Poly(3-hydroxybutyrate) (PHB), a semicrystalline polymer that is fully biodegradable and biocompatible, is obtained fromrenewable resources. From a sustainability perspective, PHB offers many attractive properties though it is more expensive toproduce than standard plastics. The accompanying data on melting point (°C) for each of 12 specimens of the polymer usinga differential scanning calorimeter appeared in an article.180.6181.7180.9181.7182.7181.7181.4182.1182.1180.2181.8180.4Compute the following. (Round your answer to two decimal places.)(a) the sample range (in °C)°C(b) the sample variance s? from the definition [Hint: First subtract 180 from each observation.]s² =(c) the sample standard deviationS =(d) s? using the shortcut methods2

As per our guidelines we can solve first three sub part of question and rest can be reposted.…

Answered: Poly(3-hydroxybutyrate) (PHB), a…

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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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?…
An EWMA chart is used to monitor a temperature in a pharmaceutical process. An automated measurement is obtained selected every minute and added to the EWMA chart. Experience indicates that o = 0.5 °F. The process target (process mean when in control) is 72 °F. The EWMA uses a lamda value 1=0.3. If the accumulated value of the EWMA at the end of measurement 47 is Z17 = 72.165 and measurement 48 is equal to X49 = 72.466, what will be the accumulated value of the EWMA at the end of measurement 48, that is ZAg = ? 72.255 72.240 72.270 72.285 72.225 not enough information
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 = 18.11 kgf/cm² for the modified mortar (m = 42) and y = 16.82 kgf/cm² for the unmodified mortar (n = 30). Let μ₁ and μ₂ 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 ₁ = 1.6 and ₂ = 1.3, test Ho: ₁ - ₂ = 0 versus H₂ : ₁ - ₂ > 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 Reject Ho. The data does not suggest 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 tension bond strengths…
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 = 18.11 kgf/cm² for the modified mortar (m = 42) and y = 16.82 kgf/cm² for the unmodified mortar (n = 32). Let μ₁ and μ₂ 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 0₁ = 1.6 and ₂ = 1.3, test Ho: ₁ - ₂ = 0 versus H₂: M₁-M₂ > 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 Reject Ho. The data does not suggest that the difference in average tension bond strengths exceeds 0. O Fail to reject Ho. The data suggests that the difference in average tension bond strengths exceeds…
Special dyes have been developed that only bind to DNA. The intensity of the color is directly related to the quantity of DNA present in the cell. A haploid cell would have a “low” amount of color, a cell in G1 would have a “medium” amount of color, and a cell in M phase would have a “high” amount of color. Is this sort of assessment quantitative, qualitative or semi-quantitative? Justify your reasoning.
A certain chemical pollutant in the Pasig River has been constant for several years with mean u = 34 ppm (parts per million). A group of industry representatives claim that their waste management programs have lowered this average significantly.
How wellmaterials conduct heat matters when designing houses, forexample. Conductivity is measured in terms of watts of heatpower transmitted per square meter of surface per degree Celsius of temperature difference on the two sides of the material.The National Institute of Standards and Technology (NIST)provides data on properties of materials. Here are 9 NISTmeasurements of the heat conductivity of a particular type offibrous-glass board: 0.0339 0.0337 0.0334 0.0334 0.03330.0333 0.0333 0.0332 0.0330 (a) We can consider this an SRS of all specimens of fibrousglass board of this type. Make a stemplot. Is there any sign ofmajor deviation from Normality? (b) Give a 95% confidence interval for the mean conductivity. (c) Is there significant evidence at the 5% level that themean conductivity of this type of fibrous-glass board is not0.0330?
Lichen has a high absorbance capacity for radiation fallout from nuclear accidents. Since lichen is a major source for Alaskan caribou, and caribou is major food source tor the Alaskan villagers, it is important to monitor the level of radioactivity in lichen. Researcher at the University of Alaska, Fairbanks, collected 9 lichen specimens and measured the amount in micro-curies per milliliter) of the radioactive element cesium-137 for each. Using the Minitab software, it was found that the mean and standard deviation for the 9 observations were X = 0.009027 аnd s = 0.004854 A) Give a point estimate for the mean amount of cesium in lichen specimens collected in Alaska. B) Give the t- value used in a small sample 95% confidence interval for the true mean amount of cesium in Alaskan lichen specimens. C) Form a 95% confidence interval for the true mean amount of cesium in Alaskan lichen specimens. D) Give a practical interpretation for the interval you obtained in part C. E) What…
A baby dolphin is born into the ocean, which is a fairly cold environment. Water has a high heat conductivity, so the thermal regulation of a newborn dolphin is quite important. It has been known for a long time that baby dolphins' blubber is different in composition and quantity from the blubber of adults. Does this make the babies better protected from the cold compared to adults? One measure of the effectiveness of blubber is its "conductance". This value as calculated on 5 newborn dolphins and 10 adults dolphins (Dunkin et al. 2005). The newborn dolphins had an average conductance of 10.44, with a standard error of the mean equal to 0.58. The adult dolphins' conductance averaged 8.44, with the standard error of this estimate equal to 1.87. All measures are given in watts per square meter per degree Celcius. Calculate the standard deviation of conductance for the baby dolphin group. Round your answer two decimal places.
Given the “data” determined by y = x^3 + (x-1)^2 with x = 0.1, 1.2, 2.3, and 2.9, calculate SSTO, SSR, and R^2. Then recalculate these using x = 0.3, 1, 2.45, and 2.8. Does where you collect your data (i.e., which values of x) appear to impact your interpretation of how good the linear model fits?.

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