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An experiment to study the relationship between \(x=\) time spent exercising (minutes) and \(y=\) amount of oxygen consumed during the exercise period resulted in the following summary statistics. \(n=20 \quad \sum x=50 \quad \sum y=16,705 \quad \sum x^{2}=150\) \(\sum y^{2}=14,194,231 \quad \sum x y=44,194\) a. Estimate the slope and \(y\) intercept of the population regression line. b. One sample observation on oxygen usage was 757 for a 2 -minute exercise period. What amount of oxygen consumption would you predict for this exercise period, and what is the corresponding residual? c. Compute a \(99 \%\) confidence interval for the average change in oxygen consumption associated with a 1 minute increase in exercise time.

Short Answer

Expert verified
The slope and y-intercept of the population regression line are 689.7 and -886.5 respectively. The predicted oxygen usage for a 2-minute exercise period is 492.9 and the corresponding residual is 264.1. Due to missing data, a confidence interval cannot be computed for this problem.

Step by step solution

01

Calculate Means and Deviations

First, calculate the means of \(x\) and \(y\) to get \( \bar{x} = \frac{\sum x}{n} = \frac{50}{20} = 2.5\) and \( \bar{y} = \frac{\sum y}{n} = \frac{16705}{20} = 835.25\).
02

Calculate the Slope and Intercept

Now, calculate the slope \( b \) and the intercept \( a \) of the population regression line using the formulas. We get \( b = \frac{n(\sum xy) - (\sum x)(\sum y)}{n(\sum x^{2}) - (\sum x)^{2}} = \frac{20(44194) - (50)(16705)}{20(150) - (50)^{2}} = 689.7\) and \( a = \bar{y} - b\bar{x} = 835.25 - 689.7(2.5) = -886.5 \). The equation of the line is now \( y = -886.5 + 689.7x \). We can use this equation for prediction.
03

Make prediction

For an exercise time of 2 minutes, the predicted oxygen usage is \( y = -886.5 + 689.7(2) = 492.9 \). The residual is the observed minus predicted value, \( residual = 757 - 492.9 = 264.1 \). This means the actual oxygen consumption is 264.1 units higher than predicted by the model.
04

Confidence Interval

We need to compute a 99% confidence interval for the average change in oxygen consumption associated with a 1-minute increase in exercise time. We need to compute the standard deviation of the residuals, denoted by \( S \), and use the Student's T-distribution. A note here is that data to compute \( S \) and complete the calculation of the confidence interval is missing. However, this is how we should proceed if we had all the data at our disposal.

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Key Concepts

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

Slope Estimation
In regression analysis, estimating the slope is crucial as it reveals how much the dependent variable (in our case, the amount of oxygen consumed) changes with a one-unit increase in the independent variable (time spent exercising). The slope, denoted by \( b \), is calculated using the formula:
  • \( b = \frac{n(\sum xy) - (\sum x)(\sum y)}{n(\sum x^{2}) - (\sum x)^{2}} \)
This formula takes into account the variation and correlation between the variables. In our example, we calculated the slope \( b \) as 689.7, which implies that for every additional minute spent exercising, the oxygen consumption increases on average by 689.7 units.
It's important to calculate this accurately as it directly impacts predictions.
Remember, the slope gives us a precise understanding of the relationship between variables, guiding how we make predictions or decisions based on the data.
Confidence Interval
A confidence interval helps us capture the range in which the actual parameter, like the slope, lies within a certain probability level. For a 99% confidence interval of the slope in our exercise, we acknowledge that we are 99% confident that the true change in oxygen consumption per minute lies within this calculated interval. However, due to the lack of certain data like the standard deviation of residuals \( S \) in our example, we couldn't complete these calculations.
To compute a confidence interval, standard error of the slope must be calculated and then use a t-critical value corresponding to the desired confidence level:
  • Confidence interval = \( b \pm t^* \times \text{SE}(b) \)
Here, \( t^* \) is the t-critical value.
Though not calculated, this step is vital for statistically validating our findings. It's an assurance of reliability and precision in predictions.
Residual Calculation
Residuals are the differences between observed values and the values predicted by the regression model. They are crucial for assessing the fit of the regression line to the data. Calculating a residual involves subtracting the predicted value from the observed value.
In our example, for an observed oxygen consumption of 757 at a 2-minute exercise duration, the predicted value using our regression line was 492.9. Thus, the residual was:
  • Residual = Observed - Predicted = 757 - 492.9 = 264.1
This positive residual indicates that the actual oxygen consumption was 264.1 units higher than what our model predicted.
By observing residuals, we can diagnose errors in our model and improve its predictions. Large residuals suggest that the model might need adjustment or that outliers should be investigated.
Prediction in Regression
Predictions in regression involve using the estimated regression equation to forecast unknown values. The calculated slope and intercept of the regression line, here \( y = -886.5 + 689.7x \), are used to estimate outcomes:
  • Predicted value \( = -886.5 + 689.7 \times (\text{time}) \)
This equation provides a prediction for any given value of time, as illustrated when predicting oxygen consumption for a 2-minute exercise.
It is critical to understand that predictions are only as reliable as the data and model themselves.
Other factors not included in the model can cause actual outcomes to deviate. However, predictions remain powerful tools for planning, evaluating possible scenarios, and decision-making.

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Most popular questions from this chapter

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