Question

How good are you EX1212 at estimating? To test a subject's ability to estimate sizes, he was shown 10 different objects and asked to estimate their length or diameter. The object was then measured, and the results were recorded in the table below. $$ \begin{array}{lrr} \text { Object } & \text { Estimated (inches) } & \text { Actual (inches) } \\\ \hline \text { Pencil } & 7.00 & 6.00 \\ \text { Dinner plate } & 9.50 & 10.25 \\ \text { Book 1 } & 7.50 & 6.75 \\ \text { Cell phone } & 4.00 & 4.25 \\ \text { Photograph } & 14.50 & 15.75 \\ \text { Toy } & 3.75 & 5.00 \\ \text { Belt } & 42.00 & 41.50 \\ \text { Clothespin } & 2.75 & 3.75 \\ \text { Book 2 } & 10.00 & 9.25 \\ \text { Calculator } & 3.50 & 4.75 \end{array} $$ a. Find the least-squares regression line for predicting the actual measurement as a function of the estimated measurement. b. Plot the points and the fitted line. Does the assumption of a linear relationship appear to be reasonable?

Short answer

Answer: The least-squares regression line for predicting actual measurements based on estimated measurements is Y = 0.67 + 0.89X. The correlation coefficient of 0.927 indicates a strong linear relationship between the estimated and actual measurements, making the assumption of a linear relationship reasonable.

Step-by-step solution

Find the least-squares regression line

The least-squares regression line is given by the formula \(Y = a + bX\), where X is the estimated size, Y is the actual size, a represents the Y-intercept, and b represents the slope. To find the values of a and b, we need to calculate the mean values of X and Y and the standard deviations of X and Y. Mean of X: \(\bar{X} = \frac{\sum{X}}{n}\) Mean of Y: \(\bar{Y}= \frac{\sum{Y}}{n}\) Standard deviation of X: \(s_{X} = \sqrt{\frac{\sum{(X - \bar{X})^2}}{n-1}}\) Standard deviation of Y: \(s_{Y} = \sqrt{\frac{\sum{(Y - \bar{Y})^2}}{n-1}}\) Slope (b): \(b = r \frac{s_{Y}}{s_{X}}\), where r is the correlation coefficient. Y-Intercept (a): \(a = \bar{Y} - b\bar{X}\) Using the given data, we can calculate the means, standard deviations, and other values.

Calculate mean and standard deviations:

Mean of Estimated (X): \(\bar{X} = \frac{1}{10} \sum{X} = 6.05\) Mean of Actual (Y): \(\bar{Y} = \frac{1}{10} \sum{Y} = 6.225\) Standard deviation of Estimated (X): \(s_{X} = \sqrt{\frac{\sum{(X - 6.05)^2}}{9}} = 2.54\) Standard deviation of Actual (Y): \(s_{Y} = \sqrt{\frac{\sum{(Y - 6.225)^2}}{9}} = 2.46\) Next, we can find the correlation coefficient (r).

Calculate the correlation coefficient:

Correlation Coefficient (r) = \(\frac{\sum{(X - \bar{X})(Y - \bar{Y})}}{\sqrt{\sum{(X - \bar{X})^2}\sum{(Y - \bar{Y})^2}}} = 0.927\) Now, we can calculate the slope (b) and Y-intercept (a).

Calculate the slope and Y-intercept:

Slope (b) = \(r \frac{s_{Y}}{s_{X}} = 0.927 \times \frac{2.46}{2.54} = 0.89\) Y-Intercept (a): \(a = \bar{Y} - b\bar{X} = 6.225 - 0.89 \times 6.05 = 0.67\) The least-squares regression line is: \(Y = 0.67 + 0.89X\)

Plot the points and the fitted line

To check if the assumption of a linear relationship appears to be reasonable, we need to plot the points and the fitted line. This can be done using any graphing or plotting software. Once the points and the line are plotted, if the data points are closely following the line, the assumption of a linear relationship can be considered reasonable. However, if the points are scattered far from the line, the linear relationship assumption might not be warranted. In this case, the correlation coefficient of 0.927 suggests a strong linear relationship between the estimated and actual measurements, making the assumption reasonable.

Key concepts

Correlation Coefficient

The correlation coefficient, commonly denoted as \( r \), is a statistical measure that describes the strength and direction of a linear relationship between two variables. In simpler terms, it tells us how closely the estimated values follow the actual measurements in a straight line.

- The value of \( r \) ranges from \(-1\) to \(1\). Where \( r = 1 \) indicates a perfect positive linear relationship, \( r = -1 \) indicates a perfect negative linear relationship, and \( r = 0 \) means no linear relationship.
- In our example, we calculated the correlation coefficient as \( r = 0.927 \), which is close to \( 1 \). This tells us that there is a strong, positive linear relationship between the estimated and actual measurements.

It’s important to note that while a high correlation suggests a strong relationship, it does not imply causation. Factors other than estimation accuracy could be influencing the measurements.

Linear Relationship

A linear relationship between two variables suggests that a change in one variable is directly proportional to a change in the other variable. When plotted on a graph, this relationship appears as a straight line.

In our context, we're exploring the linear relationship between the estimated sizes and their actual measurements. To do this, we've used the least-squares regression line, represented in the formula as \( Y = a + bX \), where:

• \( Y \) is the actual measurement
• \( X \) is the estimated measurement
• \( a \) is the Y-intercept
• \( b \) is the slope of the line

After calculating, the regression equation for our data is \( Y = 0.67 + 0.89X \).

This implies that for every additional unit of estimated size, the actual size increases by 0.89 units, given the strong linear relationship found.

Mean and Standard Deviation

Mean and standard deviation are statistical tools that help describe the data distribution, which is essential for understanding relationships between variables.

**Mean** is the average value of a dataset and gives a central tendency of the data:

• For estimated measurements, the mean \( \bar{X} \) was 6.05 inches.
• For actual measurements, the mean \( \bar{Y} \) was 6.225 inches.

Calculating the mean provides a baseline for measuring variance.

**Standard deviation** measures the spread of data points around the mean. It tells us how much variation or dispersion exists from the average:

• The standard deviation for estimated sizes \( s_{X} \) was 2.54 inches.
• The standard deviation for actual sizes \( s_{Y} \) was 2.46 inches.

A smaller standard deviation indicates that the data points are close to the mean, while a larger standard deviation suggests more variability.
By understanding mean and standard deviation, we lay the groundwork for further analysis, such as calculating the correlation coefficient and the regression line.