Minitab: Predictor Coef SE Coef P Constant 321.878 2.483 129.64 0.000 absorb 156.711 6.464 24.24 0.000 S = 3.60498 R-Sq = 98.5% R-Są (adj) - 98.3% SOURCE DF MS F P Regression 1 7639.0 7639.0 587.81 0.000 Residual Error 9 117.0 13.0 Total 10 7756.0 (a) Does the simple linear regression model appear to be…
- Wrinkle recovery angle and tensile strength are the two most important characteristics for evaluating the performance of crosslinked cotton fabric. An increase in the degree of crosslinking, as determined by ester carboxyl band absorbance, improves the wrinkle resistance of the fabric (at the expense of reducing mechanical strength). The accompanying data on x = absorbance and y = wrinkle resistance angle was read from a graph in the paper "Predicting the Performance of Durable Press Finished Cotton Fabric with Infrared Spectroscopy".t x 0.115 0.126 0.183 0.246 0.282 0.344 0.355 0.452 0.491 0.554 0.651 y 334 342 355 363 365 372 381 400 392 412 420 Here is regression output from Minitab: Predictor Constant absorb S = 3.60498 Coef 321.878 156.711 SOURCE Regression Residual Error Total R-Sq= 98.5% DF SE Coef 2.483 6.464 1 9 10 SS 7639.0 117.0 7756..0 T 129.64 24.24 P 0.000 0.000. R-Sq (adj) 98.3% MS 7639.0 13.0 F 587.81 (a) Does the simple linear regression model appear to be appropriate?…Q3) An experiment was carried out to investigate variation of solubility of chemical X in water. The quantities in kg that dissolved in 1 liter at various temperatures are show in the table (1). Table (1) Temperature C Mass of X 2.1 2.6 2.9 3.3 15 20 25 30 35 4 50 5.1 70 7 Use the proper methods to answer the following questions: a) Draw a scatter diagram to show the data. b) Estimate the temperature based on the mass of X. c) What quantity might be expected to dissolve at 42 C? Find the quantity that your cquation indicates would dissolve at 10 C and comment on your answer.An experiment to compare the tension bond strength of polymer latex modified mortar (Portland cement mortar to which polymer latex emulsións have been added during mixing) to that of unmodified mortar resulted in x = 18.11 kgf/cm2 for the modified mortar (m = 42) and y = 16.88 kgf/cm2 for the unmodified mortar (n = 31). Let ₁ and ₂ be the true average tension bond strengths for the modified and unmodified mortars, respectively. Assume that the bond strength distributions are both normal. (a) Assuming that o₁ = 1.6 and ₂ = 1.3, test Ho: ₁ - ₂ = 0 versus H₂: H₁ - H₂> 0 at 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 four decimal places.) Z = P-value = State the conclusion in the problem context. O Fail to reject Ho. The data suggests that the difference in average tension bond strengths exceeds 0. Fail to reject Ho. The data does not suggest that the difference in average tension bond strengths…
- 17.7 Butterfly wings. Researchers studied the morphological attributes of monarch butterflies (Danaus plexippus), a species that undertakes large seasonal migrations over North America. They measured the forewing weight (in milligrams, mg) of a sample of 92 monarch butterflies, all of which had been reared in captivity in identical conditions.° Figure 17.4 shows the output from the statistical software JMP. (The data are also available in the Large.Butterfly the data file if you wish to practice working with your own software.) Estimate with 95% confidence the mean forewing weight of monarch butterflies reared in captivity. Follow the four- step process as illustrated in Example 17.2. 4 STEP そMP FWweight 30 25 20 15 10 11 12 13 14 15 8 9 10 Summary Statistics Mean 11.795652 Std Dev 1.1759413 Std Err Mean 0.1226004 Upper 95% Mean Lower 95% Mean 1 FIGURE 17.4 Software output (JMP) for the forewing weight of monarch 12.039183 11.552122 92 N. butterflies. CountAccumulation of mercury (Hg) in fish is hypothesized to correlate with the fish size. Barbonymus schwanenfeldii is a species commonly found in a dam in Sarawak. Table 1 shows data of Hg concentration (mg/kg) present in three sizes of fish caught in the dam in triplicates. (i, ii, iii was already answered.)An article in Concrete Research ("Near Surface Characteristics of Concrete: Intrinsic Permeability," Vol. 41, 1989), presented data on compressive strength x and intrinsic permeability y of various concrete mixes and cures. Summary quantities are n = 14, Σy = 572, Σ.y = 23,530, x = 43, Ex = 157.42, and xy = 1697.80. Assume that the two variables are related according to the simple linear regression model. Statistical Tables and Charts Part 1 Calculate the least squares estimate of the slope. (Round your answer to 3 decimal places.)
- An experiment to compare the tension bond strength of polymer latex modified mortar (Portland cement mortar to which polymer latex emulsions have been added during mixing) to that of unmodified mortar resulted in x = 18.13 kgf/cm? for the modified mortar (m = 42) and y = 16.85 kgf/cm2 for the unmodified mortar (n = 32). Let u, and u, be the true average tension bond strengths for the modified and unmodified mortars, respectively. Assume that the bond strength distributions are both normal. (a) Assuming that o, = 1.6 and o, = 1.3, test Ho: 4, - H, = 0 versus H: u, - µ, > 0 at 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 four decimal places.) z = 3.80 P-value = 0.0001 State the conclusion in the problem context. O Fail to reject H,. The data suggests that the difference in average tension bond strengths exceeds 0. O Fail to reject Ho: The data does not suggest that the difference in average…The article "Influence of Freezing Temperature on Hydraulic Conductivity of Silty Clay" (J. Konrad and M. Samson, Journal of Geotechnical and Geoenvironmental Engineering, 2000:180–187) describes a study of factors affecting hydraulic conductivity of soils. The measurements of hydraulic conductivity in units of 108 cm/s (y), initial void ratio (x), and thawed void ratio (x2) for 12 specimens of silty clay are presented in the following table. y 1.01 1.12 1.04 1.30 1.01 1.04 0.955 1.15 1.23 1.28 1.23 1.30 0.84 0.88 0.85 0.95 0.88 0.86 0.85 0.89 0.90 0.94 0.88 0.90 X1 0.81 0.85 0.87 0.92 0.84 0.85 0.85 0.86 0.85 0.92 0.88 0.92 X2 Fit the model y = Bo + fix1 + e. For each coefficient, test the null hypothesis that it is equal to 0. Fit the model y = Bo + Bzx2 + e. For each coefficient, test the null hypothesis that it is equal to 0. Fit the model y = Bo + BzX1 + Bzxz + e. For each coefficient, test the null hypothesis that it is equal to 0. d. Which of the models in parts (a) to (c) is…A paper gives data on x = change in Body Mass Index (BMI, in kilograms/meter2) and y = change in a measure of depression for patients suffering from depression who participated in a pulmonary rehabilitation program. The table below contains a subset of the data given in the paper and are approximate values read from a scatterplot in the paper. BMI Change (kg/m²) 0.5 -0.5 0 0.1 0.7 0.8 1 1.5 1.2 1 0.4 0.4 Depression Score Change -1 9 4 4 5 8 13 14 17 18 12 14 The accompanying computer output is from Minitab. Fitted Line Plot Depression score change = 6.512 + 5.472 BMI change 20 S 5.26270 R-Sq 27.16% R-Sq (adj) 19.88% 15- : 10- -0.5 0.0 1.5 Ⓡ S 5.26270 Coefficients Term Coef VIF SE Coef 2.26 T-Value 2.88 P-Value 0.0164 Constant 6.512 BMI change 5.472 2.83 1.93 0.0823 1.00 Regression Equation Depression score change = 6.512 + 5.472 BMI change (a) What percentage of observed variation in depression score change can be explained by the simple linear regression model? (Round your answer to…