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In Exercise 13.17, we considered a regression of \(y=\) oxygen consumption on \(x=\) time spent exercising. Summary quantities given there yield $$ \begin{array}{llrl} n & =20 & \bar{x} & =2.50 & S_{x x} & =25 \\ b & =97.26 & a & =592.10 & s_{e} & =16.486 \end{array} $$ a. Calculate \(s_{a+b(2.0)}\), the estimated standard deviation of the statistic \(a+b(2.0)\). b. Without any further calculation, what is \(s_{a+b(3.0)}\) and what reasoning did you use to obtain it? c. Calculate the estimated standard deviation of the statistic \(a+b(2.8)\). d. For what value \(x^{*}\) is the estimated standard deviation of \(a+b x^{*}\) smallest, and why?

Short Answer

Expert verified
The answers are: a. \(s_{a+b(2.0)} = 16.486 \cdot \sqrt{1/20 + (2.0 - 2.5)^2/25}\) b. \(s_{a+b(3.0)} = 16.486 / \sqrt{20}\) c. \(s_{a+b(2.8)} = 16.486 \cdot \sqrt{1/20 + (2.8 - 2.5)^2/25}\) d. The smallest estimated standard deviation occurs when \(x^{*} = \bar{x} = 2.5\)

Step by step solution

01

Calculate \(s_{a+b(2.0)}\)

Use the formula \(s_{a+b(x)} = s_{e} \cdot \sqrt{1/n + (x-\bar{x})^2/S_{xx}}\). Substituting the given values: \(s_{a+b(2.0)} = 16.486 \cdot \sqrt{1/20 + (2.0 - 2.5)^2/25}\), compute to find the solution.
02

Without any further calculation, find \(s_{a+b(3.0)}\)

Notice that 3.0 is the mean of x. According to the formula, if x equals its mean, the second term in the square root vanishes, so the \(s_{a+b(x)}\) in this case simplifies to \(s_{e} /\sqrt{n}\). Hence, \(s_{a+b(3.0)} = s_{e} / \sqrt{n}\) = 16.486 / \sqrt{20}\.
03

Calculate the estimated standard deviation of the statistic \(a+b(2.8)\)

Similar to Step 1, plug in \(x = 2.8\) into the formula: \(s_{a+b(2.8)} = 16.486 \cdot \sqrt{1/20 + (2.8 - 2.5)^2/25}\) and calculate the value.
04

Determining the x for smallest estimated standard deviation

Looking at the formula \(s_{a+b(x)} = s_{e} \cdot \sqrt{1/n + (x-\bar{x})^2/S_{xx}}\), it can be observed that the part which includes x is the factor \((x-\bar{x})^2\). It reaches its smallest value (which is zero) when \(x = \bar{x}\). So, indeed, to find the smallest estimated standard deviation \(x^{*} = \bar{x} = 2.5\). Indeed, when \(x = \bar{x}\), the deviation is minimized.

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Key Concepts

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

Estimated Standard Deviation
The estimated standard deviation, often denoted as s, is a measure that gives us an idea of how spread out individual observations are in a dataset. In the context of regression analysis, this statistic helps gauge the precision of predictions made by the regression model.

For instance, in the exercise, the estimated standard deviation of the regression estimate for a specific value of x (oxygen consumption) at two time points (2.0 and 2.8 hours) was calculated. This is crucial because it informs us how much variability there is in our estimates of oxygen consumption at these times, providing insight into the reliability of our model's predictions. To calculate this for any given point, we use the formula:
\[ s_{a+b(x)} = s_{e} \cdot \sqrt{\frac{1}{n} + \frac{(x-\bar{x})^2}{S_{xx}}} \]
where se is the estimated standard error of the regression, n is the number of observations, ³æÌ„ is the mean of the independent variable, and Sxx is the sum of squared deviations of the independent variable. Simplifying this calculation for different values of x enables us to assess the precision across different levels of exercise duration.
Linear Regression
Linear regression is a powerful statistical tool for understanding the relationship between two variables. By fitting a straight line through the data points that minimizes the squared differences between the observed and predicted values, linear regression provides a way to predict one variable based on another.

In our exercise scenario, linear regression is used to predict oxygen consumption based on the time spent exercising. The equation of a linear regression line is typically written as:
\[ y = a + bx \]
where y represents the dependent variable (oxygen consumption), a is the y-intercept, b is the slope of the line which signifies how much y changes with a one-unit change in x (time spent exercising), and x is the independent variable. Understanding the dynamics of this linear relationship allows us to make inferences about how changes in exercise duration might affect oxygen consumption.
Statistical Inference
Statistical inference involves using data from a sample to draw conclusions about a larger population. It forms the basis of many decisions and predictions in both research and real life. In regression analysis, we use statistical inference to determine if the relationships we observe in our sample data are likely to exist in the broader population.

In the given exercise, statistical inference would involve determining whether the observed relationship between exercise time and oxygen consumption is statistically significant and not due to random chance. Methods such as hypothesis testing and confidence intervals are commonly used to make such inferences. These methods would leverage the estimated standard deviation of the regression coefficients to ascertain their reliability and the likelihood that similar relationships would occur in other samples or the population at large.
Oxygen Consumption
Oxygen consumption is a critical indicator of an individual's aerobic fitness and metabolic rate. In sports science and health studies, it's often measured to determine how the body responds to physical exercise.

The exercise we're discussing uses oxygen consumption as the dependent variable, meaning it's the outcome being affected by the independent variable – time spent exercising. By analyzing oxygen consumption through statistical tools like linear regression, we can predict and understand how increasing the intensity or duration of exercise affects the body's oxygen use. This kind of analysis not only demonstrates a practical application of statistics in health sciences but also provides valuable insights for developing personal or athletic training programs.
Exercise and Statistics
Combining the principles of exercise physiology with statistics creates a robust framework for understanding and quantifying the effects of physical activity on health indicators such as oxygen consumption. This intersection allows for the quantifiable analysis and interpretation of how different variables, like the duration of exercise, influence physiological outcomes.

Applying statistical methods, like regression analysis, to data collected from exercise experiments enables researchers and fitness professionals to make evidence-based decisions. It also contributes to personalized exercise regimens that optimize performance or health outcomes. By utilizing concepts like the estimated standard deviation and linear regression, we can provide scientifically grounded advice for individuals looking to improve their fitness levels or monitor their health condition.

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Most popular questions from this chapter

The authors of the paper studied a number of variables they thought might be related to bone mineral density (BMD). The accompanying data on \(x=\) weight at age 13 and \(y=\) bone mineral density at age 27 are consistent with summary quantities for women given in the paper. A simple linear regression model was used to describe the relationship between weight at age 13 and \(\mathrm{BMD}\) at age 27\. For this data: $$ \begin{array}{lll} a=0.558 & b=0.009 & n=15 \\ \mathrm{SSTo}=0.356 & \text { SSResid }=0.313 & \end{array} $$ a. What percentage of observed variation in \(\mathrm{BMD}\) at age 27 can be explained by the simple linear regression model? b. Give a point estimate of \(\sigma\) and interpret this estimate. c. Give an estimate of the average change in BMD associated with a \(1 \mathrm{~kg}\) increase in weight at age 13 . d. Compute a point estimate of the mean BMD at age 27 for women whose age 13 weight was \(60 \mathrm{~kg}\).

The figure at the top of the page is based on data from the article. It shows the relationship between aboveground biomass and soil depth within the experimental plots. The relationship is described by the estimated regression equation: biomass \(=-9.85+25.29\) (soil depth) and \(r^{2}=.65 ; P<0.001 ; n=55 .\) Do you think the simple linear regression model is appropriate here? Explain. What would you expect to see in a plot of the standardized residuals versus \(x\) ?

The authors of the paper studied the relationship between childhood environmental lead exposure and a measure of brain volume change in a particular region of the brain. Data were given for \(x=\) mean childhood blood lead level \((\mu \mathrm{g} / \mathrm{dL})\) and \(y=\) brain volume change \((\mathrm{BVC}\), in percent \() .\) A subset of data read from a graph that appeared in the paper was used to produce the accompanying Minitab output. Carry out a hypothesis test to decide if there is convincing evidence of a useful linear relationship between \(x\) and \(y\). Assume that the basic assumptions of the simple linear regression model are reasonably met.

The paper suggests that the simple linear regression model is reasonable for describing the relationship between \(y=\) eggshell thickness (in micrometers) and \(x=\) egg length \((\mathrm{mm})\) for quail eggs. Suppose that the population regression line is \(y=0.135+0.003 x\) and that \(\sigma=0.005 .\) Then, for a fixed \(x\) value, \(y\) has a normal distribution with mean \(0.135+0.003 x\) and standard deviation \(0.005\). a. What is the mean eggshell thickness for quail eggs that are \(15 \mathrm{~mm}\) in length? For quail eggs that are \(17 \mathrm{~mm}\) in length? b. What is the probability that a quail egg with a length of \(15 \mathrm{~mm}\) will have a shell thickness that is greater than \(0.18 \mu \mathrm{m}\) ? c. Approximately what proportion of quail eggs of length \(14 \mathrm{~mm}\) has a shell thickness of greater than .175? Less than . 178 ?

Exercise \(13.21\) gave data on \(x=\) nerve firing frequency and \(y=\) pleasantness rating when nerves were stimulated by a light brushing stoke on the forearm. The \(x\) values and the corresponding residuals from a simple linear regression are as follows: a. Construct a standardized residual plot. Does the plot exhibit any unusual features? b. A normal probability plot of the standardized residuals follows. Based on this plot, do you think it is reasonable to assume that the error distribution is approximately normal? Explain.

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