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 µ_{Recruits} for chapters who threw 3 parties last year, given the mean parties thrown last year is 4.5.
Clue | Insight |
---|---|
We need to find the 95% confidence interval. | We're working with confidence intervals. |
We're being asked to find the "true µ_{Recruits} 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). |
Assumption | Validate |
---|---|
Linearity | For the sake of this example, let's assume the underlying scatterplot shows a linear relationship. ✅ |
Independence | We 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 Variance | For the sake of this example, let's assume the residual plot shows equal variance. ✅ |
Normality | For the sake of this example, let's assume the residuals are normally distributed. ✅ |
y-hat_{v} is the predicted response (based on the regression) at a given x_{v} value.
t* is our critical value.
SE(µ-hat_{v}) is the standard error for the population mean at x_{v}.
Why are we not using SE(y-hat_{v})? ➡️ We want the standard error of the true population mean [a.k.a. SE(µ-hat_{v})]... 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-hat_{v})]!
y-hat_{v} = ???
t* = ???
SE(µ-hat_{v}) = ???
We need to calculate all of these!
We need to plug into the regression equation to find the predicted response at 3 parties!
y-hat = b_{0} + b_{1}x
b_{0} = 85.8312
b_{1} = 20.5313
y-hat = 85.8312 + 20.5313(x)
Considering x_{v} = 3...
y-hat_{v} = 85.8312 + 20.5313(3)
y-hat_{v} = 85.8312 + 61.5939
y-hat_{v} = 147.4251
df = 50
α = (1 - Confidence Level) / 2
α = (1 - 0.95) / 2
α = (0.05) / 2
α = 0.025
t* = 2.009
SE(b_{1}) is the standard error of our coefficient b_{1}.
x_{v} is the predictor variable value we're gathering the confidence interval of the response for.
x-bar is the mean of the x-values.
s_{e} is the standard deviation of the residuals (a.k.a. Root Mean Square Error (RMSE)).
n is the number of observations.
SE(b_{1}) = 4.16456
x_{v} = 3
x-bar = 4.5
s_{e} = 43.8180
n = 52
SE(µ_{v}) = √[(4.16456)^{2} * (3 - 4.5)^{2} + ((43.8180)^{2} / 52)]
SE(µ_{v}) = √[17.3435 * (-1.5)^{2} + (1920.017124 / 52)]
SE(µ_{v}) = √[17.3435 * 2.25 + 36.9234]
SE(µ_{v}) = √[39.022875 + 36.9234]
SE(µ_{v}) = √[75.946275]
SE(µ_{v}) = 8.7147
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 (µ_{Recruits}) they'll receive the following rush season will be between 129.9171 and 164.9331
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Your university wants to predict the “Sign-Ups” at a student organization receives at mega fair based on the social media “Posts” made by that organization throughout the semester. They randomly sample 30 student organizations, resulting in the following regression analysis.
Find the 95% confidence interval for the true µ_{Sign-Up} for student organizations that posted 6 times over the semester, given the mean posts during the semester were 8.
We are % confident that for student organizations who posted times, the true mean number of sign-ups they’ll receive will be between and .
(When calculating test statistic, round to intermediary values to 3 decimal places. Round bounds to 3 decimal places.)
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