jduwana_deliv6_121423

Magnitude Depth 3.03 7.5 0.74 2.5 0.64 14.0 Calculations 1.20 15.5 r 0.1147 0.70 3.0 slope 0.9111 2.20 2.4 y-intercept 7.5706 1.98 14.4 0.64 5.7 1.22 6.1 0.50 7.1 1.62 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 0.83 19.1 1.34 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 1.44 5.4 1.00 5.9 2.24 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

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r 0.11227 CV of r 0.21438 2a. Calculate the value of the linear correlation coefficient r and the critical value of r using α = 0.05. Show your calculations for r below. Include an explanation on how you found the critical value. Linear correlation coefficient R can be found in two ways: CORREL(): Write "=CORREL()" in the desired cell. Pass the two data ranges of Depth and Magnitude as two different arguments of the CORREL() function Press Enter. Write "=COVAR()" in the desired cell. Pass the two data ranges of Depth and Magnitude as two different arguments of the CORREL() function the covariance. Write "=STDEV()" in the desired cell. Pass the data range of Depth as an argument of the STDEV() function. This will give us the standard dev Write "=STDEV()" in the desired cell. Pass the data range of Magnitude as an argument of the STDEV() function. This will give us the standard Magnitude. The formula for correlation coefficient is: ρ ( X , Y )= σX σY COV ( X , Y ) Enter the formula in the desired cell. Press Enter. The value for the correlation coefficient found was 0.11227. The critical value for the correlation coefficient can be found by using the formula: rc = tc 2+ dftc 2 t c can be found from the Excel formula: = T.INV(1-α ,df) =(T.INV(1-α ,df))/SQRT((T.INV(1-α ,df))^2+df) 2b. Determine whether there is sufficient evidence to support the claim of a linear correlation betwee magnitudes and the depths from the earthquakes. Explain your answer. From our above calculations, we can say that there is not much evidence suggesting a correlation betwe variables. Firstly, the scatterplot is too random. The points seem to be very spread out and there is no definitive sh whole chart. Secondly, the Correlation Coefficient is 0.11227. This implies that if correlation even exists, weak but positive. Thirdly, when we compare this to the critical value of r, we can see that |0.11227| < 0 implies that our correlation coefficient is not statistically significant. Hence, there is not much evidence s

n. n. This will give us viation of Depth. d deviation of en the een the two hape to the , it is very very 0.21438. This suggesting a

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3a. Find the regression equation. Let the predictor ( x ) variable be the m your regression equation. Show your calculations below. Describe how you constructed the regress The linear regression model attempts to draw a relationship between a the explanatory variable) and a quantitative dependent variable (which modeled by the equation: slope= 0.9075 Intercept= 7.5761 y=mx+b Where: y = the outcome. x = the explanatory variable. m = the slope (or rate of change/coefficient) b = the y-intercept (the value y will assume if x=0/constant). where B1 is slope and B2 is intercept. = SLOPE(depth,magnitude) = INTERCEPT(depth, magnitude) Data > Data Analysis > Regression > Select X and Y > OK p value 4.35044E-07 he independent variable(s). rrect units.

magnitude. Identify the slope and the y-intercept within sion equation. quantitative independent variable (which is also known as is also known as the outcome). This line is mathematically