1. Corrosion of still reinforcing bars is the most important durability problem for reinforced concrete structures. Carbonation of cocrete results from a chemical reaction that lowers the pH value enough to to initiate corrosion. The table below shows a sample dependence of the strength in MPa (y) on the carbonation depth in mm (x) taken from a particular building.
Correlation
Correlation defines a relationship between two independent variables. It tells the degree to which variables move in relation to each other. When two sets of data are related to each other, there is a correlation between them.
Linear Correlation
A correlation is used to determine the relationships between numerical and categorical variables. In other words, it is an indicator of how things are connected to one another. The correlation analysis is the study of how variables are related.
Regression Analysis
Regression analysis is a statistical method in which it estimates the relationship between a dependent variable and one or more independent variable. In simple terms dependent variable is called as outcome variable and independent variable is called as predictors. Regression analysis is one of the methods to find the trends in data. The independent variable used in Regression analysis is named Predictor variable. It offers data of an associated dependent variable regarding a particular outcome.
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Previous data for reference:
(a) Hence, the fitted linear regression model is: = 27.18 – 0.2976 x
For a carbonation depth of 25.0 mm, substitute x = 25 in the above equation.
= 27.18 – 0.2976 * (25) = 19.74 MPa.
(b) The t-test statistic for the slope has P-value 2.01347 × 10^–6 or approximately 0.
Decision rule using P-value: Reject H0 at significance level α, if P-value ≤ α. Otherwise, fail to reject H0.
Since P-value (≈ 0) < α (= 0.01), reject H0.
Thus, the regression is significant.
(c) The F-test statistic for the analysis of variance has P-value 2.01347 × 10^–6 or approximately 0. Since P-value (≈ 0) < α (= 0.01), reject H0. Thus, the regression is significant.
(d) For a carbonation depth of 25.0 mm, substitute x = 25 in the above equation.
= 27.18 – 0.2976 * (25) = 19.74 MPa.
From the regression output, the standard error is Se = 2.864.
t0.025,16 = 2.1199.
Thus, the 95% prediction interval for y when x = 25.0 mm is (13.4143, 26.06385).
It is given that the measured value of y when x = 25 is 12.2. Since, the value 12.2 does not fall in the prediction interval of y when x = 25, there is no sufficient evidence to conclude that the strength 12.2 MPa is not consistent with our model.
(e) The 95% confidence interval for y when x = 25 is obtained as (17.99005, 21.49783) from the calculation given below:
The predicted value of y when x = 35 is obtained as 16.76834 from the calculation given below:
Thus, the predicted value of y when x = 35 is 16.76834.
The 95% confidence interval for y when x = 35 is obtained as (15.32984, 18.20684) from the calculation given below:
(f)
From the residual plot, it is seen that none of the data points are far away from each other.
From the normal probability plot, it is seen that the data points are forming the straight line.
From these three plots it can be concluded that the data does not contain outliers.
Hence, it can be concluded that the model is adequate.
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