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We calculate the predicted value of the dependent variable (Y) by inserting the estimated value of the independent variable (X) in the regression equation. The predicted value of the dependent variable (Y) is determined using the formula:

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

Where:

- \(\widehat{Y}\) = Predicted value of the dependent
- \(X\) = Estimated value of the independent variable

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

$$\widehat{Y}=0.070953-0.90405X_{i}$$

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

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

$$\begin{align*}\widehat{Y}&= 0.070953-0.90405\times 7.5\%\\&= 0.3\%\end{align*}$$

Because the estimated regression doesn’t perfectly describe the relationship between the independent and the dependent variables, we need an interval estimate of the forecast to reflect the uncertainty. The estimated variance of the prediction error of *Y *given *X *is:

$$s_f^2=s_e^2 1+ \frac1n+\frac{(X_f-X ̅ )^2}{((n-1) s_X^2 )}=s_e^2 1+\frac1n+\frac{(X_f-X ̅ )^2}{∑_{i=1}^n(X_i-X ̅ )^2 )}$$

And the standard error of the forecast is:

$$s_f^2=\sqrt[s_e^2]1+\frac1n+\frac{(X_f-X ̅ )^2}{∑_{i=1}^n(X_i-X ̅ )^2 )}$$

Where:

\(S_e\) = The standard error of the estimate.

\(n\) = The number of observations.

\(X_f\) = The forecasted value of the independent variable.

\(\bar{X}\) = The estimated mean.

From the standard error of the forecast, we can see that:

- A better fit of the regression model will result in a smaller standard error of estimates, translating to a smaller standard error of the forecast.
- Large sample size in the regression estimation will result in a smaller standard error of the forecast.
- The closer the mean of the independent variable (\(\bar{X}\)) is to the forecasted independent variable (\(X_f\)) the smaller the standard error of the forecast.

After we have calculated our standard error of the forecast determining the prediction interval follows. The formula for prediction interval is:

$$(\hat{Y_f} ) ± t_\text{critical for α/\(2S_f\)}$$

Where:

\(t_c\)=Two-tailed critical t-value at the given significance level with n-2 df.

\(s_f\)= The estimated variance of the prediction error.

To develop the prediction interval, we use the following steps:

- Predict the value of Y given the forecasted value of X.
- Choose a significance level for the prediction interval.
- Determine the critical value for the prediction interval based on the level of significance and degrees of freedom.
- Compute the standard error of the forecast.
- Compute the percent prediction interval for the prediction.

## Question

Given \(Y=-159+0.26X\)

Where:

- Y = Quantity supplied
- X = Price
The predicted value of the quantity supplied when the price equals 1,200 is

closest to:

- 153
- 155
- 471
## Solution (A)

The correct answer is A.$$Y=-159+0.26\times 1,200=153$$

Notice that \(|t|>t_{c}\) (i.e \(21.67>3.18\)).

Thus, the null hypothesis can be rejected. The conclusion is that the estimated slope coefficient is statistically different from zero.

The estimated value of the intercept \(\widehat{b_{}}=0.0710\).

The estimated value of the standard error for \(\widehat{b_{0}}, (\widehat{S_{b_{0}}})= 0.0094\).

The 95% confidence interval is then calculated as:

\(\text{CI}=\widehat{b_{0}}-t_{c}\widehat{S_{b_{0}}}\) to \(\widehat{b_{0}}-t_{c}\widehat{S_{b_{0}}}\)

\(\text{CI}=0.0710±2.306\times0.0094=0.0493\) to \(0.0927\)

Since 0 lies outside this confidence interval, we can reject the null hypothesis and conclude that the intercept is significantly different from zero.

## Solution (B)

We are testing

$$H_0: b_1=1$$

v.s

$$H_a:b_1≠1$$

The critical value for testing this hypothesis is the same as the critical value for testing the first hypothesis. Thus, \(t_{c}=2.306\).

The estimated value of the parameter \(\widehat{b_{1}}=-0.9041\).

The estimated value of the standard error for \(\widehat{b_{1}},(\widehat{S_{b_{1}}})\) is 0.1755.

The 95% CI for any particular hypothesized value of \(b_{1}\) is determined as follows:

$$\text{CI}=\widehat{b_{1}}±t_{c}\widehat{S_{b_{1}}}$$

$$\begin{align*}\text{CI}&= -0.9041±2.306\times0.1755\\&=-1.3088 \text{to} -0.4994\end{align*}$$

Notice that the hypothesized value of \(b_1=1\) falls outside the above range. Thus, we eject the null hypothesis at the 5% significance level.

*Reading 0: Introduction to Linear Regression *

*LOS 0 (g) Calculate and interpret the predicted value for the dependent variable, and a prediction interval for it, given an estimated linear regression model and a value for the independent variable*