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91Ó°ÊÓ

Construct a \(99 \%\) confidence interval for the mean value of \(y\) and a \(99 \%\) prediction interval for the predicted value of \(y\) for the following. a. \(\hat{y}=3.25+.80 x\) for \(x=15\) given \(s_{e}=.954, \bar{x}=18.52, \mathrm{SS}_{x x}=\) \(144.65\), and \(n=10\) b. \(\hat{y}=-27+7.67 x\) for \(x=12\) given \(s_{e}=2.46, \bar{x}=13.43, \mathrm{SS}_{x x}=\) \(369.77\), and \(n=10\)

Short Answer

Expert verified
Now, knowing the steps and methods, it is possible to find the confidence and prediction intervals. Actual numeric results would depend on the calculated t-score and these formulas applied with the given parameters. Therefore, please practice calculating the values for better understanding and improvement.

Step by step solution

01

- Calculate T-Score

First, using a t-distribution table or a t-distribution calculator, find the t-score for a \(99%\) confidence level with \(n-2\) degrees of freedom. For both \(a\) and \(b\), that would be \(10-2=8\) degrees of freedom.
02

- Calculate the Confidence Interval for a

Next, using the given parameters for \(a\), values need to be substituted in the confidence interval formula \[CI = \hat{y} \pm t_{score} * s_{e} * \sqrt{1/n + (x-\bar{x})^2/SS_{xx}}\] In this formula: \(\hat{y}\) would be the predicted value of \(y\) using the given regression equation with \(x=15\); \(t_{score}\) is the t-score calculated in step 1; \(s_{e}\) is the standard error (\(.954\)); \(n\) is the sample size (\(10\)); \(x\) is the value for which prediction is being made (\(15\)); \(\bar{x}\) is the sample mean (\(18.52\)); \(SS_{xx}\) is the sum of squares (\(144.65\)). The calculated values gives the confidence interval for mean value of \(y\).
03

- Calculate the Prediction Interval for a

Similar to Step 2, but apply the prediction interval formula:\[PI = \hat{y} \pm t_{score} * s_{e} * \sqrt{1 + 1/n + (x-\bar{x})^2/SS_{xx}}\]All parameters remain the same as in step 2. The result would be the prediction interval.
04

- Calculate the Confidence Interval for b

Apply the formulas from Step 2 with the parameters for \(b\), to determine the \(99%\) confidence interval. The parameters are \(\hat{y}\) for \(x=12\), \(s_{e}=2.46\), \(\bar{x}=13.43\), \(SS_{xx}=369.77\), and \(n=10\). This will provide the \(99%\) confidence interval for \(b\).
05

- Calculate the Prediction Interval for b

Apply the formulas from Step 3 with parameters for \(b\). This will give you the \(99%\) prediction interval for \(b\).

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Key Concepts

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

Prediction Interval
A prediction interval provides a range where we expect a future observation to fall, given certain conditions. It is wider than the confidence interval because it accounts for the uncertainty not only in the mean estimate but also the individual variability of future observations. To calculate a prediction interval, we use the formula: \[PI = \hat{y} \pm t_{score} \times s_e \times \sqrt{1 + \frac{1}{n} + \frac{(x-\bar{x})^2}{SS_{xx}}}\]Here:
  • \(\hat{y}\) is the predicted value from the regression equation.
  • \(t_{score}\) corresponds to the t-score for specified confidence level and degrees of freedom.
  • \(s_e\) is the standard error of the estimate.
  • \(n\) is the sample size.
  • \(x\) is the value for which prediction is made, and \(\bar{x}\) is the mean of the \(x\) values.
  • \(SS_{xx}\) is the sum of squared deviations of \(x\) values.
In practice, if you're making predictions about future events or measurements using your model, understanding the prediction interval is essential because it indicates the confidence range that captures the variability in individual responses.
Regression Analysis
Regression analysis is a statistical method used for estimating relationships among variables. It allows us to understand how the typical values of the dependent variable change when any one of the independent variables is varied.In linear regression, we try to fit a line that best describes the data with an equation of the form:\[\hat{y} = a + bx\]Here:
  • \(\hat{y}\) represents the predicted (dependent) variable.
  • \(a\) is the y-intercept of the regression line.
  • \(b\) is the slope of the line, showing the relationship between \(x\) and \(\hat{y}\).
  • \(x\) is the independent variable.
Through regression analysis, we can make predictions, identify correlations, and understand the strength of the effect of the independent variable(s) on the dependent variable. However, caution must be used in interpreting these relationships since they don't necessarily imply causation.
T-Distribution
The t-distribution is a probability distribution used in statistics when the sample size is small, and the population standard deviation is unknown. It is similar to the normal distribution but has heavier tails.The t-distribution is used to calculate the t-score, which is a critical component in constructing confidence and prediction intervals. Key points include:
  • As the sample size increases, the t-distribution approaches the normal distribution.
  • The shape of the t-distribution is determined by its degrees of freedom, calculated as the sample size minus one (minus\(n-1\)).
  • T-distribution takes into account variability in small samples, providing more conservative estimates of uncertainty.
In the calculation of both confidence and prediction intervals, the t-score from the t-distribution table is vital because it adjusts the interval width to account for sample variability and confidence level. This adjustment is critical for providing realistic estimates.
Standard Error
The standard error quantifies how much the sample mean’s estimated value likely differs from the true population mean. It indicates the average distance that the observed values fall from the regression line, implying the precision of the regression predictions.Mathematically, the standard error of the estimate is calculated by:\[s_e = \sqrt{\frac{\sum (y_i - \hat{y}_i)^2}{n-2}}\]Where:
  • \(s_e\) denotes the standard error.
  • \(y_i\) are the observed values.
  • \(\hat{y}_i\) are the predicted values.
  • \(n\) is the sample size.
The standard error is crucial because a smaller standard error signifies that the model predicts the data more accurately, whereas a larger standard error indicates less reliable predictions. In both confidence and prediction intervals, the standard error influences the width of the interval, reflecting the precision of the prediction or estimate.

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Most popular questions from this chapter

The following table gives the 2015 total payroll (in millions of dollars) and the percentage of games won during the 2015 season by each of the American League baseball teams. $$ \begin{array}{lcc} \hline \text { Team } & \begin{array}{c} \text { Total Payroll } \\ \text { (millions of dollars) } \end{array} & \begin{array}{c} \text { Percentage of } \\ \text { Games Won } \end{array} \\ \hline \text { Baltimore Orioles } & 110 & 50 \\ \text { Boston Red Sox } & 187 & 48 \\ \text { Chicago White Sox } & 115 & 47 \\ \text { Cleveland Indians } & 86 & 50 \\ \text { Detroit Tigers } & 174 & 46 \\ \text { Houston Astros } & 71 & 53 \\ \text { Kansas City Royals } & 114 & 59 \\ \text { Los Angeles Angels } & 151 & 53 \\ \text { Minnesota Twins } & 109 & 51 \\ \text { New York Yankees } & 219 & 54 \\ \text { Oakland Athletics } & 86 & 42 \\ \text { Seattle Mariners } & 120 & 47 \\ \text { Tampa Bay Rays } & 76 & 49 \\ \text { Texas Rangers } & 142 & 54 \\ \text { Toronto Blue Jays } & 123 & 57 \\ \hline \end{array} $$ Compute the linear correlation coefficient, \(\rho\). Does it make sense to make a confidence interval and to test a hypothesis about \(\rho\) here? Explain.

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 \(n=10\)

An economist is studying the relationship between the incomes of fathers and their sons or daughters. Let \(x\) be the annual income of a 30 -year-old person and let \(y\) be the annual income of that person's father at age 30 years, adjusted for inflation. A random sample of 300 thirty-year-olds and their fathers yields a linear correlation coefficient of \(.60\) between \(x\) and \(y .\) A friend of yours, who has read about this research, asks you several questions, such as: Does the positive value of the correlation coefficient suggest that the 30 -year-olds tend to earn more than their fathers? Does the correlation coefficient reveal anything at all about the difference between the incomes of 30 -year-olds and their fathers? If not, what other information would we need from this study? What does the correlation coefficient tell us about the relationship between the two variables in this example? Write a short note to your friend answering these questions.

The following data give the experience (in years) and monthly salaries (in hundreds of dollars) of nine randomly selected secretaries. $$ \begin{array}{l|rrrrrrrrr} \hline \text { Experience } & 14 & 3 & 5 & 6 & 4 & 9 & 18 & 5 & 16 \\ \hline \begin{array}{l} \text { Monthly } \\ \text { salary } \end{array} & 62 & 29 & 37 & 43 & 35 & 60 & 67 & 32 & 60 \\ \hline \end{array} $$ Construct a \(90 \%\) confidence interval for the mean monthly salary of all secretaries with 10 years of experience. Construct a \(90 \%\) prediction interval for the monthly salary of a randomly selected secretary with 10 years of experience.

The following data give information on the ages (in years) and the number of breakdowns during the last month for a sample of seven machines at a large company. $$ \begin{array}{l|rrrrrrr} \hline \text { Age (years) } & 12 & 7 & 2 & 8 & 13 & 9 & 4 \\ \hline \begin{array}{l} \text { Number of } \\ \text { breakdowns } \end{array} & 10 & 5 & 1 & 4 & 12 & 7 & 2 \\ \hline \end{array} $$ a. Taking age as an independent variable and number of breakdowns as a dependent variable, what is your hypothesis about the sign of \(B\) in the regression line? (In other words, do you expect \(B\) to be positive or negative?) b. Find the least squares regression line. Is the sign of \(b\) the same as you hypothesized for \(B\) in part a? c. Give a brief interpretation of the values of \(a\) and \(b\) calculated in part b. d. Compute \(r\) and \(r^{2}\) and explain what they mean. e. Compute the standard deviation of errors. f. Construct a \(99 \%\) confidence interval for \(B\). g. Test at a \(2.5 \%\) significance level whether \(B\) is positive. h. At a \(2.5 \%\) significance level, can you conclude that \(\rho\) is positive? Is your conclusion the same as in part g?

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