# Help: Prediction intervals and comparisons of two interval widths

Here's my situation:

The equation for a simple linear regression model (below) was calculated and the summary output for the model was produced (below). The task is to calculate all equations by hand; however, for the sake of other users, I incorporated the data in R format too.

Linear Regression Model:

                     y = -0.76 + 1.21x


Data in R format structure(list(x = c(74L, 77L, 79L, 84L, 95L, 98L, 112L, 120L,
128L, 129L, 132L, 135L, 136L, 147L, 148L, 149L, 150L, 153L, 159L,
161L, 161L, 163L, 197L), y = c(85L, 91L, 95L, 104L, 110L, 107L,
134L, 146L, 169L, 156L, 165L, 166L, 169L, 172L, 173L, 189L, 182L,
180L, 195L, 186L, 196L, 195L, 231L)), .Names = c("x", "y"), class =     "data.frame", row.names = c(NA,
-23L))


Question

Based on the linear regression model fitted, calculate estimates and 95% prediction intervals for the weight after 35 days for larvae weighing 128 mg initially and 200 mg initially. State which of the prediction intervals for x = 128 and x = 200 is the wider, and explain why it is wider.

Problem

I have attempted to calculate the prediction intervals for when x = 128 mg and x = 200 mg by hand, which will be used for further data analysis; however, my answers appear to be rather spurious. My main concern is that I feeling unsure about weather I plugged the correct values from the model summary output (below) into the prediction interval equations (below).

More specifically, I am unsure whether I should be multiplying either the standard error prediction or the standard deviation of the x-value with the t-value based on the 95 % prediction level with 21 degrees of freedom and what does the notation Sxx mean?

Inquiries

1. What does Sxx stand for?
2. Do you multiply the t-value with the standard prediction value or the standard deviation for the x-values?
3. Does the notation for S in the prediction equation stand for standard prediction error or the standard deviation for the x values
4. Which interval for x = 128 or x = 200 is wider?

If anyone can help by please providing advice, I would be deeply appreciative.

Equations

The answers (below) in the prediction interval equations appear to be wrong; therefore, the upper and lower boundary limits were used from the model summary output table to denote the margins above and below the y value. Model output: • Looks pretty good. Double check your arithmetic just to be sure. Finally, you might want to confirm your estimates using your favourite statistics program. Well done. – user140401 Dec 28 '16 at 10:26

We can use the regression estimates to predict the value of the dependent variable $\hat{y}$ for a given value of the dependent variable $x_g$. In introductory expositions of simple linear regression, the $100(1-\alpha/2)$ percent prediction interval is given by

$$\hat{y} \pm t_{(1-\alpha/2)}s_{y|x}\sqrt{1+\frac{1}{n}+\frac{(x_g-\bar{x})^2}{\sum_i (x_i-\bar{x})^2}}$$

where $t_{(1-\alpha/2)}$ is the t score at $n-2$ degrees of freedom and $s_{y|x}$ is $\sqrt{MSE}=\sqrt{SSE/(n-2)}$.

What does Sxx stand for?

In some textbooks, you will see the same formula described as

$$\hat{y} \pm t_{(1-\alpha/2)}s_{y|x}\sqrt{1+\frac{1}{n}+\frac{(x_g-\bar{x})^2}{S_{xx}}}$$

If so, then $S_{xx}=\sum_i (x_i-\bar{x})^2$. This might be the case in your notation, although you need to check this for yourself.

Do you multiply the t-value with the standard prediction value or the standard deviation for the x-values?

As seen above, you multiply the t-value with two terms, $s_{y|x}$ and the beast under the radical. In your question, I think that you are asking what $s_{y|x}$ means. If so, you can interpret $s_{y|x}$ as the square root of the estimated variance of $y$ given $x$. Thus, it is neither the standard prediction value not the standard deviation of the $x$ values. As given above, $s_{y|x}=\sqrt{MSE}=\sqrt{SSE/(n-2)}$.

1. Check $s_{y|x}$. In the formula given at the start, $s_{y|x}$ does not change. However, in the two equations of "E" in your worked example, you seem to have used two different values of $s_{y|x}$: 6.1102 and 6.69088. HINT: It is neither of the values you've entered. You're looking in the wrong place in your output.
2. Check $\sum_i (x_i-\bar{x})^2$. In your worked example, you've used the value 32.79. This is waaaaaaayyy off. HINT: This value is not shown anywhere in your output. You will need to manipulate some values by yourself.