The data from exercise I follow. (xi)/(yi)

step 1 So we're given the following information. I went ahead and computed the mean for the xes, which

The data from exercise I follow. $$ \frac{x_{i}}{y_{i}} \left| \begin{array}{ccccc}{1} & {2} & {3} & {4} & {5} \\ \hline y_{i} & {3} & {7} & {5} & {11} & {14}\end{array}\right. $$ $$ \begin{array}{l}{\text { a. Use equation }(12.23) \text { to compute an estimate the standard deviation of } \hat{y}^{*} \text { when } x=4 \text { . }} \\ {\text { b. Use expression }(12.24) \text { to develop a } 95 \% \text { confidence interval for the expected value }} \\ {\text { of } y \text { when } x=4 .}\end{array} $$ $$ \begin{array}{l}{\text { c. Use equation }(12.26) \text { to estimate the standard deviation of an individual value of } y} \\ {\text { when } x=4 .} \\ {\text { d. Use expression }(12.27) \text { to develop a } 95 \% \text { prediction interval for } y \text { when } x=4 \text { . }}\end{array} $$

The data from exercise I follow.
$$
\frac{x_{i}}{y_{i}} \left| \begin{array}{ccccc}{1} & {2} & {3} & {4} & {5} \\ \hline y_{i} & {3} & {7} & {5} & {11} & {14}\end{array}\right.
$$
$$
\begin{array}{l}{\text { a. Use equation }(12.23) \text { to compute an estimate the standard deviation of } \hat{y}^{*} \text { when } x=4 \text { . }} \\ {\text { b. Use expression }(12.24) \text { to develop a } 95 \% \text { confidence interval for the expected value }} \\ {\text { of } y \text { when } x=4 .}\end{array}
$$
$$
\begin{array}{l}{\text { c. Use equation }(12.26) \text { to estimate the standard deviation of an individual value of } y} \\ {\text { when } x=4 .} \\ {\text { d. Use expression }(12.27) \text { to develop a } 95 \% \text { prediction interval for } y \text { when } x=4 \text { . }}\end{array}
$$

Key Concepts

t-Distribution in Regression Analysis

In regression analysis, especially when the sample size is small or the population variance is unknown, the t-distribution is used to account for the extra uncertainty in parameter estimates. It is employed in constructing both confidence intervals for the mean response and prediction intervals for individual observations, with the degrees of freedom determined by the number of observations minus the number of estimated parameters.

Confidence Interval for the Mean Response

A confidence interval for the mean response provides a range within which the expected value of y is likely to fall at a given x. It is constructed using the standard error of the predicted mean response (often involving the t-distribution) and reflects the uncertainty in estimating the true regression line.

Prediction Interval for an Individual Observation

Unlike the confidence interval for the mean response, a prediction interval estimates the range within which a future individual observation is expected to fall. It is wider than the confidence interval because it incorporates not only the estimation error of the mean response but also the additional variability inherent in individual responses.

Simple Linear Regression

This concept involves modeling the relationship between a dependent variable and an independent variable with a linear function, typically expressed as y = ?0 + ?1x + ?. In this framework, the parameters ?0 and ?1 are estimated from data, and the error term ? captures the variability of observations around the regression line.

Standard Error of Prediction

This concept refers to the estimation of the standard deviation of the predicted value of the response variable at a specific level of the predictor. It takes into account both the variability due to the estimation of the regression line parameters and the spread of the data around the line. Equations like (12.23) or (12.26) present formulas for calculating these standard errors for the mean response or an individual predicted value.

Transcript

00:01 So we're given the following information.

00:02 I went ahead and computed the mean for the xes, which is 1 plus 2 plus 3 plus 4 plus 5 divided by 5.

00:07 So we get an x bar of 3.

00:09 And because we know this is from exercise number 1, in exercise number 1, we figured out the estimated regression equation is that y hat equals 0 .2 plus 2 .6x.

00:18 So i found out each of the individual yhats over here.

00:24 And with this, we can go forward in the problem.

00:26 The first thing we have to compute is the estimate of the standard deviation of y hat star when x is equal to 4.

00:33 So here's the formula for the estimate of the standard deviation for y hat.

00:39 So the first thing we need to compute is this s over here.

00:42 We have to figure out what that is.

00:44 And the formula for s is equal to the square root of the mean squares of the error.

00:49 The mean square of the error is equal to the sum of squares of the error divided by n minus 2.

00:55 N is the number of elements in our set.

00:57 And the sum of squares of the error is equal to the difference between, or the sum of the difference, some of the difference between each individual y value that we have and the expected y value based on our regression equation squared.

01:20 So let's find this sum of squares of the error first.

01:25 And we get, let me open the new page, the sum squares of the error is equal to 3 minus 2 .8 squared plus 7 minus 5 .4 squared plus the difference between all the individual y values and their corresponding y -hats.

01:52 And eventually we get some of the squares of the error equaling 12 .4.

01:58 And now we need to come up with the mean squared.

02:01 Of the error, which is equal to the sum of squares of the error, divided by the number of elements we have minus two.

02:08 We have five elements here, five elements, and five minus two is equal to three, so 12 .4 divided by three is equal to 4 .13 repeating.

02:21 And now we can come up with the standard deviation, which is equal to the square root of 4 .13 repeating, is equal to 2 .033 .1.

02:33 And now, we have one part of our equation.

02:37 So this is equal to 2 .0331.

02:42 And we also know that our n is equal to 5.

02:44 And we know that this x star represents the value that we are given, the independent value.

02:50 And we are estimating y hat when x is equal to 4.

02:54 So this is equal to 4.

02:56 And our x bar, we already calculated to be 3.

02:59 Our x sub i is each of our individual x value.

03:02 So we're given that and our x bar is equal to three.

03:05 So using this information we can come up with the estimated standard deviation for y hat when x is equal to four.

03:13 So this is equal to 2 .0331 times the square root of 1 over our n which is 5 plus the difference between our given value which is 4 minus our mean which is 3 squared divided by the sum of each of our individual x values and three squared.

03:38 So this over here, if we expand it, is equal to 1 minus 3 squared, 2 minus 3 squared, 4 minus 3 squared, 4 minus 3 squared, 5 minus 3 squared, all sum together.

03:54 So 1 minus 3 squared plus 2 minus 3 squared plus 3 minus 3 squared plus all the way to 5 minus 3 squared.

04:04 And we're going to take the sum of this.

04:07 And with that, we get a total of 10 down here.

04:13 So let me erase all of this.

04:19 I don't want to move that...

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