Question

. Your answer is partially correct. Two different methods can be used for measuring the temperature of the solution of a Hall cell used in aluminum smelting, a thermocouple implanted in the cell and an indirect measurement produced from an IR device. The indirect method is preferable because the thermocouples are eventually destroyed by the solution. Consider the following 10 measurements: Thermocouple, x 920935 916 920 940|939|925 940|933 927|| 915|934|924|921|945|927|919943|932|935 IR. Y (a) Fit a simple linear regression model to the data. û = 69.971 +1 0.926 x (Round your answers to 3 decimal places.) (b) Test for significance of regression using a = 0.05 and calculate R. Calculate fo: i 21.72 (Round your answer to 2 decimal places.) Is the simple linear regression model significant? Yes. Y R2 = i 73.08 % (Round your answer to 2 decimal places.) (c) Test the hypothesis Ho: Bi = 1 versus H :B1 * 1. Calculate to: i 0.374 .(Round your answers to 3 decimal places.) Is there evidence to support a claim that both devices produce equivalent temperature measurements? No.

Answer 1

Answer Date: 02/12/2019 a) The simple linear regression of IR (v) on thermocouple (x) is, By using Excel software, find regression equation with help of following steps: 1) Import the data. Select Data Analysis from Data menu. Select Regression and click Ok. Input Y Range and Input X range. Click Ok.

SUMMARY OUTPUT Regression Statistics Multiple R 0.779845264 R Square 0.608158636 Adjusted R Square 0.559178466 Standard Error 6.670718466 Observations 10 ANOVA SS 1 Regression Residual Total MS F Significance F 552.5121212 552.5121212 12.41643 0.007803612 355.9878788 44.49848485 908.5 9 Intercept Thermocouple Coefficients Standard Error t Stat p-value Lower 95% Upper 95% 127.6888889 227.5581728 0.561126359 0.590078 -397.0611982 652.438976 0.862626263 0.244807311 3.523694862 0.007804 0.298099593 1.42715293 From the Excel output, the simple linear regression of IR (V) on thermocouple (x) is, = Be + B. x = 127.689 +0.863x

The F-test statistics is, From the Excel output in part (a), the F-test statistics is 12.42 The R-squared is, From the Excel output in part (a), the R-squared is 60.82%. To test the hypothesis is that the slop is different from 1 at 5% significance level. The null and alternative hypothesis is, H:B2 = 1 H.B. 1 The t-test statistics is,

t =- Coefficient -0 S.E.Coefficients 0.8626-1 0.2448 = -0.561 The t-test statistics is -0.561. The t critical value for the 95% confidence is, The sample size is small and two-tailed test. Look in the column headed a = 0.05 and the row headed in the t distribution table by using degree of freedom is, d.f.=n-2 = 10-2 = 8 The t critical value for the 95% confidence is † 2.306. Decision

The conclusion is that the t value corresponds to sample statistics is not fall in the critical region, so the null hypothesis is not rejected at 5% level of significance. There is insufficient evidence to indicate that the slop is different from 1. The result is not statistically significant. that both devices produce equivalent So, there is an evidence to support a claim temperature measurements.