Question

Use the data set below to answer the questions. $$ \begin{array}{r|rrrrr} x & -2 & -1 & 0 & 1 & 2 \\ \hline y & 1 & 1 & 3 & 5 & 5 \end{array} $$ Find a \(90 \%\) prediction interval for some value of \(y\) to be observed in the future when \(x=1\).

Short answer

Answer: The 90% prediction interval for some value of y to be observed when x=1 is (0.889, 6.111).

Step-by-step solution

Calculate Mean and Variance

The mean of x: $$ \bar{x} = \frac{(-2) + (-1) + 0 + 1 + 2}{5} = 0 $$ The variance of x: $$ Var(x) = \frac{(-2)^2 + (-1)^2 + 0^2 + 1^2 + 2^2}{5} - \bar{x}^2 = 2 $$ The mean of y: $$ \bar{y} = \frac{1 + 1 + 3 + 5 + 5}{5} = 3 $$ The variance of y: $$ Var(y) = \frac{1^2 + 1^2 + 3^2 + 5^2 + 5^2}{5} - \bar{y}^2 = 2 $$

Calculate Covariance

The covariance of x and y: $$ Cov(x, y) = \frac{(-2)(1) + (-1)(1) + (0)(3) + (1)(5) + (2)(5)}{5} - \bar{x}\bar{y} = 2 $$

Find the Best-Fit Linear Regression Line

The slope (b) and y-intercept (a) of the best-fit line: $$ b = \frac{Cov(x, y)}{Var(x)} = \frac{2}{2} = 1 $$ $$ a = \bar{y} - b\bar{x} = 3 - 1\cdot0 = 3 $$ So, the best-fit line equation is: $$ y = a + bx = 3 + 1\cdot x $$

Calculate the Standard Error

The standard error of the observed values of y: $$ SE = \sqrt{\frac{1}{n-2}\sum(y_i - (a + bx_i))^2} $$ Substituting the values, we get: $$ SE = \sqrt{\frac{1}{5-2}[(1 - (3 + 1 \cdot (-2)))^2 + (1 - (3 + 1 \cdot (-1)))^2 + (3 - (3 + 1 \cdot 0))^2 + (5 - (3 + 1 \cdot 1))^2 + (5 - (3 + 1 \cdot 2))^2]} $$ $$ SE = \sqrt{\frac{1}{3}(2^2 + 1^2 + 0^2 + 1^2 + 0^2)} = 1 $$

Find the Appropriate t-Score

Using a t-distribution with 5-2 = 3 degrees of freedom and a 90% confidence interval, we find the t-score: $$ t_{0.05, 3} = 2.353 $$

Calculate the Width of the Prediction Interval

The width of the prediction interval when x=1: $$ W = t_{0.05, 3} \cdot SE \cdot \sqrt{1 + \frac{1}{n} + \frac{(x - \bar{x})^2}{\sum(x_i - \bar{x})^2}} $$ Substituting the values, we get: $$ W = 2.353 \cdot 1 \cdot \sqrt{1 + \frac{1}{5} + \frac{(1 - 0)^2}{(-2)^2 + (-1)^2 + 0^2 + 1^2 + 2^2}} $$ $$ W = 2.353 \cdot \sqrt{1 + 0.2 + 0.2} = 2.353 \cdot \sqrt{1.4} = 3.111 $$

Calculate the Confidence Interval

The 90% confidence interval for y when x=1: $$ CI = (a + bx \pm W) = (3 + 1 \cdot 1 \pm 3.111) = (0.889, 6.111) $$ The 90% prediction interval for some value of y to be observed in the future when x=1 is (0.889, 6.111).

Key concepts

Understanding Mean and Variance

When working with a dataset, two of the most fundamental statistical concepts are the mean and the variance. The mean of a set of numbers is simply the average, calculated by summing all the values and then dividing by the count of the values. It serves as a measure of the central tendency of the data.

The variance, on the other hand, measures the spread or dispersion of the data around the mean. It's calculated by averaging the squared differences from the mean. High variance indicates that data points are spread out widely around the mean, while low variance signifies that they are clustered closely.

In the given exercise, we first find the average values (means) of both x and y datasets, which are essential when determining the relationship between the two variables in subsequent steps, like linear regression.

Exploring Covariance

Covariance is a measure that expresses the extent to which two variables change together. If the values of one variable tend to be high when the values of the other variable are high, and similarly low when they are low, then the covariance will be positive, indicating a positive relationship between the variables.

On the contrary, if one variable tends to be high when the other is low, the covariance will be negative, reflecting a negative relationship. In the context of the exercise, calculating the covariance between x and y allows us to understand the direction of their linear relationship, which is a cornerstone for linear regression analysis.

Linear Regression Basics

Linear regression is a method used for modeling the relationship between a dependent variable and one (simple linear regression) or more (multiple linear regression) independent variables. The goal is to find the straight line, known as the regression line, that best fits the data.

The equation of a simple linear regression line is usually given by \( y = a + bx \), where \( a \) is the y-intercept and \( b \) is the slope of the line. The slope indicates how much y changes for a one-unit change in x. In the exercise, this line helps us predict the value of y for any given value of x, which is especially useful when creating prediction intervals.

Importance of Standard Error

The standard error (SE) measures the accuracy with which a sample represents a population. In regression, the standard error of the estimate is a measure of the variability of the actual data points from the estimated regression line. A smaller standard error implies that the observed data points are closer to the fitted line.

This concept is crucial in the calculation of prediction intervals, as it influences the width of the interval. A larger standard error would lead to a wider prediction interval, indicating less certainty about where the true value of y will fall for a given x. Understanding how to calculate and interpret the standard error is key to effective data analysis.

T-Distribution and Confidence Intervals

The t-distribution is a probability distribution that is symmetrical and bell-shaped like the normal distribution but has heavier tails. It is used instead of the normal distribution when the sample size is small. One of its main applications in statistics is to estimate the mean of a normally distributed population when the sample size is limited and the population standard deviation is unknown.

In the context of our exercise, the t-distribution is used to find the t-score, which then helps to define the width of a prediction or confidence interval. The creation of these intervals is a way to express uncertainty in predictions and estimates. The interval has a specified level of confidence, such as 90%, signifying that the interval is expected to contain the true parameter with that probability.