Question

3. A plastics producer is interested changing the composition of one of its materials to improve durability. They need to increase the strength of an existing material to a specific standard to meet the demands of a customer. They decide that introducing a new chemical (chemical) is the cheapest solution to increasing the strength. Their R&D guys test the breaking strengths of a plastic produced with several different concentrations of chemical. The values are reported below. Concentration 0.01 0.006 0.031 0.005 0.021 0.0110.036 0.01 0.005 0.027 Break 9.85 9.02 11.075 9.487 10.443 9.716 10.53 9.537 8.935 10.41 a) The mean concentration is .0163 and the mean of the breaking tensions is 9.9, similarly. Sc-.0012, Sky - 0654 and Syy = 4.37. Use this information to find the least squares regression line b) Find the correlation between chemical X composition and breaking tension c) Find the residual standard deviation ie.se d) Find a 95% interval for the expected mean breaking tension when the concentration is .025.

Answer 1

a)

het x represents concentration and press Y represents break strength A sample af size 10 is given Griver N=0.0.163 Sca=0.0012 J = a.a 0.0654 sxy Syy- 4.37 We want to fit the regression line. y=a+bx know Say Sax 0.0654 ooola 54.5 â ý bã 9.q- 54.58 0.0163 -9.9.-0.88835 9.0116 the regression line is ŷ = 9:0116 + 54.5% + €

b) c)

correlation between X and y is 0.0654 sxy Jsxx Syy O-0012 X 4.37 0:0654 Jo.005244 0.9031 g=0.9031 c) residual standard deviation, is se SSRES D-2 sbres Ssanal -b sxy Syy-bosay Syy=SSTA - 18.37 - 54.5 x 0.0654 O. 6057 Se 0.4057 10-2 Jo-1007125 0:31735

d)

e) have to find the confidence Posterval for E(Y/x=0.025) 100(1-2% confidence interval for Ecylko) is for Elyo) - topg-2 0-2 MS Res (hot (26-zija Sxx 2 Ely/%) < El E [Slot MS ha D-2 Res (+5+(%-2 Sxx Elŷlxo) = 9 = 9:0116 + 54.5*00025 10.3741 Mores ssres Oo805 D-2 0.10071 toos, o 2.3060 I from table

Substituting all these values in the equation for confidence interval we get,

$\tiny 10.3741-2.306\sqrt{0.10071(0.1 +\frac{0.0087^{2}}{0.0012})} \leq E(y/x=0.025) \leq 10.3741+2.306\sqrt{0.10071(0.1 +\frac{0.0087^{2}}{0.0012})}$

10.3741 - 0.2955 $\tiny \leq E(y/x=0.025)\leq$ 10.3741 +0.2955

10.0516 $\tiny \leq E(y/x=0.025)\leq$ 10.6426 is the 95% confidence interval.