12 - Inference for Linear Regression
Hypothesis Testing
Alex John Quijano
12/01/2021
Previously on Statistics…
- Hypothesis testing and confidence interval for paired means.
Inference on Single Mean
Today, we will discuss the following:
Sandwich Store Linear Model (1/2)
Sandwich Store Linear Model (2/2)
The population model is: \[y_{revenue} = \beta_0 + \beta_1 x_{advertising} + e\] where \(y\) is the response, \(x\) is the predictor, \(\beta_0\) is the intercept, \(\beta_1\) is the slope, and \(e\) is the error term.
The least squares regression model uses the data to find a sample linear fit: \[\hat{y}_{revenue} = b_0 + b_1 x_{advertising}.\] where \(b_0 = 11.23\), \(b_1 = 4.8\).
Variability of the Statistic (1/5)
Variability of the Statistic (2/5)
A second sample of size 20 also shows a positive trend!
Variability of the Statistic (3/5)
Variability of the Statistic (4/5)
Variability of the Statistic (5/5)
Baby Weights
Consider data births gathered originally from the US Department of Health and Human Services. The births14 data can be found in the openintro R package. We will work with a random sample of 100 observations from these data.
We want to predict the baby weight based on number of weeks. The population linear model is \[y_{weight} = \beta_0 + \beta_1 x_{weeks} + e\]
The relevant hypotheses for the linear model setting can be written in terms of the population slope parameter. Here the population refers to a larger population of births in the US.
- \(H_0: \beta_1= 0\), there is no linear relationship between
weight and weeks.
- \(H_A: \beta_1 \ne 0\), there is some linear relationship between
weight and weeks.
Technical Conditions
Linearity. The scatterplot of the explanatory and response must be nearly linear.
Independent Observations. The samples must be independent.
Normally Distributed Residuals. The errors must show a nearly normal distribution.
Constant or equal variability. The error must exhibit homoscedasticity.
Least Squares Approximation
The least squares estimates of the intercept and slope are given in the estimate column. The observed slope is 0.335.
|
term
|
estimate
|
std.error
|
|
(Intercept)
|
-5.72
|
1.61
|
|
weeks
|
0.34
|
0.04
|
The least squares regression model uses the data to find a sample linear fit: \[\hat{y}_{weight} = -5.72 + 0.34 x_{weeks}.\]
R code:
lm(weight ~ weeks, data = births14_100)
where the data frame births14_100 is a subset of the original births14 data.
Hypothesis Testing - Randomization Method (1/2)
Two different permutations of the weight variable with slightly different least squares regression lines.
Hypothesis Testing - Randomization Method (2/2)
Histogram of slopes given different permutations of the weight variable. The vertical red line is at the observed value of the slope, 0.335.
Hypothesis Testing - Theoretical Method
The least squares estimates of the intercept and slope are given in the estimate column. The observed slope is 0.335.
|
term
|
estimate
|
std.error
|
statistic
|
p.value
|
|
(Intercept)
|
-5.716
|
1.6137
|
-3.54
|
6e-04
|
|
weeks
|
0.335
|
0.0416
|
8.07
|
<0.0001
|
- Computing the test statistic.
\[T = \frac{b_1 - \text{null value}}{SE} = \frac{0.335 - 0}{0.0416} = 8.0529\]
- Degrees of Freedom: \(df = n - k - 1\), where \(n\) is the sample size and \(k\) is the number of predictors. So, in this case, \(df = 100 - 2 = 98\).
10.10-Minute Activity
Consider the following least squares output.
|
term
|
estimate
|
std.error
|
statistic
|
p.value
|
|
(Intercept)
|
-29.90
|
7.79
|
-3.84
|
0.0012
|
|
perc_pov
|
2.56
|
0.39
|
6.56
|
<0.0001
|
Here, we model the murders per mile based on the poverty level.
Write the linear equation for the population model and the estimated linear model.
What are the hypotheses for evaluating whether the slope of the model predicting annual murder rate from poverty percentage is different than 0?
State the conclusion of the hypothesis test from part (2) in context of the data. What does this say about whether poverty percentage is a useful predictor of annual murder rate?
