The accompanying data on x = current density (mA/cm2) and y = rate of deposition (µm/min) appeared in an article. Do you agree with the claim by the article's author that "a linear relationship was obtained from the tin-lead rate of deposition as a function of current density"?
x	20	40	60	80
y	0.29	1.35	1.81	2.22
Find the value of r2. (Round your answer to three decimal places.)
r2 =

To find the coefficient of determination (\( r^2 \)), we can follow these steps:

1. **Calculate the means of \( x \) and \( y \)**:
\[
\bar{x} = \frac{20 + 40 + 60 + 80}{4} = \frac{200}{4} = 50
\]
\[
\bar{y} = \frac{0.29 + 1.35 + 1.81 + 2.22}{4} = \frac{5.67}{4} = 1.4175
\]

2. **Calculate the slope (\( b \)) and intercept (\( a \)) of the regression line \( y = a + bx \)**:
\[
b = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2}
\]

First, compute the individual terms:
\[
\sum (x_i - \bar{x})(y_i - \bar{y}) = (20 - 50)(0.29 - 1.4175) + (40 - 50)(1.35 - 1.4175) + (60 - 50)(1.81 - 1.4175) + (80 - 50)(2.22 - 1.4175)
\]
\[
= (-30)(-1.1275) + (-10)(-0.0675) + (10)(0.3925) + (30)(0.8025)
\]
\[
= 33.825 + 0.675 + 3.925 + 24.075 = 62.5
\]

\[
\sum (x_i - \bar{x})^2 = (20 - 50)^2 + (40 - 50)^2 + (60 - 50)^2 + (80 - 50)^2
\]
\[
= (-30)^2 + (-10)^2 + (10)^2 + (30)^2
\]
\[
= 900 + 100 + 100 + 900 = 2000
\]

\[
b = \frac{62.5}{2000} = 0.03125
\]

\[
a = \bar{y} - b\bar{x} = 1.4175 - (0.03125 \times 50) = 1.4175 - 1.5625 = -0.145
\]

So, the regression line is:
\[
y = -0.145 + 0.03125x
\]

3. **Calculate the total sum of squares (\( SS_{\text{total}} \)), the regression sum of squares (\( SS_{\text{reg}} \)), and the residual sum of squares (\( SS_{\text{res}} \))**:

\[
SS_{\text{total}} = \sum (y_i - \bar{y})^2 = (0.29 - 1.4175)^2 + (1.35 - 1.4175)^2 + (1.81 - 1.4175)^2 + (2.22 - 1.4175)^2
\]
\[
= (-1.1275)^2 + (-0.0675)^2 + (0.3925)^2 + (0.8025)^2
\]
\[
= 1.2713 + 0.0046 + 0.1541 + 0.6440 = 2.0739
\]

\[
SS_{\text{reg}} = \sum (\hat{y_i} - \bar{y})^2
\]

Where \(\hat{y_i} = a + bx_i\):

\[
\hat{y_1} = -0.145 + 0.03125 \times 20 = 0.48
\]
\[
\hat{y_2} = -0.145 + 0.03125 \times 40 = 1.11
\]
\[
\hat{y_3} = -0.145 + 0.03125 \times 60 = 1.74
\]
\[
\hat{y_4} = -0.145 + 0.03125 \times 80 = 2.37
\]

\[
SS_{\text{reg}} = (0.48 - 1.4175)^2 + (1.11 - 1.4175)^2 + (1.74 - 1.4175)^2 + (2.37 - 1.4175)^2
\]
\[
= (-0.9375)^2 + (-0.3075)^2 + (0.3225)^2 + (0.9525)^2
\]
\[
= 0.8789 + 0.0946 + 0.104 + 0.9073 = 1.9848
\]

\[
SS_{\text{res}} = SS_{\text{total}} - SS_{\text{reg}} = 2.0739 - 1.9848 = 0.0891
\]

4. **Calculate \( r^2 \)**:
\[
r^2 = \frac{SS_{\text{reg}}}{SS_{\text{total}}} = \frac{1.9848}{2.0739} = 0.957
\]

Thus, the \( r^2 \) value is approximately 0.957 (rounded to three decimal places).

Therefore, the claim that "a linear relationship was obtained from the tin-lead rate of deposition as a function of current density" is supported by this \( r^2 \) value, indicating a very strong linear relationship.

Do you want details or have any questions? Here are some questions you might ask next:

1. How do you interpret the slope and intercept in this context?
2. What is the significance of the \( r^2 \) value?
3. Can you show me a plot of the data and the regression line?
4. How would you test if the linear model is statistically significant?
5. What other types of regression models could be considered for this data?

**Tip:** When interpreting the slope of a regression line, it indicates how much the dependent variable (y) changes for a one-unit increase in the independent variable (x).

Discover how to calculate the coefficient of determination (r²) for the tin-lead rate of deposition based on current density values. This solution includes data points: current density (20, 40, 60, 80 mA/cm²) and deposition rates (0.29, 1.35, 1.81, 2.22 µm/min), illustrating a strong linear relationship with r² approximately 0.957.

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