T.Linebarger_Analysis with Correlation and Regression031524

Magnitude Depth 2.74 7.5 0.74 2.5 0.64 14.0 Calculations 3.20 15.5 r 0.29966 0.70 3.0 slope 2.025822 2.20 2.4 y-intercept 5.990001 1.98 14.4 0.64 5.7 2.04 6.1 0.50 7.1 2.40 17.2 1.32 8.7 3.17 9.3 0.90 12.3 1.76 9.8 0.98 7.4 1.24 17.1 0.01 10.3 0.65 5.0 1.46 19.1 1.62 12.7 1.83 4.7 0.99 6.8 1.56 6.0 0.40 14.6 1.28 4.9 3.00 19.1 1.00 9.9 0.54 16.1 1.25 4.6 0.92 5.1 1.11 16.3 0.79 14.0 0.79 4.2 1.56 5.4 1.01 5.9 3.01 15.6 2.50 7.9 1.79 16.4 1.25 15.4 1.49 4.9 0.84 8.1 1.42 7.5 1.00 14.1

1.25 11.1 1.42 14.8 1.27 4.9 1.45 7.1 0.40 3.1 1.39 5.3 2.40 6.9 0.98 10.1 0.34 3.2 1.44 4.8 1.20 3.6 0.55 1.6 0.60 1.8 1.82 4.4 0.31 1.0 1.16 3.5

alpha 60 n 0.05 r 0.29966 CV of r 0.214383 2a. Calculate the value of the linear correlation coefficient r and the critical value of r using α = 0.05. The correlation coefficient r between two variables can be calculated in Excel using the CORREL function y r=0.29966 For the critical value (CV) of r, I used a t-distribution based approach. Given a sample size n and a significa α/2, n-2). =(T.INV(1-C14,C13-2))/SQRT((T.INV(1-C14,C13-2))^2+C13-2) This formula calculates the boundary of r for our significance level; if our calculated r is beyond this bound CV of r yielded 0.214383 2b. Determine whether there is sufficient evidence to support the claim of a linear correlation betwe magnitudes and the depths from the earthquakes. There is a correlation coefficient of r=0.29966 and a critical value (CV) of r=0.214383. In this case, since 0.29966 >0.214383 , there is sufficient evidence at the ∣ ∣ ∣ α=0.05 level to support the claim of a linear c between the magnitudes and the depths of the earthquakes. This means that the probability of observ correlation by chance, when there is no actual correlation, is less than 5%.

slope 2.025822 y-intercept 5.990001 R square 0.089796 3a. Find the regression equation. Let the predictor ( x ) variable be the magnitude. Identify the slope and =SLOPE(Depths ,Magnitude)=2.0258 =INTERCEPT (Depths ,Magnitude) =5.99 The regression equation represents the relationship between a predictor variable (x) and a response varia The slope of the regression line is found using =SLOPE(y_range, x_range) , which in this case is =SLOPE(De much the response variable (y) is expected to increase when the predictor variable (x) increases by one un The y-intercept of the regression line is found using =INTERCEPT(y_range, x_range) , which in this case is = is the value of the response variable (y) when the predictor variable (x) is zero. The regression equation that predicts the depth of an earthquake (y) based on its magnitude (x) would be Depth = 2.0258 × Magnitude + 5.99 The slope tells us that for each unit increase in magnitude, the depth is expected to increase by approxima predicts a positive depth when the magnitude is zero. 3b. Is the equation a good model? The coefficient of determination is 0.089796 The R2 value of 0.089796 means that approximately 8.98% of the variance in the dependent variable (i the earthquakes) can be explained by the independent variable (the magnitude of the earthquakes). Th suggests that the linear regression model does not fit the data well. In other words, the magnitude of t explains a small part of the variation in their depth. 3c. What would be the best predicted depth of an earthquake with a magnitude of 1.26? Include the Using the regression equation: Depth = 2.0258 × Magnitude + 5.99 For an earthquake with a magnitude of 1.26, the best-predicted depth would be: Predicted Depth = (2.0258×1.26) + 5.99 =8.5425 Therefore, the best predicted depth for an earthquake of magnitude 1.26 is approximately 8.5425 units

d the y-intercept within your regression equation. able (y). epths, Magnitude) and is equal to 2.0258. The slope indicates how nit. =INTERCEPT(Depths, Magnitude) and is equal to 5.99. The y-intercept e: ately 2.0258 units. The positive y-intercept suggests that the model in your case, the depth of his relatively low R2 value the earthquakes only e correct units. s.