Prediction Intervals - Excel: Videos & Practice Problems

Prediction intervals extend regression analysis by providing a confidence interval for predicted y values, incorporating the margin of error calculated using the critical t-value, standard error of the estimate ($s_e$), and variability of x. Valid predictions require strong linear correlation (high R²) and x values within the data range. This method enhances inferential statistics by quantifying uncertainty in regression predictions, crucial for accurate forecasting and decision-making in business analytics.

Prediction Intervals - Excel

Regression analysis allows us to predict values of a dependent variable y based on an independent variable x, but these predictions are estimates rather than exact values. To quantify the uncertainty around these predictions, we use prediction intervals, which are similar to confidence intervals but specifically apply to predicted y values from a regression model.

Consider a scenario where a public transportation association studies the relationship between temperature and the number of bus riders. The data shows a strong linear correlation with an r² value of 0.874 and a standard error of the estimate, s_e, equal to 2.97. The regression line equation is given, allowing us to predict the number of riders for a specific temperature.

To predict the number of riders when the temperature is 35 degrees, substitute x₀ = 35 into the regression equation:

\[\hat{y}_0 = b_0 + b_1 x_0\]

For example, if the regression equation is \(\hat{y} = 79.143 - 1.0459x\), then:

\[\hat{y}_0 = 79.143 - 1.0459 \times 35 = 42.54\]

This means the best estimate for the number of riders at 35 degrees is approximately 42.54.

Before constructing a 95% prediction interval for this estimate, two conditions must be verified: there must be a strong linear correlation between x and y, and the prediction point x₀ must lie within the range of observed data to avoid unreliable extrapolation. In this case, both conditions are satisfied.

The prediction interval is calculated as:

\[\hat{y}_0 \pm t_{\alpha/2, n-2} \times s_e \sqrt{1 + \frac{1}{n} + \frac{(x_0 - \bar{x})^2}{\sum (x_i - \bar{x})^2}}\]

Here, \(t_{\alpha/2, n-2}\) is the critical value from the t-distribution with \(n-2\) degrees of freedom, corresponding to the desired confidence level (e.g., 95%), \(s_e\) is the standard error of the estimate, \(n\) is the number of data points, \(\bar{x}\) is the mean of the observed x values, and the denominator is the sum of squared deviations of x values from their mean.

To find the critical t-value for a 95% prediction interval with 13 data points, calculate:

\[t_{0.025, 11} \approx 2.201\]

Using the data, the mean temperature \(\bar{x}\) is approximately 29.92, the sum of the x values is 389, and the sum of squares of the x values is 12,255. The numerator and denominator inside the square root are computed as:

\[\text{Numerator} = n (x_0 - \bar{x})^2 = 13 \times (35 - 29.92)^2 = 335.1\]\[\text{Denominator} = n \sum x_i^2 - (\sum x_i)^2 = 13 \times 12,255 - 389^2 = 7,994\]

Substituting these values into the margin of error formula:

\[E = t_{\alpha/2, n-2} \times s_e \times \sqrt{1 + \frac{1}{n} + \frac{\text{Numerator}}{\text{Denominator}}} = 2.201 \times 2.97 \times \sqrt{1 + \frac{1}{13} + \frac{335.1}{7,994}} \approx 6.914\]

Finally, the 95% prediction interval for the number of riders at 35 degrees is:

\[( \hat{y}_0 - E, \hat{y}_0 + E ) = (42.54 - 6.914, 42.54 + 6.914) = (35.62, 49.45)\]

This interval means we are 95% confident that the actual number of bus riders when the temperature is 35 degrees will fall between approximately 35.62 and 49.45.

Understanding prediction intervals enhances the interpretation of regression predictions by quantifying the expected variability around predicted values. Tools like Excel simplify the calculation process, especially for complex formulas involving sums and critical values, making it easier to apply these concepts in practical data analysis.

A small business tracks how advertising spending relates to weekly sales.
$\left(A\right)$ Plot Advertising Spending (x) vs Sales (y) & find the regression line & correlation coefficient.

• A. $y=20.587x+2015.6$; $0.971$
• B. $y=20.587x+2015.6$; $0.985$
• C. $y=24.373x$; $0.985$
• D. $y=24.373x$; $0.971$

Answer: (B)

A small business tracks how advertising spending relates to weekly sales.
$\left(B\right)$ Predict what the business would make in weekly sales if they spent \$400 in advertising.

• A. 10,200
• B. 10,250
• C. 8,235
• D. 9,749

Answer: (B)

A small business tracks how advertising spending relates to weekly sales.
$\left(C\right)$ Create a prediction interval for the value in part $\left(B\right)$.

• A. $(9016.22,11383.78)$
• B. $(8755.90,11644.10)$
• C. $(9066.62,11434.18)$
• D. $(8806.30,11694.50)$

Answer: (C)

A small business tracks how advertising spending relates to weekly sales.
$\left(D\right)$ Are the actual sales numbers for weeks with \$400 ad-spending within the interval?

• A. Yes
• B. No
• C. More info is required.

Answer: (A)

Here’s what students ask on this topic:

What is a prediction interval in regression analysis and how is it different from a confidence interval?

A prediction interval in regression analysis provides a range within which we expect a single new predicted value of the dependent variable (y) to fall, given a specific value of the independent variable (x). It accounts for both the uncertainty in estimating the regression line and the natural variability of individual data points around that line. In contrast, a confidence interval estimates the range for the mean value of y at a given x, focusing on the average response rather than individual predictions. Prediction intervals are wider than confidence intervals because they include extra variability from individual observations. This distinction is important when making forecasts or decisions based on regression models, as prediction intervals give a more realistic range for future observations.

How do you calculate a 95% prediction interval for a predicted y value using Excel?

To calculate a 95% prediction interval in Excel, first find the predicted y value ($\mathrm{\backslash hat\{y\}\_0}$) by plugging the x value into the regression equation. Then, calculate the margin of error (e) using the formula involving the critical t-value ($\mathrm{t\_\{\backslash alpha/2\}}$), standard error of the estimate ($\mathrm{s\_e}$), and the variability of x values. Use Excel functions like T.INV.2T for the t critical value, AVERAGE for the mean of x, SUM and SUMSQ for sums and sums of squares of x. The margin of error formula is:
$$e=\mathrm{t\_\{\backslash alpha/2\}}\backslash times\mathrm{s\_e}\backslash times\sqrt{1+\frac{1}{n}+\frac{{\mathrm{x\_0}}^2}{\mathrm{\backslash sum\; x\_i^2}}}$$
Finally, the prediction interval is $\mathrm{\backslash hat\{y\}\_0\; \backslash pm\; e}$. This process ensures you account for uncertainty in your prediction.

What conditions must be met before creating a prediction interval in Excel?

Before creating a prediction interval in Excel, two key conditions must be met: First, there should be a strong linear correlation between the independent variable (x) and dependent variable (y). This is often verified by a high coefficient of determination ($\mathrm{R^2}$) close to 1 and a scatterplot showing a linear pattern. Second, the x value for which you want to predict y must lie within the range of the observed data (no extrapolation). Predicting outside this range can lead to unreliable intervals. Meeting these conditions ensures the prediction interval is valid and meaningful, reflecting the true uncertainty in the regression prediction.

How do you find the critical t-value for a prediction interval in Excel?

To find the critical t-value ($\mathrm{t\_\{\backslash alpha/2\}}$) in Excel for a prediction interval, use the function T.INV.2T. The inputs are the significance level alpha (which is 1 minus the confidence level, e.g., 0.05 for 95%) and the degrees of freedom (df), which for linear regression is $\mathrm{n\; -\; 2}$, where $n$ is the number of data points. The formula looks like this:
=T.INV.2T(0.05, n-2)
This returns the two-tailed critical t-value needed to calculate the margin of error in the prediction interval.

Why is the margin of error in a prediction interval larger than in a confidence interval?

The margin of error in a prediction interval is larger than in a confidence interval because it accounts for two sources of variability: the uncertainty in estimating the mean response at a given x and the natural variability of individual observations around the regression line. Confidence intervals estimate the range for the average y value at x, which has less variability. Prediction intervals, however, predict where a single new observation will fall, which is inherently more uncertain. This additional variability is reflected in the formula by the extra term inside the square root, making the margin of error—and thus the interval—wider.