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# Predicted Value 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:

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

Where:

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

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

#### Example: Calculating the Predicted Value of a Dependent Variable

The estimated linear regression equation for inflation and unemployment rates for a certain country is:

$$\hat{Y}=0.070953-0.90405X_i$$

Calculate the predicted value of the inflation rate if the forecasted value of the unemployment rate is 7.5%.

Solution

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

\begin{align}\hat{Y}&=\widehat{b}_0+\widehat{b}_1X\\ &=0.070953-0.90405\times 7.5\%=0.3\%\end{align}

# 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.

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

Where:

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

$$n$$ = Nnumber of observations.

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

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

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

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

#### Example: Calculating the Confidence Interval of the Predicted Value

Consider the results of the regression analysis of inflation on unemployment performed by an analyst:

$$\small{\begin{array}{llll}\hline{}& \textbf{Regression Statistics} & & \\ \hline {}& \text{Multiple R} & 0.8766 & {}\\ {}& \text{R Square} & 0.7684 & {}\\ {}& \text{Adjusted R Square} & 0.7394 & {}\\ {}& \text{Standard Error} & 0.0063 & {}\\ {}& \text{Observations} & 10 & {}\\ \hline{}& \textbf{Coefficients} & \textbf{Standard Error} & \text{t-Stat} \\ \hline\text{Intercept} & 0.0710 & 0.0094 & 7.5160\\ \text{Forecast (Slope)} & -0.9041 & 0.1755 & -5.1516\\ \hline\end{array}}$$

Calculate the 95% confidence interval of the predicted value of the inflation rate given that the forecasted unemployment rate is 7.5%.

Solution

$$\text{Prediction Interval} = \widehat{Y}±t_{c}S_{f}$$

The estimated variance of the prediction error is:

$$S_{f}^{2}=S^{2}\bigg[1+\frac{1}{2}+\frac{(X-\bar{X})^2}{(n-1)S_{x}^{2}}\bigg]$$

\begin{align*}S_f^{2}&=0.0063^2\bigg[1+\frac{1}{10} +\frac{(0.075-0.0526)^2}{(10-1)0.0013}\bigg]\\&=0.006733^2\end{align*}

From the regression results, the estimated regression equation is:

$$\widehat{Y}=0.0710-0.9041X$$

The predicted value of the inflation rate given an unemployment rate of 7.5% is:

\begin{align*}\widehat{Y}&= 0.0710-0.9041\times7.5\%\\&=0.3\%\end{align*}

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:

$$\text{Prediction Interval (PI)} = \widehat{Y}±t_{c}S_{f}$$

$$\text{PI} = 0.3\%±2.306\times 0.6735\% = -1.25\% \text{ to } 1.85\%$$

Interpretation

Given an unemployment rate of 7.5%, we are 95% confident that the inflation rate will lie between -1.25% and 1.85%.

## Question 1

The regression equation of 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.

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

## Question 2

Consider the following regression equation of the quantity supplied on price per unit of a particular product:

$$Y=-159+0.26X$$

Where:

$$Y$$ = Quantity supplied.

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

Assume that the forecasted value of the price is $1,200, and the estimated variance of the prediction error is 19. The 95% confidence interval of the predicted value of the quantity supplied is closest to: A. 139.14 to 166.86. B. 141.36 to 164.61. C. 92.58 to 213.42. ### Solution The correct answer is A. The predicted value of the quantity supplied, given the price per unit of the product of$1,200, is:

$$\widehat{Y}=\widehat{b_{0}} +\widehat{b_{1}}X$$

$$\widehat{Y} =-159+0.26\times 1,200=153$$

The two-tailed critical t-value with 3 (n-2) degrees of freedom at the 5% significance level is 3.18. The prediction interval at the 95% confidence level is:

$$\text{Prediction Interval (PI)} =\widehat{Y}±t_{c}S_{f}$$

\begin{align*}\text{PI} &=153±3.18\times 19^{0.5}\\&= 153±13.86\\&=139.14 \ to \ 166.86\end{align*}

Interpretation

Given an estimated price of \$1,200, we are 95% confident that the quantity supplied will lie between 139.14 and 166.86 units.

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