A study is made to determine the force required to pull apart pieces of glued plastic. Three types of plastic were tested using 4different levels of humidity. The results, in kilograms, are given as follows:Plastic Type...Humidity70%33.829.726.730%39.050%33.190%33.028.530.936.927.427.229.2Use a 0.05 level of significance to test the hypothesis that there is no difference in the mean force required to pull the gluedplastic apart(a) When the different types of plastic are used(b) For different humidity conditionsI. Hypothesis:II. Level of significance:III. ANOVA TĀBLE:Source ofVariationSum of SquaresDegrees ofFreedomMean SquareComputed (f)Tabulated(f)RowColumnErrorTotalComputations: PRINT-SCREEN YOUR EXCEL SOLUTION)IV. DECISION:V. ConclusionABC

Given We need to find the anova table and decision , conclusion from the data below Plastic type…

Answered: A study is made to determine the force…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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You investigate the relationship between height and dbh (diameter at breast height) of oak trees. You get the following results: dbh (cm): 26 49 52 32 15 15 39 41 20 30 46 42 21 40 17 38 13 26 49 26height(m): 10.2 13.5 16.3 10.8 8.9 10.0 14.2 11.3 9.7 11.4 13.5 13.6 11.6 11.2 7.7 13.1 7.9 10.4 15.6 9.0 Read the data into R. (a) Calculate the following using R: ̄x , ̄y , SSx, SSy. (Note that to get SSy or SSx, you can just ask R for the variance (var) or standard deviation, and then do the appropriate calculation).(b) Perform a complete test of the hypothesis that there is no difference in height as dbh increases. Write out all the appropriate steps of a regular hypothesis test (give H0, H1, α, your decision, etc.)(c) Give the equation of the least squares line
The following are the weight losses of certain machine parts due to friction (in milligrams) when used with three different lubricants: Lubricant 1: 13 11 10 13 Lubricant 2: 9. 11 Lubricant 3: 7 6. Test at the 0,01 level of significance whether the type of lubricant effects the weight loss of the machine parts due to friction. While carrying out the test, follow the steps below and answer the questions. 1- Determine the null and alternative hypotheses. Ho: H: 2-Fill in the following ANOVA Table. ANOVA Table Source of Variation Degrees of Freedom Sum of Squares Mean Sum of Squares Treatment Error Total 3-State your decision and conclusion.
4) Blood calcium levels are measured in mg/dL. In patients over 30, μ = 9.75 mg/dL, and σ = 2.25 mg/dL. Again, use R. a) Give the blood calcium values for the middle 30% of patients. b) Give the blood calcium values for the middle 75% of patients. c) Give the 87.5th percentile.
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Convert to raw-scores μ = 3,0 = 2 Z 0.50 -1.00 2.37 -0.75 -1.49 2.64 -1.47 2.64 Blank 1; Blank 2; Blank 3; Blank 4; Blank 5; Blank 6; Blank 7; Blank 8 Blank 1 Blank 2 Blank 3 Blank 4 Blank 5 Blank 6 Blank 7 Blank 8 Add your answer Add your answer Add your answer Add your answer Add your answer Add your answer Add your answer Add your answer Br
View the attatched photo and select the correct answer for A. Then look at the questions below (B and C) and answer those using the data oe information in the photo. A) Answer A which is in the photo. B) Calculate the standardized test statistic. T = ? C) Calculate the P-value. P = ?
Values of modulus of elasticity (MOE, the ratio of stress, i.e., force per unit area, to strain, i.e., deformation per unit length, in GPa) and flexural strength (a measure of the ability to resist failure in bending, in MPa) were determined for a sample of concrete beams of a certain type, resulting in the following data: MOE 29.7 33.1 33.6 35.2 35.4 36.2 36.2 36.3 37.4 37.8 Strength 5.8 7.1 7.2 6.2 8.0 6.9 7.1 7.7 6.7 6.4 МОЕ 38.6 38.7 39.7 40.9 42.7 42.7 43.6 45.7 45.9 46.8 Strength 7.1 6.4 8.0 8.9 8.3 8.8 7.9 9.6 7.3 7.6 МОЕ 47.9 49.2 51.6 62.7 69.7 79.6 80.1 Strength 9.8 7.7 7.8 11.5 11.4 11.9 10.6 Fitting the simple linear regression model to the n = 27 observations on x = modulus of elasticity and y = flexural strength given in the data above resulted in ý = 7.589, s, = 0.180 when x = 40 and ý = 9.749, s, = 0.255 for x = 60. The simple linear regression model also resulted in a value of s = 0.8718. (a) Explain why s, is larger when x = 60 than when x = 40. O The closer x is to…
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