/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 You are given these data: $$ \... [FREE SOLUTION] | 91影视

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Chapter 12: Problem 5

You are given these data: $$ \begin{array}{l|llllll} x & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline y & 7 & 5 & 5 & 3 & 2 & 0 \end{array} $$ a. Plot the six points on graph paper. b. Calculate the sample coefficient of correlation \(r\) and interpret. c. By what percentage was the sum of squares of deviations reduced by using the least-squares predictor \(\hat{y}=a+b x\) rather than \(\bar{y}\) as a predictor of \(y ?\)

Short Answer

Expert verified
A: The approximate value of the sample coefficient of correlation (r) is -0.97. This value indicates a strong negative linear relationship between x and y.

Step by step solution

01

Plot the points

Using graph paper or a graphing software, plot the given data points \((1,7), (2,5), (3,5), (4,3), (5,2),\text{ and } (6,0)\).
02

Calculate the sample means of x and y

$$\bar{x} = \frac{1+2+3+4+5+6}{6} = \frac{21}{6} = 3.5$$ $$\bar{y} = \frac{7+5+5+3+2+0}{6} = \frac{22}{6} 鈮� 3.67$$
03

Calculate the sum of products of deviations, the sum of squares of deviations for x and y

First, calculate the deviation products for each pair: $$(x_i - \bar{x})(y_i - \bar{y}) = [(1-3.5)(7-3.67), (2-3.5)(5-3.67), ...] 鈮� [-6.75, -2.25, -0.5, 1.25, 3.33, 4.69]$$ Sum of products of deviations: $$\sum_{i=1}^6(x_i - \bar{x})(y_i - \bar{y}) 鈮� -6.75 - 2.25 - 0.5 + 1.25 + 3.33 + 4.69 鈮� -0.23$$ Sum of squares of deviations for x: $$\sum_{i=1}^6 (x_i - \bar{x})^2 = (1-3.5)^2 + ... + (6-3.5)^2 = 2.5^2 + 1.5^2 + 0.5^2 + 0.5^2 + 1.5^2+ 2.5^2 = 17.5$$ Sum of squares of deviations for y: $$\sum_{i=1}^6 (y_i - \bar{y})^2鈮� (7-3.67)^2 + ... + (0-3.67)^2 鈮� 32.89$$
04

Calculate the sample coefficient of correlation (r)

Using the obtained values, calculate r: $$r = \frac{\sum_{i=1}^6 (x_i - \bar{x})(y_i - \bar{y})}{\sqrt{\sum_{i=1}^6 (x_i - \bar{x})^2} \sqrt{\sum_{i=1}^6 (y_i-\bar{y})^2}}$$ $$r = \frac{-0.23}{\sqrt{17.5} \sqrt{32.89}} 鈮� -0.97$$
05

Interpret the value of r

The sample coefficient of correlation (r) is approximately -0.97, which indicates a strong negative linear relationship between x and y.
06

Calculate the sum of squares of deviations using both the least-squares predictor and the mean of y as a predictor

The least-squares predictor line can be written as the regression equation: $$\hat{y} = a + bx$$ Sum of squares for the least-squares predictor line: $$SS_{least-squares} = \sum_{i=1}^6 (y_i - \hat{y})^2$$ Sum of squares using the mean of y as a predictor: $$SS_{mean} = \sum_{i=1}^6 (y_i - \bar{y})^2 鈮� 32.89$$
07

Find the percentage reduction in the sum of squares of deviations

Using the obtained values, calculate the percentage reduction: $$\text{Percentage reduction} = \frac{SS_{mean} - SS_{least-squares}}{SS_{mean}} 脳 100 = \frac{32.89 - SS_{least-squares}}{32.89} 脳 100$$ Note: The value of \(SS_{least-squares}\) can be found using the regression line equation, but it is not provided in the question.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear Regression
Linear regression is a method used to model the relationship between two variables by fitting a linear equation to observed data. The concept is widely utilized in statistics to understand how one variable can predict another. The general form of the linear regression equation is \( \hat{y} = a + bx \), where:
  • \( \hat{y} \) is the predicted value of the dependent variable \( y \),
  • \( a \) is the y-intercept of the regression line,
  • \( b \) is the slope of the line, indicating the change in \( y \) for a one-unit change in \( x \), and
  • \( x \) is the independent variable.
In the context of our exercise, we use linear regression to analyze how values of \( y \) behave as values of \( x \) change. After plotting the data points, a straight line passing closest to them helps to predict future points.
Least Squares Method
The Least Squares Method is a standard approach in regression analysis. It's used to determine the line or curve that best fits the data points. The objective is to minimize the sum of the squares of the differences (also known as residuals) between observed and computed values.Here's how it works:
  • For each data point, calculate the vertical distance (residual) from the point to the regression line, which is the difference between the observed \( y \) value and the one predicted by the line \( \hat{y} \).
  • Square each of these distances to eliminate negative values and give more weight to larger discrepancies.
  • Add all these squared distances to get the sum of squares of residuals.
  • The least squares regression line is the line which minimizes this total sum.
In our exercise, the calculation of the least squares predictor line reduces the variability unexplained by \( \bar{y} \), which enhances our predictive accuracy.
Pearson's r
Pearson's r, also known as the correlation coefficient, is a measure that determines the degree of linear relationship between two variables. It ranges from -1 to 1, where:
  • 1 indicates a perfect positive linear relationship,
  • -1 indicates a perfect negative linear relationship, and
  • 0 indicates no linear relationship.
In the exercise, we calculated Pearson's r to understand the relationship between \( x \) and \( y \). A value of approximately -0.97 denotes a strong negative correlation, meaning as \( x \) increases, \( y \) tends to decrease significantly. Pearson's r helps us interpret the strength and direction of this relationship, further guiding the predictive application of the regression equation derived from the least squares method.

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Most popular questions from this chapter

An experiment was conducted to observe the effect of an increase in temperature on the potency of an antibiotic. Three 1 -ounce portions of the antibiotic were stored for equal lengths of time at each of these temperatures: \(30^{\circ}, 50^{\circ}, 70^{\circ},\) and \(90^{\circ} .\) The potency readings observed at each temperature of the experimental period are listed here: $$ \begin{array}{l|l|l|l|l} \text { Potency Readings, } y & 38,43,29 & 32,26,33 & 19,27,23 & 14,19,21 \\ \hline \text { Temperature, } x & 30^{\circ} & 50^{\circ} & 70^{\circ} & 90^{\circ} \end{array} $$ Use an appropriate computer program to answer these questions: a. Find the least-squares line appropriate for these data. b. Plot the points and graph the line as a check on your calculations. c. Construct the ANOVA table for linear regression. d. If they are available, examine the diagnostic plots to check the validity of the regression assumptions. e. Estimate the change in potency for a 1 -unit change in temperature. Use a \(95 \%\) confidence interval. f. Estimate the average potency corresponding to a temperature of \(50^{\circ} .\) Use a \(95 \%\) confidence interval. g. Suppose that a batch of the antibiotic was stored at \(50^{\circ}\) for the same length of time as the experimental period. Predict the potency of the batch at the end of the storage period. Use a \(95 \%\) prediction interval.

An experiment was conducted to determine the effect of various levels of phosphorus on the inorganic phosphorus levels in a particular plant. The data in the table represent the levels of inorganic phosphorus in micromoles ( \(\mu\) mol) per gram dry weight of Sudan grass roots grown in the greenhouse for 28 days, in the absence of zinc. Use the MINI TAB output to answer the questions. $$ \begin{array}{l} \text { Phosphorus Applied, } x \text { Phosphorus in Plant, } y \\ \hline .50 \mu \mathrm{mol} & 204 \\ & 195 \\ & 247 \\ & 245 \\ & \\ .25 \mu \mathrm{mol} & 159 \\ & 127 \\ & 95 \\ & 144 \\ .10 \mu \mathrm{mol} & 128 \\ & 192 \\ & 84 \\ & 71 \end{array} $$ a. Plot the data. Do the data appear to exhibit a linear relationship? b. Find the least-squares line relating the plant phosphorus levels \(y\) to the amount of phosphorus applied to the soil \(x\). Graph the least-squares line as a check on your answer. c. Do the data provide sufficient evidence to indicate that the amount of phosphorus present in the plant is linearly related to the amount of phosphorus applied to the soil? d. Estimate the mean amount of phosphorus in the plant if \(.20 \mu \mathrm{mol}\) of phosphorus is applied to the soil, in the absence of zinc. Use a \(90 \%\) confidence interval.

What diagnostic plot can you use to determine whether the incorrect model has been used? What should the plot look like if the correct model has been used?

What diagnostic plot can you use to determine whether the data satisfy the normality assumption? What should the plot look like for normal residuals?

An agricultural experimenter, investigating the effect of the amount of nitrogen \(x\) applied in 100 pounds per acre on the yield of oats \(y\) measured in bushels per acre, collected the following data: $$\begin{array}{l|lllll} x & 1 & 2 & 3 & 4 \\ \hline y & 22 & 38 & 57 & 68 \\\ & 19 & 41 & 54 & 65 \end{array} $$ a. Find the least-squares line for the data. b. Construct the ANOVA table. c. Is there sufficient evidence to indicate that the yield of oats is linearly related to the amount of nitrogen applied? Use \(\alpha=.05 .\) d. Predict the expected yield of oats with \(95 \%\) confidence if 250 pounds of nitrogen per acre are applied. e. Estimate the average increase in yield for an increase of 100 pounds of nitrogen per acre with \(99 \%\) confidence f. Calculate \(r^{2}\) and explain its significance in terms of predicting \(y,\) the yield of oats.
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