The 95% confidence interval for the forecasted values ŷ of *x* is

This means that there is a 95% probability that the true linear regression line of the population will lie within the confidence interval of the regression line calculated from the sample data.

**Figure 1 – Confidence vs. prediction intervals**

In the graph on the left of Figure 1, a linear regression line is calculated to fit the sample data points. The confidence interval consists of the space between the two curves (dotted lines). Thus there is a 95% probability that the true best-fit line for the population lies within the confidence interval (e.g. any of the lines in the figure on the right above).

There is also a concept called **prediction interval**. Here we look at any specific value of *x*, *x _{0}*, and find an interval around the predicted value ŷ

_{0}for

*x*such that there is a 95% probability that the real value of y (in the population) corresponding to

_{0}*x*is within this interval (see the graph on the right side of Figure 1).

_{0}The 95% prediction interval of the forecasted value ŷ_{0} for *x _{0}* is

where the **standard error of the prediction** is

For any specific value *x*_{0} the prediction interval is more meaningful than the confidence interval.

**Example 1**: Find the 95% confidence and prediction intervals for the forecasted life expectancy for men who smoke 20 cigarettes in Example 1 of Method of Least Squares.

**Figure 2 – Confidence and prediction intervals**

Referring to Figure 2, we see that the forecasted value for 20 cigarettes is given by FORECAST(20,B4:B18,A4:A18) = 73.16. The confidence interval, calculated using the standard error 2.06 (found in cell E12), is (68.70, 77.61).

The prediction interval is calculated in a similar way using the prediction standard error of 8.24 (found in cell J12). Thus life expectancy of men who smoke 20 cigarettes is in the interval (55.36, 90.95) with 95% probability.

**Example 2**: Test whether the y-intercept is 0.

We use the same approach as that used in Example 1 to find the confidence interval of ŷ when *x* = 0 (this is the y-intercept). The result is given in column M of Figure 2. Here the standard error is

and so the confidence interval is

Since 0 is not in this interval, the null hypothesis that the y-intercept is zero is rejected.

Charles,

What’s the rationale for the added 1 under the square root of the standard error of the prediction? Theoretically, why 1?

Thank you

Ryan,

The standard error for a prediction interval adds the variability of the points around the predicted mean. mathematically this is where the 1 comes from.

I plan to eventually show the proofs of the formulas for the confidence and prediction intervals. You will then be able to see from the proofs more precisely where the 1 comes from.

Charles

Hi Charles,

This post has helped me so much already, really very insightful and easy to follow!

I am trying to do a prediction interval for some metal fatigue test data but I am trying to find 90% confidence in the 98% reliability data. In order to do this is just using 0.1 as the probability in TINV sufficient for the method to change it to the 90% probability or do I need to make more changes please?

Thanks for your help!

Tristan

Tristan,

TINV(.1,df) (i.e. alpha = .1) is part of the formula for a 90% prediction interval. But I don’t understand what the 98% reliability data means.

Charles

Hi Charles,

I am using a exponential/hyperbolic function to fit my data. The model is (a*X+b)-SQRT[(a*X+b)^2-((4*a*X*b*c)/(2*c))].

I wonder if I can calculate prediction intervals in the way you show, or if there is any parameter that is different for this type of models. If so, can you tell me how I should calculate it?

Thank you very much

Elisa

Elisa,

I can’t think of a way of doing this. Perhaps someone else has suggestion.

Charles

Hi Charles,

Is Syx = Sres = STEYX(Y,X)? Is it same as Syx = SQRT((SUM(yi – Yi)^2)/(degrees of freedom)), where (xi,yi) are given data and Y is any nonlinear model (not a straight line, say a sigmoidal or logistic curve), which fits the data.

Rizwan,

Sorry for the long delay in responding to you. I just realized that I overlooked responding to you.

It is true that Syx = Sres = STEYX(Y,X) = SQRT((SUM(yi – Yi)^2)/df) for linear models. I am not sure what these terms would even mean for non linear models such as logistic regression models.

Charles

Hi Charles,

Thanks for your contributions on this site. I’m a bit confused by your base formula, though. Where you use the sum of squared deviations of x (SSx, calculated as DEVSQ(x) or DEVSQ(A4:A:18), I’ve learned to use the standard deviation of x times (n-1), or STDEV.S(A4:A:18)*(n-1) in Excel speak. This would yield a value of CI SE of 2.090695467 and a PI SE of 8.244184143. The difference is small in your dataset, but where deviations are larger the use of sums of squared deviations instead of the method I’ve heard will yield very different results. That said, I’m not at all certain which method is correct–can you point to some references for your formula, please?

Thanks in advance for taking the time to clarify this issue for me.

David

Hi David,

Note that for any range R1, the square root of DEVSQ(R1) is STDEV.S(R1)*(n-1). In fact, in the formula for cell E12 in Figure 2 of the referenced page I do take the square root of SSx, and so it seems that we should get the same answer. Can you send me an example of your calculation so that I can see why the results are not the same?

Charles

Thanks for your contribution. I would like to know how to calculate CI and PI if there are two independent variables. Thanks.

Zhang,

Sorry but I haven’t had enough time to figure out or find an answer to your question.

Charles

Please help how u got value of SSx which I suppose to be:-271.6

Anu,

SSx (cell E11) is calculated by the formula =DEVSQ(A4:A18). It has the value 2171.6.

Charles

Hi Charles,

Great. Thank u.

/Kristian

Hi,

Whats the formula in J12? Cannot get the same results…

Thanks

/Kristian

Hi Kristian,

J12 contains the same value as cell E9. The formula in E9 is =FORECAST(E8,B4:B18,A4:A18).

Charles

Hi Charles,

I’m refering to J12, not J11 J12 contains the formula for se (prediction standard error) and formula result i 8.236857, which I cannot get by using the exact same numbers you do.

What formula is in cell J12??

I think it is in the (x – x_)^2 that something is wrong!

Thanks

/ristian

Hi Kristian,

The formula in cell J12 is =E10*SQRT(1+1/E5+(E8-E7)^2/E11).

Charles

Dr. Zaiontz,

Very neat and concise example. I’m particularly interested in a one sided C.I. (lower bound)

Would you agree to use

\hat{y} – t_{crit} s.e.

where t_{crit} should be calculated in Excel using =TINV(2*\alpha,df),

where \alpha = 1-p?

Regards,

Joaquin

Joaquin,

I believe that what you wrote is correct.

Charles

Hi all,

It would be good to point out that the function TINV gives a two-sided confidence interval. The new function T.INV (with a dot) gives a one-sided confidence interval. In case of using the new function, you should take \alpha/2; furthermore, it uses the 1-\alpha/2 value, thus, T.INV(0.975,df).

Emiel

Emiel,

Actually, more simply you should use the T.INV.2T function for the two-sided critical value. It is equivalent to TINV.

This is explained on the webpage http://www.real-statistics.com/excel-capabilities/built-in-statistical-functions/

I will shortly update this information to better explain the various usages of the functions.

Charles

Hi Charles,

Your post has been of tremendous help to me.

Although it took me some time I have applied most of the solutions and they have worked just fine.

I need clarification on example 2. Why did you have to eliminate the # 1 before 1/15?

Would really appreciate your prompt response.

Thanks.

The test uses the confidence interval and not the prediction interval. This is why a 1 is not inserted before 1/15. Shortly I will update the webpage to explain better when the prediction interval is used and when the confidence interval is used.

Charles