Math Problem Statement

The following sample observations were randomly selected: (Round the final answer to 4 decimal places.)

x:

6

12

16

8

6

18

4

5

y:

13

7

25

14

12

17

10

7

a. Determine the standard error of estimate.

b. Suppose a large sample is selected (instead of just eight). About 95% of the predictions would be between what two values? Assuming that the Standard Error of Estimate does not change. Use z = 2.

y' ±

Solution

Let's start by solving the problem step by step.

Part (a): Determine the Standard Error of Estimate (SEE)

The standard error of estimate (SEE) is a measure of the accuracy of predictions made with a regression line. It is calculated using the following formula:

SEE=(yiy^i)2n2SEE = \sqrt{\frac{\sum (y_i - \hat{y}_i)^2}{n - 2}}

Where:

  • yiy_i = observed values of yy
  • y^i\hat{y}_i = predicted values of yy using the regression equation
  • nn = number of observations

To calculate the SEE, we need to first determine the regression equation (y^=a+bx\hat{y} = a + bx) and then use it to find the predicted values (y^i\hat{y}_i).

Steps:

  1. Calculate the means of x and y: xˉ=xin,yˉ=yin\bar{x} = \frac{\sum x_i}{n}, \quad \bar{y} = \frac{\sum y_i}{n}
  2. Find the slope (b) of the regression line: b=(xixˉ)(yiyˉ)(xixˉ)2b = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2}
  3. Find the y-intercept (a) of the regression line: a=yˉbxˉa = \bar{y} - b\bar{x}
  4. Use the regression equation to find the predicted values y^i\hat{y}_i.
  5. Calculate the SEE using the formula provided.

Let's go through these calculations now.The Standard Error of Estimate (SEE) is approximately 4.83174.8317.

Part (b): 95% Prediction Interval

For a large sample, the 95% prediction interval can be estimated using the formula:

Prediction Interval=y^±Z×SEE\text{Prediction Interval} = \hat{y} \pm Z \times \text{SEE}

Given that Z=2Z = 2 for a 95% confidence level:

y^±2×SEE\hat{y} \pm 2 \times \text{SEE}

Substituting the value of SEE we found:

y^±2×4.8317\hat{y} \pm 2 \times 4.8317

Let's calculate this range.The 95% prediction interval suggests that about 95% of the predictions would fall between y^±9.6634\hat{y} \pm 9.6634, assuming the Standard Error of Estimate does not change.

If you have any further questions or would like more details, feel free to ask!


Here are some related questions you might find interesting:

  1. How do you calculate the correlation coefficient rr from a data set?
  2. What does a high or low SEE indicate about a regression model?
  3. How is the coefficient of determination R2R^2 related to the SEE?
  4. What assumptions are made in linear regression analysis?
  5. How can outliers affect the SEE and regression line?
  6. What is the difference between SEE and standard deviation?
  7. How do confidence intervals differ from prediction intervals?
  8. How would the prediction interval change if the sample size increased significantly?

Tip: Always check the residuals to ensure that they are randomly distributed, which validates the assumptions of linear regression.

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Math Problem Analysis

Mathematical Concepts

Regression Analysis
Standard Error of Estimate
Prediction Interval

Formulas

SEE formula: SEE = sqrt(sum((yi - yhati)^2) / (n - 2))
Prediction Interval: y' ± Z * SEE

Theorems

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Suitable Grade Level

Advanced High School