Question

Question 1 A factory investigated four different methods of preparing the superconducting compound PbMo.Ss. They contend that the presence of oxygen during the preparation process affects the material's superconducting transition temperature Te. Preparation methods 1 and 2 use techniques that are designed to eliminate the presence of oxygen, and methods 3 and 4 allow oxygen to be present. Five observations on Te (in °K) were made for each method, and the results are as follows: Preparation Transition Temperature Te(°K) Method 1 14.8 14.8 14.7 14.8 14.9 2 14.6 14.8 14.7 12.4 12.7 11.6 12.7 12.1 12.2 11.7 14.2 14.4 14.4 (a) Is there evidence to support the claim that the presence of oxygen during preparation affects the mean transition temperature? Use a = 0.05. (b) What is the P-value for the F-test in part (a)? (c) Analyze the residuals from this experiment. (d) Find a 95% confidence interval on mean T, when method 1 is used to prepare the material.

Answer 1

fathe information observed that the parenting Pasay atract's too madziads Supoconducting a melhor temperature te raportion mittals landa we techniques that are designed to atirinate the pletence daygen while methods sand 4 allows the dela is as folland Dygen to be present pluparation method Tebrition Temperature 14.8 14814.7 14.8 14.9 14.6 15149 148 14,3 16 12.4 112-7112.1 12-7 14.2 14.4 14.4 12.7 1-7 % check whether Here is omgh evidence to infor Hol the differences est belucer to for Arperation mellads or not Corrider the null and alterative hypotheses as follows pell hypothesis Ho: 4 = 42 = Uz o MM Allomalive hypothesis Has at least one mean is differ for at

MSE the significance levd & this tot al sy ③ the formula for the test stasies is f= MST where msis Su 1 and MsE: SSE litere SST < En; 65- where SST ki ( 3; -)?cmd SSES (23 -712 - (-1) s? +604-1) +..-.f (mx-1) set Her kis He number of tootments nj. J = 1,...k is the side & the sample with respect to ji poatment and no no Now compute the require tegl stdistic using MEGASTAT by following the instructions ster 1 : Type o import the data in the spreadsheet step-2 - click Add-ins button and sleet MEGASTAT bulton step-3 AROU clicking on MEGASTAT button and then select ANOTA and again aick on the way ANOVA JO1

stept altor diking on one way a dialog box will open. acted the core tonge in the impordat dita alumn and click on ok bulton one fador ANOTA MEAN n 14.80 5 14.80 12.30 13.38 13.82 std. Der 0.071 Group 2 0.158 GRUP-2 0.464 Group. 3 1.320 Gauge 1.258 Total s 20 A NOVA table 1. Source df 3 16 F P. volle 0.0001 Tool Elod? is 22.114 7.498 30.072 MS 7.3747 0.4968 14.85 - Total 19 Calculate the cátical value os falaas F10.05.316) = 324 (using Exay finr(Pocbabity. dg-Fixadomas, dag - fdezome)) Here Pobabity -0.05 dog - Feedom 1 -3. dog farbom2-16

INTERPRET : The value of the test statute se the calculated value & the tgl stastistic is glater than the cathal value thom dejeds to hell hypothesis - There for there is enough evidence to infor that the mean fargition temperature are equal for the facer Paeparation methods * Glulate the p-values as follows Bon port ca) MEGASTAT odret. observed that to praleis 0.0001 p-value0.000 (using seel's foist(x dong - Bandol. dag sedany Here 8-1485 dog foredom 1:3 dog foredom 2:16

Analyze the residuals from the experiment Normal Probability Plot (response is Response) 8 8 - 8 -- 8 8 Percent % % % % 8 in -1.5 -1.0 -0.5 0.0 Residual 0.5 1.0 1.5

Versus Fits (response is Response) Residual 12.0 12.5 13.0 13.5 Fitted Value 14.0 14.5 15.0 The normal probability plot of the residuals plot of transition temperature is not normal. And the residual versus fits reveals model adequacy.

a arable the as a confidence intored on the mean - Te uton method is used to prepare the motorice we have to formal for the loo (1.8) pdecent confidence interval on Ho man & the ith palmant miis giu t talsiahn-) * JmSE from the dla we have the following values 14.8. MSE: 0.797.75 0.797.7-5.94 quote that toons.4(5-1) = ko. 025.16 = 9.12 there fore the 95% confidence interval on the mante whon method is used to prove the modria is Vit to 095.36 VMSELN = 148 £9.12 * 079779 =74.87 0.8464 al (13.9536 L M, L 15.6464]