Question

A Chemical Experiment A chemist measured SET the peak current generated (in microamperes) DS1205 when a solution containing a given amount of nickel (in parts per billion) is added to a buffer: $$\begin{array}{cc}\hline x=\mathrm{Ni}(\mathrm{ppb}) & y=\text { Peak } \text { Current }(\mathrm{mA}) \\\\\hline 19.1 & .095 \\\38.2 & .174 \\\57.3 & .256 \\\76.2 & .348 \\\95 & .429 \\\114 & .500 \\\131 & .580 \\\150 & .651 \\\170 & .722 \\\\\hline\end{array}$$ a. Use the data entry method for your calculator to calculate the preliminary sums of squares and crossproducts, \(S_{x x}, S_{y},\) and \(S_{x y}\) b. Calculate the least-squares regression line. c. Plot the points and the fitted line. Does the assumption of a linear relationship appear to be reasonable? d. Use the regression line to predict the peak current generated when a solution containing 100 ppb of nickel is added to the buffer. e. Construct the ANOVA table for the linear regression.

Short answer

Based on the given dataset and calculations, the ANOVA table for linear regression can be constructed as follows: | Source of Variance | Degree of Freedom (df) | Sum of Squares (SS) | Mean Squares (MS) | F-value (F) | p-value (P) | |--------------------|-----------------------|---------------------|-------------------|-------------|-------------| | Regression | 1 | 0.9625 mA² | 0.9625 mA² | 132 | < α | | Error | 7 | 0.0363 mA² | 0.00726 mA² | N/A | N/A | | Total | 8 | 0.9988 mA² | N/A | N/A | N/A | This ANOVA table represents the variance in the dataset and helps us assess the goodness of fit and significance of the linear relationship between the nickel concentration and peak current.

Step-by-step solution

Calculate Preliminary Sums of Squares and Crossproducts

Calculate the sum of \(x\) values, sum of \(y\) values, sum of squared \(x\) and \(y\) values, and sum of the product of \(x\) and \(y\) values for the dataset. By calculating, - \(S_x = \sum x_i = 871\) - \(S_y = \sum y_i = 3.055\,\mathrm{mA}\) - \(S_{xx} = \sum x_i^2 = 115240.7\) - \(S_y = \sum y_i^2 = 1.420527\,\mathrm{mA^2}\) - \(S_{xy} = \sum x_i y_i = 500.661\,\mathrm{mA}\)

Compute the Least-Squares Regression Line

Based on the given data, we will calculate the coefficients of the least-squares regression line, which follows the equation: $$y = a + bx,$$ where: - \(y\) is the peak current in mA. - \(x\) is the concentration of nickel in ppb. - \(a\) is the intercept. - \(b\) is the slope. Using the formula: $$b = \frac{nS_{xy} - S_xS_y}{nS_{xx} - (S_x)^2},$$ and $$a = \frac{S_y - bS_x}{n},$$ where: - \(n\) is the number of data points. We can calculate \(a\) and \(b\). For this dataset - \(n = 9\), We can substitute our values, $$b = \frac{9(500.661) - (871)(3.055)}{9(115240.7) - (871)^2} \approx 0.00425\,\mathrm{mA/ppb},$$ $$a = \frac{3.055 - 0.00425(871)}{9} \approx -0.01786\,\mathrm{mA}.$$ Thus, the least-squares regression line can be represented as: $$y = -0.01786 + 0.00425x.$$

Plot the Points and Check for Linearity

By plotting each data point and our least-squares regression line on a graph, check whether the assumption of a linear relationship between nickel concentration and peak current seems reasonable. Upon visual inspection, if the data points are close to the fitted line, a linear relationship can be considered reasonable.

Predict the Peak Current for 100 ppb Nickel Concentration

Use the regression line equation to predict the peak current generated when the nickel concentration is 100 ppb. $$y = -0.01786 + 0.00425(100) \approx 0.40964\,\mathrm{mA}.$$ Hence, the predicted peak current for a 100 ppb nickel solution is approximately 0.40964 mA.

Construct the ANOVA Table for Linear Regression

For constructing the ANOVA table for this linear regression, compute the sums of squares for Regression (SSR), Error (SSE), and Total (SST). These values are used in dividing variation in data into different parts. The ANOVA table consists of the following information: - Source of Variance (Regression, Error, Total) - Degree of Freedom (df) - Sum of Squares (SS) - Mean Squares (MS) - F-value (F, only for Regression) - p-value (P, only for Regression) Use the formulas: - $$SSR = b^2\frac{S_{xx}}{n}$$ - $$SST = S_y^2 - (\frac{S_y^2}{n})$$ - $$SSE = SST - SSR$$ - $$MSR = \frac{SSR}{df_{Regression}}$$ - $$MSE = \frac{SSE}{df_{Error}}$$ - $$F = \frac{MSR}{MSE}$$ On calculating, - \(SSR = 0.9625\,\mathrm{mA^2}\) - \(SST = 0.9988\,\mathrm{mA^2}\) - \(SSE = 0.0363\,\mathrm{mA^2}\) - \(MSR = 0.9625\,\mathrm{mA^2}\) - \(MSE = 0.00726\,\mathrm{mA^2}\) - \(F = 132\) Put all the calculated values neatly into an ANOVA table.

Key concepts

Sums of Squares

In the realm of statistics, particularly when we deal with linear regression, 'sums of squares' plays a pivotal role in quantifying variations. Essentially, it breaks down the variance in observed data into components that can be attributed to different sources. For students, think of it as a method to express how spread out data points are around the mean or a fitted line.

There are three primary 'sums of squares' that you’ll often encounter: Total Sum of Squares (SST), Regression Sum of Squares (SSR), and Error Sum of Squares (SSE). SST measures the total variance in the response variable (in our case, the peak current produced). SSR tells us how much of that variance is explained by the linear relationship with the explanatory variable (nickel concentration). Meanwhile, SSE quantifies the variance that is not explained by the regression model, or in simpler terms, it's the discrepancy between the observed and predicted values.

To calculate these 'sums of squares', you start with SST. Using the formula \(SST = \sum(y_i - \bar{y})^2\), where \(y_i\) is the observed value and \(\bar{y}\) is the mean of the response variable. SSR is then calculated by \(SSR = b^2(\sum(x_i - \bar{x})^2)\) if \(b\) is the slope of the regression line. Lastly, we derive SSE by subtracting SSR from SST, represented by the formula \(SSE = SST - SSR\).

This compartmentalization of variance is crucial for determining the goodness-of-fit for our regression model, showing us just how much of the changes in our dependent variable can be explained by its relationship with the independent variable.

Least-Squares Regression Line

The least-squares regression line is a fundamental concept in statistics for predictive analysis. When you look at a scatterplot of bivariate data, this line shows the best-fit trendline through the points, minimizing the sum of the squares of the vertical distances of the points from the line.

In the chemist's experiment, we're searching for the best line that describes how peak current (\(y\)) is associated with the nickel concentration (\(x\)). To find this line, we adopt the classic formula \(y = a + bx\), where \(a\) is the y-intercept and \(b\) represents the slope. The slope tells us how much we expect the dependent variable to change for each one-unit change in the independent variable.

To obtain the slope (\(b\)), one would use the formula \(b = \frac{nS_{xy} - S_xS_y}{nS_{xx} - (S_x)^2}\). The y-intercept (\(a\)) is computed from \(a = \frac{S_y - bS_x}{n}\). By inputting the calculated slope and intercept into the regression line formula, we end up with a precise equation that can be used for predictions, like estimating future peak currents for given nickel concentrations.

By engaging with the least-squares regression line, students will grasp how statistical modelling translates into a powerful predictive tool, enabling us to anticipate outcomes and understand relationships within the data we analyze.

ANOVA Table

The Analysis of Variance (ANOVA) table is an indispensable statistical tool used to assess the degree to which a linear model explains the variance in observed data. It provides a systematic layout to help interpret the regression analysis results. Through the ANOVA table, we can understand more deeply the distribution and sources of variation in our data.

The table consists of several components. First, there's the 'Source of Variation' which includes Regression and Error, essentially breaking down the total variance from the 'sums of squares' concept. Each source has its own 'Sum of Squares (SS)' which measures the variation it accounts for. For students, it's like separating the total fluctuations into parts that are caused by the model and parts that are just random fluctuations.

Next, we tally the 'Degrees of Freedom' for each variance source, followed by the 'Mean Square' values which are the SS divided by their respective degrees of freedom. The 'F-value' is the ratio of these mean squares, and it’s used to determine whether the relationship between the response and predictor variables is statistically significant. High F-values typically indicate that the regression model provides a better explanation of variability compared to the null hypothesis, which suggests no linear relationship.

By creating an ANOVA table and interpreting its results, students can quantitatively support the usefulness (or lack thereof) of the linear regression model, thus proving its significance in the study. It's a gateway to understanding the confidence we have in our predictive models and a way to validate our insights mathematically.