Pooled Linear Regression Coefficient Analysis
This problem examines the properties of a pooled least squares estimate ($\hat{\beta_0}$) for the slope coefficient in a simple linear regression model. We are given two independent samples, yielding slope estimates $\hat{\beta_1}$ and $\hat{\beta_2}$ respectively, with the condition $\hat{\beta_1} > \hat{\beta_2} > 0$. We need to determine the possible behavior of $\hat{\beta_0}$, the estimate obtained from the combined (pooled) sample.
Understanding Least Squares Estimates
The simple linear regression model is given by $Y = \alpha + \beta X + \epsilon$. The least squares estimate for the slope coefficient ($\beta$) in a sample of size $n$ is calculated as:
$ \hat{\beta} = \frac{\sum_{i=1}^{n} (X_i - \bar{X})(Y_i - \bar{Y})}{\sum_{i=1}^{n} (X_i - \bar{X})^2} $A positive slope estimate ($\hat{\beta} > 0$) indicates that, on average, $Y$ tends to increase as $X$ increases within that specific sample. The pooled estimate $\hat{\beta_0}$ is calculated from the combined dataset containing $n_1 + n_2$ observations.
Analyzing the Pooled Estimate ($\hat{\beta_0}$)
While individual sample estimates $\hat{\beta_1}$ and $\hat{\beta_2}$ are positive, the pooled estimate $\hat{\beta_0}$ is not necessarily bounded between $\hat{\beta_2}$ and $\hat{\beta_1}$. The pooled estimate reflects the overall relationship between $X$ and $Y$ across all observations. $\hat{\beta_0}$ can be negative if the covariance between $X$ and $Y$ in the pooled sample is negative. This occurs when the distribution of $X$ values and $Y$ values in the combined dataset results in $\sum (X_i - \bar{X})(Y_i - \bar{Y}) < 0$.
Scenario Illustrating a Negative Pooled Estimate
To demonstrate that $\hat{\beta_0}$ can be negative, consider the following hypothetical data points satisfying $\hat{\beta_1} > \hat{\beta_2} > 0$ (with $n_1, n_2 > 2$):
- Sample 1 ($n_1=3$): Data points $(1, 10), (1.5, 11.5), (2, 13)$. The least squares slope estimate is $\hat{\beta_1} = 3$.
- Sample 2 ($n_2=3$): Data points $(9, 10), (9.5, 10.5), (10, 11)$. The least squares slope estimate is $\hat{\beta_2} = 1$.
Here, the condition $\hat{\beta_1} = 3 > \hat{\beta_2} = 1 > 0$ is met.
Now, let's pool these six observations: $(1, 10), (1.5, 11.5), (2, 13), (9, 10), (9.5, 10.5), (10, 11)$.
The pooled sample means are $\bar{X} = 5.5$ and $\bar{Y} = 11$.
We calculate the pooled slope estimate $\hat{\beta_0}$ using the formula:
$ \hat{\beta_0} = \frac{\sum_{i=1}^{n_1+n_2} (X_i - \bar{X})(Y_i - \bar{Y})}{\sum_{i=1}^{n_1+n_2} (X_i - \bar{X})^2} $The sum of the products $(X_i - \bar{X})(Y_i - \bar{Y})$ for the pooled sample is:
- For points from Sample 1: $(1-5.5)(10-11) + (1.5-5.5)(11.5-11) + (2-5.5)(13-11) = (-4.5)(-1) + (-4)(0.5) + (-3.5)(2) = 4.5 - 2 - 7 = -4.5$.
- For points from Sample 2: $(9-5.5)(10-11) + (9.5-5.5)(10.5-11) + (10-5.5)(11-11) = (3.5)(-1) + (4)(-0.5) + (4.5)(0) = -3.5 - 2 + 0 = -5.5$.
The total sum of products is $-4.5 + (-5.5) = -10$.
The sum of the squared deviations $(X_i - \bar{X})^2$ for the pooled sample is:
- From Sample 1 points: $(-4.5)^2 + (-4)^2 + (-3.5)^2 = 20.25 + 16 + 12.25 = 48.5$.
- From Sample 2 points: $(3.5)^2 + (4)^2 + (4.5)^2 = 12.25 + 16 + 20.25 = 48.5$.
The total sum of squares is $48.5 + 48.5 = 97$.
Therefore, the pooled slope estimate is $\hat{\beta_0} = \frac{-10}{97}$.
Conclusion
The example clearly shows that even with positive individual regression coefficients ($\hat{\beta_1} > \hat{\beta_2} > 0$), the pooled regression coefficient ($\hat{\beta_0}$) can be negative. This demonstrates that the pooled estimate's sign and value depend on the aggregate distribution of the data, not just the relationships within subgroups. Thus, the statement "$\hat{\beta_0}$ can be negative" is true.
Important Questions from Ratio And Regression
- Consider the problem of drawing a sample of size 2 from a finite population of size 20. The sampling is done with replacement using probability proportional to size sampling scheme. The normed size measures $p_1, \cdots, p_{20}$ are given by $p_i = \frac{1}{40}$, $i = 1, \cdots, 10, \; p_i = \frac{3}{40}$, $i = 11, \cdots, 20$. The expected number of distinct units drawn is
- Consider a finite population of size $N$. Let $T_1$ be the sample mean based on a sample of size $n$ under simple random sampling with replacement (SRSWR) scheme. Let $T_2$ be the sample mean based on a stratified random sample of size $n$ where the samples are drawn from each of 4 strata using SRSWR scheme under proportional allocation. Then which of the following are sufficient conditions for $\text{Var}(T_1) = \text{Var}(T_2)$ to hold?
Suppose there are $k$ strata of $N = kM$ units each with size $M$. Draw a sample of size $n_i$ with replacement from the $i^{\text{th}}$ stratum and denote by $\bar{y}_i$ the sample mean of the study variable selected in the $i^{\text{th}}$ stratum, $i = 1, 2, \dots, k$. Define
$$ \bar{y}_s = \frac{1}{k}\sum_{i=1}^k \bar{y}_i \text{ and } \bar{y}_w = \frac{\sum_{i=1}^k n_i \bar{y}_i}{n} $$
Which of the following is necessarily true?- Suppose we draw a random sample of size $n$ from a population of size $N$, where $1 < n < N$, using simple random sampling without replacement scheme. Let $P$ be the population proportion of units possessing a particular attribute and $p$ be the corresponding sample proportion. Which of the following is an unbiased estimator for $P(1 - P)$?
Suppose there are $k$ groups each consisting of $N$ boys. We want to estimate the mean age $\mu$ of these $kN$ boys. Fix $1 < n < N$ and consider the following two sampling schemes.
I. Draw a simple random sample without replacement of size $kn$ out of all $kN$ boys.
II. From each of the $k$ groups draw a simple random sample with replacement of size $n$.
Let $\bar{Y}$ and $\bar{Y}_G$ be the respective sample mean ages for the two schemes. Which of the following are true?
