Questions: Question 15 6 Points Consider the following time series data: t Yt -------- 1 4 2 7 3 9 4 13 a). The estimated regression equation is ŷ = Blank 1 + Blank 2(t) Calculate a 95% prediction interval for the value of Y at time period t=6 (i.e., h=2 periods ahead). Hint: You will first need to fit the model using Excel to obtain the regression output, which will give you some of the values you need for the prediction interval. Then, you will need to calculate the average and standard deviation of the values of t (using Excel will make this easier). Take all calculations and the final answer to three (3) decimal places. Lower Bound = Blank 3 Upper Bound = Blank 4

Solution Steps

To find the estimated regression equation, we first calculate the means of \( t \) and \( Y \):

\[ \bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i = 2.5 \] \[ \bar{y} = \frac{1}{n} \sum_{i=1}^{n} y_i = 8.25 \]

Next, we compute the correlation coefficient \( r \):

\[ r = 0.992 \]

We then calculate the numerator and denominator for the slope \( \beta \):

\[ \text{Numerator for } \beta: \sum_{i=1}^{n} x_i y_i - n \bar{x} \bar{y} = 97 - 4 \cdot 2.5 \cdot 8.25 = 14.5 \] \[ \text{Denominator for } \beta: \sum_{i=1}^{n} x_i^2 - n \bar{x}^2 = 30 - 4 \cdot (2.5)^2 = 5.0 \]

Thus, the slope \( \beta \) is calculated as:

\[ \beta = \frac{14.5}{5.0} = 2.9 \]

The intercept \( \alpha \) is given by:

\[ \alpha = \bar{y} - \beta \bar{x} = 8.25 - 2.9 \cdot 2.5 = 1.0 \]

The estimated regression equation is:

\[ \widehat{y} = 1.0 + 2.9t \]

The mean of \( t \) is:

\[ \mu = \frac{\sum_{i=1}^N x_i}{N} = \frac{10}{4} = 2.5 \]

The variance \( \sigma^2 \) is calculated as:

\[ \sigma^2 = \frac{\sum (x_i - \mu)^2}{n-1} = 1.667 \]

Thus, the standard deviation of \( t \) is:

\[ \text{Standard Deviation} = \sqrt{1.667} = 1.291 \]

Using the regression equation, we predict \( Y \) at \( t = 6 \):

\[ Y_{\text{pred}} = \alpha + \beta \cdot t_{\text{pred}} = 1.0 + 2.9 \cdot 6 = 18.4 \]

The standard error of the estimate is calculated as follows:

\[ \text{Standard Error} = 0.5916 \]

To calculate the prediction interval, we first find the critical value \( t \) for a 95% confidence level with \( n-2 \) degrees of freedom:

\[ t_{\text{value}} = t_{0.975, n-2} \]

The margin of error is given by:

\[ \text{Margin of Error} = t_{\text{value}} \cdot \text{Standard Error} \cdot \sqrt{1 + \frac{1}{n} + \frac{(t_{\text{pred}} - \mu)^2}{(n-1) \cdot \sigma^2}} \]

Calculating the lower and upper bounds of the prediction interval:

\[ \text{Lower Bound} = Y_{\text{pred}} - \text{Margin of Error} = 13.5037 \] \[ \text{Upper Bound} = Y_{\text{pred}} + \text{Margin of Error} = 23.2963 \]

Final Answer