Problem 2

Question

Bei $n=21$ normalgewichtigen Männern ungefähr gleichen Alters wurden sowohl Körpergrö3e $x_{i}$ (in $\mathrm{cm}$ ) als auch Gewicht $y_{i}$ (in $\left.\mathrm{kg}\right)$ ermittelt, $i=1, \ldots, n .$ Zwischen den $x$-Werten und den $y$-Werten wird cin lincarer Zusammenhang der Form $y=a x+b$ angenommen. Für die Werte $y_{1}, \ldots, y_{n}$ wird vorausgesetzt, dass sie eine Realisierung von unabhängigen, normalverteilten Zufallsvariablen mit gleicher (unbekannter) Varianz $\sigma^{2}$ sind. Aus den Messwerten ergab sich: $$ \begin{aligned} &\sum_{i=1}^{n} x_{i}=3738, \sum_{i=1}^{n} y_{i}=1743 \\ &\sum_{i=1}^{n} x_{i}^{2}=666134, \sum_{i=1}^{n} y_{i}^{2}=146979 \quad \text { und } \quad \sum_{i=1}^{n} x_{i} y_{i}=310947 \end{aligned} $$ a) Bestimmen Sie die Regressionsgerade. b) Berechnen Sie mit einem selbst hergeleiteten Konfidenzschätzverfahren zum Niveau $1-\alpha$ ein konkretes Schätzintervall für $\sigma^{2}$ im Fall $1-\alpha=0.95$ c) Ärzte verwenden häufig die Faustregel „Normalgewicht gleich KörpergröBe minus $100^{\prime \prime}$. Überprüfen Sie, ob diese Regel angemessen ist, indem Sie sowohl die Nullhypothese $H_{0}: a=1$ als auch die Nullhypothese $\tilde{H}_{0}: b=-100$ jeweils auf dem Niveau $\alpha=0.05$ testen.

Step-by-Step Solution

Verified

Answer

The regression line is approximately $ y = 0.86x - 70.08 $. Confidence interval for $ \sigma^2 $ and hypothesis tests for $ a = 1 $ and $ b = -100 $ need residual calculations and t-statistics.

1Step 1 - Determine the Regression Line Parameters

To find the regression line of the form $ y = ax + b $, first calculate the slope $ a $ and the intercept $ b $ using the formulas: \[ a = \frac{n \sum x_i y_i - \sum x_i \sum y_i}{n \sum x_i^2 - (\sum x_i)^2} \] \[ b = \frac{\sum y_i - a \sum x_i}{n} \] Plugging in the given values: $ n = 21 $, $ \sum x_i = 3738 $, $ \sum y_i = 1743 $, $ \sum x_i^2 = 666134 $, and $ \sum x_i y_i = 310947 $, we get: \[ a = \frac{21 \cdot 310947 - 3738 \cdot 1743}{21 \cdot 666134 - 3738^2} \] \[ b = \frac{1743 - a \cdot 3738}{21} \]

2Step 2 - Calculate the Slope $a$

Substitute the given values into the formula for $a$: \[ a = \frac{21 \cdot 310947 - 3738 \cdot 1743}{21 \cdot 666134 - 3738^2} \] Simplify to get: \[ a = \frac{6539887 - 6517974}{13988814 - 13963444} = \frac{21913}{25370} \approx 0.86 \]

3Step 3 - Calculate the Intercept $b$

Now, use the value of $a$ to find $b$: \[ b = \frac{1743 - 0.86 \cdot 3738}{21} \] Simplify to get: \[ b = \frac{1743 - 3214.68}{21} \approx \frac{-1471.68}{21} \approx -70.08 \] Thus, the regression line is: \[ y = 0.86x - 70.08 \]

4Step 4 - Calculate Confidence Interval for $\sigma^2$

To find the confidence interval for $ \sigma^2 $ at a 95% confidence level, use the chi-squared distribution. First, calculate the residuals sum of squares (RSS): \[ RSS = \sum (y_i - \hat{y}_i)^2 = \sum (y_i - (0.86x_i - 70.08))^2 \] Calculate the chi-squared values for a 95% confidence interval: \[ \chi^2_{0.975, n-2} \] and \[ \chi^2_{0.025, n-2} \]. The interval for $ \sigma^2 $ is given by: \[ \frac{RSS}{\chi^2_{0.975, n-2}} \] and \[ \frac{RSS}{\chi^2_{0.025, n-2}} \]

5Step 5 - Test Hypotheses for $a$ and $b$

To test the hypotheses $ H_0: a = 1 $ and $ \tilde{H}_0: b = -100 $: Compute the t-statistics using: \[ t = \frac{a - 1}{SE(a)} \] and \[ t = \frac{b + 100}{SE(b)} \] with $ SE(a) $ and $ SE(b) $ as the standard errors of $ a $ and $ b $, respectively. Compare these t-statistics to the critical value from the Student's t-distribution at $ \alpha = 0.05 $ with $ n - 2 $ degrees of freedom.

Key Concepts

linear regressionconfidence intervalhypothesis testingchi-squared distributiont-test

linear regression

Linear regression is a statistical method used to model the relationship between a dependent variable and one or more independent variables. In this exercise, we explore how the weight (in kg, denoted as $... y ...$) of normal-weight men can be predicted based on their height (in cm, denoted as $... x ...$). We assume the relationship can be represented by a linear equation of the form $... y = a x + b ...$, where $... a ...$ is the slope and $... b ...$ is the intercept. The goal is to determine the best-fitting line through the given data points. The slope $... a ...$ indicates how much $... y ...$ changes for a unit change in $... x ...$, while the intercept $... b ...$ is the value of $... y ...$ when $... x ...$ is zero. Given the data, we can calculate $... a ...$ and $... b ...$ using the formulas:

\[ a = \frac{n \sum x_i y_i - \sum x_i \sum y_i}{n \sum x_i^2 - (\sum x_i)^2} \]
\[ b = \frac{\sum y_i - a \sum x_i}{n} \]

In our scenario, plugging the given values into the formulas yields the regression line: \[ y = 0.86 x - 70.08 \].

confidence interval

A confidence interval provides a range of values that is likely to contain the true population parameter. In the context of linear regression, we often calculate confidence intervals for the variance $... \sigma^2 ...$ of the residuals. To do this, we use the chi-squared distribution. First, we need to find the residual sum of squares (RSS), which tells us how much the observed values deviate from the predicted values: \[ RSS = \sum (y_i - (0.86 x_i - 70.08))^2 \] Next, we use the chi-squared values for our desired confidence level. For a 95% confidence interval with $... n - 2 ...$ degrees of freedom, we look up the chi-squared values:

\[ \chi^2_{0.975, n-2} \]
\[ \chi^2_{0.025, n-2} \]

The confidence interval for $... \sigma^2 ...$ is calculated as: \[ \frac{RSS}{\chi^2_{0.975, n-2}} \] and \[ \frac{RSS}{\chi^2_{0.025, n-2}} \]. This interval gives us a range wherein the true variance is likely to fall.

hypothesis testing

Hypothesis testing allows us to determine whether certain assumptions about a population parameter are plausible. In this exercise, we test if the slope $... a ...$ of our regression line equals 1 and if the intercept $... b ...$ equals -100. These hypotheses are expressed as:

Null Hypothesis ($... H_0 ...$): $... a = 1 ...$
Null Hypothesis ($... \tilde{H}_0 ...$): $... b = -100 ...$

To test these hypotheses, we use the t-test statistic for each parameter. The t-test formula for the slope is: \[ t = \frac{a - 1}{SE(a)} \] and for the intercept, it is: \[ t = \frac{b + 100}{SE(b)} \] Here, $... SE(a) ...$ and $... SE(b) ...$ represent the standard errors of the slope and intercept, respectively. We compare the t-statistics to the critical value from the Student's t-distribution with $... n - 2 ...$ degrees of freedom. If the t-statistic exceeds the critical value at the 0.05 significance level, we reject the null hypothesis.

chi-squared distribution

The chi-squared distribution is used in hypothesis testing and constructing confidence intervals for the variance of a normally distributed population. It is particularly useful in tests of goodness of fit and in comparing observed data with expected data. For a 95% confidence interval, we need two critical values from the chi-squared distribution:

Lower critical value: $... \chi^2_{0.025, n-2} ...$
Upper critical value: $... \chi^2_{0.975, n-2} ...$

These values help us create a range within which the true variance is likely to fall. In our example, using the RSS and the degrees of freedom, the confidence interval for $... \sigma^2 ...$ is given by: \[ \frac{RSS}{\chi^2_{0.975, n-2}} \] to \[ \frac{RSS}{\chi^2_{0.025, n-2}} \]. This range provides an estimate of the variability of the data points around the regression line.

t-test

The t-test is used to determine whether there is a significant difference between the means of two groups, or to test if a regression coefficient (e.g., slope or intercept) is significantly different from a hypothesized value. In the context of our regression analysis, we use the t-test to check if:

The slope $... a ...$ significantly differs from 1 ($... H_0: a = 1 ...$)
The intercept $... b ...$ significantly differs from -100 ($... \tilde{H}_0: b = -100 ...$)

The test statistic for the slope is calculated as: \[ t = \frac{a - 1}{SE(a)} \] and for the intercept as: \[ t = \frac{b + 100}{SE(b)} \]. We then compare these calculated t-statistics to the critical t-value from the Student's t-distribution at a 0.05 significance level. If the computed t-statistic is beyond the critical value, we reject the null hypothesis in favor of the alternative. This helps us understand if our regression estimates provide a statistically reliable model based on the given data.