An investigation of the properties of bricks used to line aluminum smelter pots

Question

An investigation of the properties of bricks used to line aluminum smelter pots was published in an article. Six different commercial bricks were evaluated. The life span of a smelter pot depends on the porosity of the brick lining (the less porosity, the longer the life span), consequently, the researchers measured the apparent porosity of each brick specimen, as well as the mean pore diameter of each brick. See the table. Click the icon to view the table. a. Find the least squares line relating porosity (y) to mean pore diameter x) | | + | |x(Round to the nearest thousandth as needed.) b. Interpret the y-intercept of the line. Choose the correct answer below. A The y-meroept is This value has no meaning because o is not in the observed range of the independent variable mean pore diameter. O B The y-intercept is For each unit increase in mean pore diameter, the mean porosity is estimated to increase by Po C. There is not enough information to answer this question. O D. c. Interpret the slope of the line. Choose the correct answer below. O The slope of the line is 1 For each unit increase in mean pore diameter, the mean porosity is estimated to increase by 1 The y-intercept is 0. The mean pore diameter is estimated to be Po when the apparent porosity is zero O B. There is not enough information to answer this question. The slope of the line is 1. The mean pore diameter is estimated to be 1 when the apparent porosity is zero The slope of the line is 1. This value has no meaning because o is not in the observed range of the independent variable mean pore diameter. C. D. d. Predict the apparent percentage of porosity for a brick with a mean pore diameter of 10 micrometers. Data Table y | % (Round to the nearest thousandth as needed.) Brick Apparent Mean Pore Diameter Porosity (%)(micrometers) 18.9 18.3 18.3 8.9 17.4 12.0 9.8 7.3 Click to select your answer(s). 10.9 16.7

Explanation / Answer

a) y = 6.233+0.984x

b) Option A is Correct because intercept has no practical significance

c) Option A is Correct

d) y = 6.233+0.984*10

= 16.072

ANOVA df SS MS F Significance F Regression 1 77.92099 77.92099 7.952676 0.047826 Residual 4 39.19234 9.798085 Total 5 117.1133 Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Intercept 6.233 3.813895 1.634288 0.177537 -4.35607 16.82207 X Variable 1 0.984 0.348877 2.820049 0.047826 0.015212 1.952489

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