# Predicted Value and Prediction Interval of a Dependent Variable

We calculate the predicted value of the dependent variable, $$Y$$, by inserting the estimated value of the independent variable, $$X$$, into the regression equation. The predicted value of the dependent variable, $$Y$$, is determined using the following formula:

$$\hat{Y}=\hat{b}_0+\hat{b}_1X$$

Where:

$$\hat{Y}$$ = Predicted value of the dependent variable.

$$X$$ = Estimated value of the independent variable.

Example: Calculating the Predicted Value of a Dependent Variable

Refer to the example of regressed inflation rates against unemployment rates from 2011 to 2020.

$$\begin{array}{c|c|c|c|c|c|c|c} \text{Year} & \text{Unemployment} & \text{Inflation} & \text{Predicted} & \text{Variation} & \text{Variation} & \text{Variation} & (X_i \\ & {\text{Rate } \% (X_i)} & {\text{Rate }\%} & \text{Unemployment} & \text{to be} & \text{Unexplained} & \text{Explained} & -\bar{X})^2 \\ & & ({{Y}}_i) & {\text{rate } (\hat Y_i)} & \text{Explained.} & & & \\ & & & & \left(Y_i-\bar{Y}\right)^2 & \left(Y_i- \hat{Y}_i\right)^2 & \left({\hat{Y}}_i-\bar{Y}\right)^2 & \\ \hline 2011 & 6.1 & 1.7 & 1.610 & 0.410 & 0.008 & 0.533 & 0.656 \\ \hline 2012 & 7.4 & 1.2 & 0.437 & 1.300 & 0.582 & 3.621 & 4.452 \\ \hline \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ \hline 2019 & 4.0 & 4.7 & 3.504 & 5.570 & 1.430 & 1.355 & 1.664 \\ \hline 2020 & 3.9 & 3.6 & 3.594 & 1.588 & 0.000 & 1.573 & 1.932 \\ \hline \textbf{Sum} & \bf{52.90} & \bf{23.4} & & \bf{13.704} & \bf{3.136} & \bf{10.568} & \bf{12.989} \\ \hline \textbf{Arithmetic} & \bf{5.29} & \bf{2.34} & & & & & \\ \textbf{Mean} & & & & & & & \\ \end{array}$$

The estimated regression model is illustrated below.

$$\hat{Y}=7.112-0.9020X_i+\varepsilon_i$$

Calculate the predicted inflation rate value if the forecasted value of the unemployment rate is 4.5%.

Solution

The predicted value of the inflation rate is determined as follows:

$$\hat{Y}=7.112-0.9020\times4.5=3.053\%$$

## Confidence Interval for Predicted Values

The calculation of the confidence interval for the predicted value of a dependent variable is the same as that of the confidence interval for regression coefficients. The confidence interval for a predicted value of the dependent variable is given by:

$$\text{Prediction Interval}=\ \hat{Y}\pm t_cs_f$$

Where:

$$t_c$$= Two-tailed critical t-value at the given significance level with $$n – 2$$ df.

$$\hat{Y}$$ = Predicted value of a dependent variable.

$$s_f^2$$= The estimated variance of the prediction error.

$$s_f^2=s_e^2\left[1+\frac{1}{n}+\frac{\left(X_f-\bar{X}\right)^2}{\left(n-1\right)s_x^2}\right]=s_e^2\left[1+\frac{1}{n}+\frac{\left(X_f-\bar{X}\right)^2}{\sum_{i\ =\ 1}^{n}\left(X_i-\bar{X}\right)^2}\right]$$

Where:

$$s_e^2$$ = The squared standard error of the estimate.

$$n$$ = Number of observations.

$$s_X^2$$= Variance of the independent variable.

$$X_f$$ = Value of the independent variable.

We can, therefore, calculate the standard error of forecast as shown below:

$$s_f=s_e\sqrt{1+\frac{1}{n}+\frac{\left(X_f-\bar{X}\right)^2}{\sum_{i=1}^{n}\left(X_i-\bar{X}\right)^2}}$$

From the formula above, we can observe that:

• A better fit of the regression analysis leads to a smaller standard error of the estimate $$(s_e)$$, subsequently resulting in a lower standard error of the forecast.
• When the sample size $$(n)$$ in the regression calculation increases, it directly corresponds to a reduction in the standard error of the forecast.
• If the forecasted independent variable $$(X_f)$$ approaches the mean of the independent variable $$(\bar{X})$$ utilized in the regression analysis, it decreases the standard error of the forecast.

Example: Calculating the Confidence Interval of the Predicted Value

Refer to the example of regressed inflation rates against unemployment rates from 2011 to 2020.

$$\begin{array}{c|c|c|c|c|c|c|c} \text{Year} & \text{Unemployment} & \text{Inflation} & \text{Predicted} & \text{Variation} & \text{Variation} & \text{Variation} & (X_i \\ & {\text{Rate } \% (X_i)} & {\text{Rate }\%} & \text{Unemployment} & \text{to be} & \text{Unexplained} & \text{Explained} & -\bar{X})^2 \\ & & ({{Y}}_i) & {\text{rate } (\hat Y_i)} & \text{Explained.} & & & \\ & & & & \left(Y_i-\bar{Y}\right)^2 & \left(Y_i- \hat{Y}_i\right)^2 & \left({\hat{Y}}_i-\bar{Y}\right)^2 & \\ \hline 2011 & 6.1 & 1.7 & 1.610 & 0.410 & 0.008 & 0.533 & 0.656 \\ \hline 2012 & 7.4 & 1.2 & 0.437 & 1.300 & 0.582 & 3.621 & 4.452 \\ \hline \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ \hline 2019 & 4.0 & 4.7 & 3.504 & 5.570 & 1.430 & 1.355 & 1.664 \\ \hline 2020 & 3.9 & 3.6 & 3.594 & 1.588 & 0.000 & 1.573 & 1.932 \\ \hline \textbf{Sum} & \bf{52.90} & \bf{23.4} & & \bf{13.704} & \bf{3.136} & \bf{10.568} & \bf{12.989} \\ \hline \textbf{Arithmetic} & \bf{5.29} & \bf{2.34} & & & & & \\ \textbf{Mean} & & & & & & & \\ \end{array}$$

Consider the results of the regression analysis of inflation rates on unemployment rates:

$$\begin{array}{lcccc} \bf{\textit{Regression Statistics} } & & & & \\ \hline \text{R Square} & 0.7711 & & & \\ \text{Standard Error} & 0.6261 & & & \\ \text{Observations} & 10 & & & \\ \hline \\ \hline \text{ANOVA} & & & & \\ \hline & \textbf{df} & \textbf{Sum of} & \textbf{Mean} & \textbf{F} \\ & & \textbf{Squares} & \textbf{Square} & \\ \hline \text{Regression} & 1 & 10.568 & 10.568 & 26.9565 \\ \text{Residual} & 8 & 3.136 & 0.392 & \\ \text{Total} & 9 & 13.704 & & \\ \hline \\ \hline & \textbf{Coefficients} & \textbf{Standard} & \textbf{t Stat} & \textbf{p-value} \\ & & \textbf{Error} & & \\ \hline \text{Intercept} & 7.112 & 0.940 & 7.565 & 0.000 \\ \text{Unemployment} & -0.902 & 0.174 & -5.192 & 0.001 \\ \text{rate (%)} & & & & \\ \hline \end{array}$$

Calculate the 95% confidence interval of the predicted value of the inflation rate, given that the forecasted unemployment rate is 4.5%.

Solution

$$\text{Prediction Interval} = \hat{Y} \pm t_{c}s_{f}$$

The estimated variance of the prediction error is:

\begin{align*} s_f^2 & =s_e^2\left[1+\frac{1}{n}+\frac{\left(X_f-\bar{X}\right)^2}{\left(n-1\right)s_X^2}\right] \\ & =s_e^2\left[1+\frac{1}{n}+\frac{\left(X_f-\bar{X}\right)^2}{\sum_{i=1}^{n}\left(X_i-\bar{X}\right)^2}\right] \\ & ={0.6261}^2\left[1+\frac{1}{10}+\frac{\left(4.5-5.29\right)^2}{12.989}\right]=0.450 \end{align*}

As such, the standard error of forecast is:

$$s_f=\sqrt{0.450}=0.6708$$

The predicted value of the inflation rate given an unemployment rate of 4.5% is:

$$\hat{Y}=7.112-0.9020\times4.5=3.05\%$$

The two-tailed critical t-value with 8 $$(n-2)$$ degrees of freedom at the 5% significance level is 2.306.

The prediction interval at the 95% confidence level is:

$$\text{Prediction Interval (PI)} = \hat{Y} \pm t_{c}s_{f}$$

$$\text{PI} = 3.05 \pm 2.306\times 0.6708= 1.50\% \text{ to } 4.60\%$$

Interpretation

Given an unemployment rate of 4.5%, we are 95% confident that the inflation rate will lie between 1.50% and 4.60%.

## Question 1

The regression equation of the quantity of goods against the price is given by:

$$Y =-159+0.26X$$

Where:

$$Y$$ = Quantity supplied.

$$X$$ = Price per unit of the product.

The predicted value of the quantity supplied when the price equals 1,200 is closest to:

1. 153.
2. 155.
3. 471.

$$Y = -159 + 0.26\times1,200=153$$

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