Objectives

Case Study I

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.

Case Study I: The Linear Model

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.

Let’s set the significance value to be \(\alpha = 0.01\).

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.

Case Study 1: Residual Analysis (1/2)

• The residuals appear approximately normal, supporting the normality assumption.
• The residual plot shows roughly “constant” variance, suggesting homoskedasticity.
• Slight clustering near the center may require further investigation but otherwise the residuals are independent of the model (no pattern).

Case Study 1: Residual Analysis (2/2)

Case Study 1: Least Squares Approximation

• The least squares estimates of the intercept and slope are given in the estimate column

The least squares regression model uses the data to find a sample linear fit: \[\hat{y}_{weight} = -3.5980 + 0.2792 \times x_{weeks}.\]

R code:

lm(weight ~ weeks, data = births14)

where the data is stored as a data frame named births14 and weight and weeks are two numerical variables in the data frame.

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, 0.28.

Hypothesis Testing - Theoretical Method (1/2)

• The least squares estimates of the intercept and slope are given in the estimate column

• Computing the test statistic. \[t = \frac{b_1 - \text{null value}}{SE} = \frac{0.2792 - 0}{0.0135} = 20.69\]
• Degrees of Freedom: \(df = n - k - 1\), where \(n\) is the sample size and \(k\) is the number of predictors. So, in this case, \(df = 1000 - 2 = 998\).

Hypothesis Testing - Theoretical Method (2/2)

• We multiply the p-value by 2 since it’s a two sided test, but it’s still 0.
• Since the p-value is less than \(\alpha = 0.01\), then we reject the null hypothesis, meaning we have enough evidence to support that the true slope (relationship) between weeks and weight is non-zero.
• More specifically, since the test statistic is positive, then we can further conclude that the relationship is strongly positive.

Using R to compute the p-value

df <- 998 # degrees of freedom
t <- 20.69 # test statistic
2*(1-pt(t,998))

## [1] 0

Confidence Interval (1/2)

99% Confidence interval for the slope

\[b_1 \pm t_{df}^* \text{SE}_{b_1}\]

• Critical t-star for a 0.99 confidence level: \(t_{df}^* = t_{998}^* = 2.5808\). Note that we use \(1-\alpha\) as the confidence level, where \(alpha = 0.01\).

Using R to compute the critical t-star

df <- 998 # degrees of freedom
lt <- (1-0.99)/2 # lower tail probability
qt(lt,df)

## [1] -2.580765

\[ \begin{aligned} 0.2792 & \pm 2.5808 \times 0.0135 \end{aligned} \] \[(0.2444,0.3140)\]

We are 99% confident that the true slope is in between 0.2444 and 0.3140. Note that the null value of 0 (no slope) is not within the interval.

Confidence Interval (2/2)