91Ó°ÊÓ

In a least-squares regression model. the residuals are assumed to be random. The following data represent the life expectancy of a female born in the given year. $$ \begin{array}{lc|cc} \text { Year, } x & \text { Life Expectancy, } y & \text { Year, } x & \text { Life Expectancy, } y \\ \hline 1999 & 79.4 & 2005 & 79.9 \\ \hline 2000 & 79.3 & 2006 & 80.2 \\ \hline 2001 & 79.4 & 2007 & 80.4 \\ \hline 2002 & 79.5 & 2008 & 80.6 \\ \hline 2003 & 79.6 & 2009 & 80.9 \\ \hline 2004 & 79.9 & 2010 & 81.0 \\ \hline \end{array} $$ The least-squares regression equation treating the year as the independent variable is \(\hat{y}=0.1633 x-247.2999 .\) The residuals from left to right are $$ \begin{array}{rrrrrr} \hline 0.290 & 0.026 & -0.037 & -0.100 & -0.163 & -0.027 \\ \hline-0.190 & -0.053 & -0.017 & 0.020 & 0.157 & 0.094 \\ \hline \end{array} $$ (a) Denote residuals above 0 with an \(A\) and those below 0 with a \(\mathrm{B}\) to form a sequence. (b) Test the assumption that the residuals are random at the \(\alpha=0.05\) level of significance.

Step by step solution

01

- Organize the residuals

List the given residuals in the order they are presented: [0.290, 0.026, -0.037, -0.100, -0.163, -0.027, -0.190, -0.053, -0.017, 0.020, 0.157, 0.094].

02

- Label residuals

Label each residual above 0 with an 'A' and each residual below 0 with a 'B': ['A', 'A', 'B', 'B', 'B', 'B', 'B', 'B', 'B', 'A', 'A', 'A'].

03

- Count the runs

Identify and count the number of runs (consecutive sequences of the same letter): There are 3 runs of 'A' and 2 runs of 'B'. This totals to 5 runs.

04

- Calculate expectations and variances

Use the following formulas to calculate the expected number of runs (E(R)) and the variance (Var(R)) for a sample of size n1 and n2 with total n: \[ E(R) = \frac{2 n_1 n_2}{n} + 1 \] \[ \text{Var}(R) = \frac{2 n_1 n_2 (2 n_1 n_2 - n)}{n^2 (n - 1)} \] \( n_1 = 5 \) (number of 'A'), \( n_2 = 7 \) (number of 'B'), and \( n = 12 \) (total number of observations). Substituting the values: \[ E(R) = \frac{2(5)(7)}{12} + 1 = \frac{70}{12} + 1 = 5.8333 \] \[ \text{Var}(R) = \frac{2(5)(7) (2(5)(7) - 12)}{12^2 (11)} = \frac{2 \times 35 \times 58}{144 \times 11} = \frac{4060}{1584} = 2.5641 \]

05

- Calculate the standardized test statistic

The standardized test statistic is calculated by \( Z = \frac{R - E(R)}{\text{Var}(R)} \). Here \( R = 5 \), so substituting we get: \[ Z = \frac{5 - 5.8333}{1.6013} = \frac{-0.8333}{1.6013} = -0.5204 \]

06

- Compare with critical value

Check the critical value for \( \alpha = 0.05 \) in a standard normal distribution table. The critical values for a two-tailed test are approximately \( \pm 1.96 \). Since \(-0.5204\) is within the range \([-1.96, 1.96]\), we fail to reject the null hypothesis.

07

- Conclusion

Based on the test, we conclude that the residuals are random at the \( \alpha = 0.05 \) level of significance.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Residual Analysis

Residual analysis is a critical aspect of evaluating a least-squares regression model. Residuals are the differences between observed and predicted values. These differences help us determine how well the model fits the data.
By examining residuals, we can identify any patterns that suggest whether or not our model is accurate.
Ideally, residuals should appear random when plotted, showing no discernible pattern.
If residuals have patterns, it indicates that the model's assumptions might be violated.
This could mean that the model is not capturing certain aspects of the data.
In this exercise, the provided residuals are analyzed to determine if they exhibit random behavior, supporting the validity of the regression model.

Randomness Test

The randomness test is an essential method for checking if residuals in a regression model are random. It involves several steps:
- **Labeling Residuals**: Residuals above 0 are labeled as 'A' and those below 0 as 'B'.
- **Counting Runs**: A run is a sequence of consecutive 'A's or 'B's.
- **Calculating Expectations and Variances** using formulas:
\[ E(R) = \frac{2 n_1 n_2}{n} + 1 \] and \[ \text{Var}(R) = \frac{2 n_1 n_2 (2 n_1 n_2 - n)}{n^2 (n - 1)} \]
These calculations help to find the expected number of runs and the variance in those runs.
Next, the standardized test statistic, \( Z \), is computed:
\[ Z = \frac{R - E(R)}{\text{Var}(R)} \]
If the computed Z-value falls within the range of critical values for the chosen significance level, we conclude the residuals are random.
This implies the model assumptions hold and are reliable.

Standardized Test Statistic

The standardized test statistic helps quantify how much our sample data deviates from what is expected under the null hypothesis.
In this context, it checks if the number of runs aligns with what would be expected by chance.
The formula to calculate the standardized test statistic is:
\[ Z = \frac{R - E(R)}{\text{Var}(R)} \]
Where
- \( R \): observed number of runs
- \( E(R) \): expected number of runs
- \( \text{Var}(R) \): variance of the number of runs
Here, \( R \) was found to be 5, and based on calculations, \( E(R) \) was 5.8333 and \( \text{Var}(R) \) was 2.5641.
The resulting Z-value is compared against critical values from the standard normal distribution.
For a significance level of 0.05 and a two-tailed test, the critical values are approximately \( \pm 1.96 \).
Because the calculated Z-value lies within this range, the assumption of randomness is supported.
This confirms the model's validity under the given significance level.