Confidence intervals with predicted mean response

Scenario: Crammer Nation University wants to develop a regression equation to predict the "Number of Recruits" a given fraternity will receive this rush season given the "Parties" that fraternity threw last year. They took a sample of 52 fraternities on campus, resulting in the regression output below.

Find the 95% confidence interval for the true µRecruits for chapters who threw 3 parties last year, given the mean parties thrown last year is 4.5.

ClueInsight
We need to find the 95% confidence interval.We're working with confidence intervals.
We're being asked to find the "true µRecruits for chapters who threw 3 parties last year".We're working with the response variable (given a predictor variable value).
The RMSE is derived from the sample... not the population.We'll need to settle for t-scores (instead of z-scores).
AssumptionValidate
LinearityFor the sake of this example, let's assume the underlying scatterplot shows a linear relationship. ✅
IndependenceWe can assume that each chapter's parties thrown and recruits received don't impact one another. ✅ (Ex: Delta Apple Pi's parties and recruits don't impact Alpha Blueberry Pi's.)
Equal VarianceFor the sake of this example, let's assume the residual plot shows equal variance. ✅
NormalityFor the sake of this example, let's assume the residuals are normally distributed. ✅

y-hatv is the predicted response (based on the regression) at a given xv value.
t* is our critical value.
SE(µ-hatv) is the standard error for the population mean at xv.

Why are we not using SE(y-hatv)? ➡️ We want the standard error of the true population mean [a.k.a. SE(µ-hatv)]... as we're working with a confidence interval for that true population mean. We don't want the standard error of the predicted response [a.k.a. SE(y-hatv)]!

y-hatv = ???
t* = ???
SE(µ-hatv) = ???

We need to calculate all of these!

How to calculate y-hatv

We need to plug into the regression equation to find the predicted response at 3 parties!

y-hat = b0 + b1x

b0 = 85.8312
b1 = 20.5313

y-hat = 85.8312 + 20.5313(x)

Considering xv = 3...

y-hatv = 85.8312 + 20.5313(3)
y-hatv = 85.8312 + 61.5939
y-hatv = 147.4251

How to locate t*

df = 50

α = (1 - Confidence Level) / 2
α = (1 - 0.95) / 2
α = (0.05) / 2
α = 0.025

How to calculate SE(µv)

SE(b1) is the standard error of our coefficient b1.
xv is the predictor variable value we're gathering the confidence interval of the response for.
x-bar is the mean of the x-values.
se is the standard deviation of the residuals (a.k.a. Root Mean Square Error (RMSE)).
n is the number of observations.

SE(b1) = 4.16456
xv = 3
x-bar = 4.5
se = 43.8180
n = 52

SE(µv) = √[(4.16456)2 * (3 - 4.5)2 + ((43.8180)2 / 52)]
SE(µv) = √[17.3435 * (-1.5)2 + (1920.017124 / 52)]
SE(µv) = √[17.3435 * 2.25 + 36.9234]
SE(µv) = √[39.022875 + 36.9234]
SE(µv) = √[75.946275]
SE(µv) = 8.7147

Conclusion

CI = 147.4251 ± 2.009(8.7147)
CI = 147.4251 ± 17.5078323
CI = (129.9171, 164.9331)

Answer: We are 95% confident that for fraternities who threw 3 parties last year, the true mean number of recruits (µRecruits) they'll receive the following rush season will be between 129.9171 and 164.9331

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