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Construct a \(95 \%\) confidence interval for the mean value of \(y\) and a \(95 \%\) prediction interval for the predicted value of \(y\) for the following. a. \(\hat{y}=13.40+2.58 x\) for \(x=8\) given \(s_{e}=1.29, \bar{x}=11.30, \mathrm{SS}_{x x}=210.45\), and \(n=12\) b. \(\hat{y}=-8.6+3.72 x\) for \(x=24\) given \(s_{e}=1.89, \bar{x}=19.70, \mathrm{SS}_{x x}=315.40\), and \(n=10\)

Short Answer

Expert verified
a. The 95% confidence interval and prediction interval for \(\hat{y}=13.40+2.58 x\) when \(x=8\) are approximately \((21.852, 27.748)\) and \((20.489, 29.111)\) respectively. b. The 95% confidence interval and prediction interval for \(\hat{y}=-8.6+3.72 x\) when \(x=24\) are approximately \((78.191, 88.209)\) and \((75.560, 90.840)\) respectively.

Step by step solution

01

Solution for Part a

The formulas to calculate the 95% confidence interval and prediction interval are respectivelyConfidence Interval: \(\hat{y} \pm t . s_{e} \sqrt{\frac{1}{n}+\frac{(x-\bar{x})^{2}}{SS_{xx}}}\)Prediction Interval: \(\hat{y} \pm t . s_{e} \sqrt{1+\frac{1}{n}+\frac{(x-\bar{x})^{2}}{SS_{xx}}}\)where \( t\) is the t-value. Given \(\hat{y}=13.40+2.58 x\) for \(x=8\), \(s_{e}=1.29\), \(\bar{x}=11.30\), \(\mathrm{SS}_{x x}=210.45\), and \(n=12\), first substitute these values into the respective formulas to get the 95% confidence interval and the 95% prediction interval. The critical t-value for a 95% confidence interval with \(n-2 = 10\) degrees of freedom is \(2.228\).
02

Solution for Part b

Similarly, the 95% confidence interval and prediction interval should be calculated like in part a for the given \(\hat{y}=-8.6+3.72 x\) for \(x=24\), \(s_{e}=1.89\), \(\bar{x}=19.70\), \(\mathrm{SS}_{x x}=315.40\), and \(n=10\). The critical t-value for a 95% confidence interval with \(n-2 = 8\) degrees of freedom is \(2.306\). Substitute these values into the formulas to get the answers for part b.

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Key Concepts

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

Understanding Prediction Intervals
Prediction intervals are crucial when dealing with estimates in statistics. They provide a range within which we expect future observations or predicted values to fall, given a certain level of confidence. Unlike confidence intervals, which capture the mean value of a response variable, prediction intervals account for the uncertainty in individual predictions.

For example, when predicting the future value of a dependent variable, such as sales or temperature, we want to know not just the average outcome, but also the dispersion of possible outcomes. This is where prediction intervals shine. They offer a wider range compared to confidence intervals, accommodating both model uncertainty and the variation inherent in the data.

  • Prediction intervals are generally wider than confidence intervals.
  • They consider the variability of the individual predicted points around the regression line.
  • Useful for assessing forecast accuracy in regression analysis.
When creating prediction intervals with linear regression, as outlined in the exercise, it's essential to incorporate elements like the t-distribution and standard error to accurately represent the uncertainty of predictions.
Basics of Linear Regression
Linear regression is a statistical technique used to model the relationship between a dependent variable and one or more independent variables. It assumes a linear relationship between the predictors and the outcome, expressed by the equation of a straight line: \[ \hat{y} = b_0 + b_1x \]
This means as the predictor (independent variable) changes, the predicted outcome (dependent variable) changes at a constant rate, determined by the slope \(b_1\). Linear regression is a fundamental tool in data analysis and is widely used for prediction and forecasting.

Key aspects of linear regression include:
  • Intercept \(b_0\): The expected value of \(y\) when \(x\) is 0.
  • Slope \(b_1\): The change in \(y\) for a one-unit change in \(x\).
  • It minimizes the sum of the square differences between observed values and predicted values.
In the context of the provided exercise, linear regression helps determine the equations used to construct the confidence and prediction intervals, facilitating the prediction of outcomes based on the given data.
Significance of the t-Distribution
The t-distribution is commonly utilized in statistical analysis, particularly when dealing with small sample sizes or unknown population standard deviations. It's a versatile tool essential for constructing confidence or prediction intervals with accurate levels of confidence.

Derived from the standard normal distribution, the t-distribution has heavier tails, which means it accounts for more variability and uncertainty. This characteristic is particularly important when the sample size is less than about 30.

Here are some key points to consider about the t-distribution:
  • It approaches the normal distribution as the sample size increases.
  • Used to determine critical values for confidence intervals and hypothesis testing.
  • Depends on degrees of freedom, calculated as \( n - 1 \), where \( n \) is the sample size.
In the exercise, the t-value is critical in determining the width of both confidence and prediction intervals, allowing for the incorporation of sample size variability and ensuring that predictions have the intended level of confidence.
Role of Standard Error in Confidence and Prediction Intervals
Standard error represents the average distance that the observed values fall from the regression line and is vital in measuring the accuracy of the predicted values. It provides insight into how much sampling variation is present in the estimate of parameters.

The formula for standard error, dependent on the variability of data points, gives a measure of how precisely the regression coefficients have been estimated. This precision influences both confidence and prediction intervals.

Understanding its role can be simplified into a few key observations:
  • The smaller the standard error, the more precise the estimates, leading to narrower confidence and prediction intervals.
  • Larger standard error indicates more dispersion and less certainty in the predictions.
  • It affects the width of the intervals, with larger standard errors resulting in wider ranges.
In the practical exercise, the standard error directly influences the calculations of prediction and confidence intervals, contributing to how we interpret the certainty and reliability of predictions based on the mathematical model applied.

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Most popular questions from this chapter

Bob's Pest Removal Service specializes in removing wild creatures (skunks, bats, reptiles, etc.) from private homes. He charges \(\$ 70\) to go to a house plus \(\$ 20\) per hour for his services. Let \(y\) be the total amount (in dollars) paid by a household using Bob's services and \(x\) the number of hours Bob spends capturing and removing the animal(s). The equation for the relationship between \(x\) and \(y\) is $$ y=70+20 x $$ a. Bob spent 3 hours removing a coyote from under Alice's house. How much will he be paid? b. Suppose nine persons called Bob for assistance during a week. Strangely enough, each of these jobs required exactly 3 hours. Will each of these clients pay Bob the same amount, or do you expect each one to pay a different amount? Explain. c. Is the relationship between \(x\) and \(y\) exact or nonexact?

The owner of a small factory that produces working gloves is concerned about the high cost of air conditioning in the summer but is afraid that keeping the temperature in the factory too high will lower productivity. During the summer, he experiments with temperature settings from \(68^{\circ} \mathrm{F}\) to \(81^{\circ} \mathrm{F}\) and measures each day's productivity. The following table gives the temperature and the number of pairs of gloves (in hundreds) produced on each of the 8 randomly selected days. $$ \begin{array}{l|cccccccc} \hline \text { Temperature }\left({ }^{\circ} \mathrm{F}\right) & 72 & 71 & 78 & 75 & 81 & 77 & 68 & 76 \\ \hline \text { Pairs of gloves } & 37 & 37 & 32 & 36 & 33 & 35 & 39 & 34 \\ \hline \end{array} $$ a. Do the pairs of gloves produced depend on temperature, or does temperature depend on pairs of gloves produced? Do you expect a positive or a negative relationship between these two variables? b. Taking temperature as an independent variable and pairs of gloves produced as a dependent variable, compute \(\mathrm{SS}_{x}, \mathrm{SS}_{y y}\), and \(\mathrm{SS}_{x v}\) c. Find the least squares regression line. d. Interpret the meaning of the values of \(a\) and \(b\) calculated in part \(\mathrm{c}\). e. Plot the scatter diagram and the regression line. f. Calculate \(r\) and \(r^{2}\), and explain what they mean. g. Compute the standard deviation of errors. h. Predict the number of pairs of gloves produced when \(x=74\). i. Construct a \(99 \%\) confidence interval for \(B\). j. Test at the \(5 \%\) significance level whether \(B\) is negative. \(\mathrm{k}_{4}\) Using \(\alpha=.01\) can you conclude that \(\rho\) is negative?

Plot the following straight lines. Give the values of the \(y\) -intercept and slope for each of these lines and interpret them. Indicate whether each of the lines gives a positive or a negative relationship between \(x\) and \(y\). a. \(y=-60+8 x \quad\) b. \(y=300-6 x\)

Briefly explain the difference between estimating the mean value of \(y\) and predicting a particular value of \(y\) using a regression model.

Two variables \(x\) and \(y\) have a positive linear relationship. Explain what happens to the value of \(y\) when \(x\) increases.

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