/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 Exercise 13.10 presented \(y=\) ... [FREE SOLUTION] | 91Ó°ÊÓ

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Exercise 13.10 presented \(y=\) hardness of molded plastic and \(x=\) time elapsed since the molding was completed. Summary quantities included $$ n=15 \quad b=2.50 \quad \text { SSResid }=1235.470 $$ \(\sum(x-\vec{x})^{2}=4024.20\) a. Calculate the estimated standard deviation of the statistic \(b\) b. Obtain a \(95 \%\) confidence interval for \(\beta,\) the slope of the population regression line. c. Does the interval in Part (b) suggest that \(\beta\) has been precisely estimated? Explain.

Short Answer

Expert verified
The short answer depends on the results of the calculations. The estimated standard deviation of the statistic \(b\) (SEb) and the confidence interval for \(\beta\) can be calculated using the provided formula and data. The precision of estimation can be inferred based on whether zero is included in the confidence interval.

Step by step solution

01

Calculate the standard error of the slope (SEb)

The SEb can be calculated using the formula: \(SEb = \sqrt{SSResid/(n-2)}/\sqrt{\sum(x-\vec{x})^{2}}\) Where: \(SSResid = 1235.470\), \(n = 15\), and \(\sum(x-\vec{x})^{2} = 4024.20\). Insert these values and calculate to get the SEb.
02

Calculate the Confidence Interval for \(\beta\)

The 95% confidence interval for \(\beta\) is calculated as \(b \pm t_{(n-2)}*\) SEb. Where \(b=2.5\) is the estimated slope and \(t_{(n-2)}\) the critical value from t-distribution table for the confidence level of 95% with \(n - 2 = 13\) degrees of freedom. Calculate the confidence interval by substituting all the known values to this equation.
03

Interpretation of the Confidence Interval

The confidence interval calculated gives an estimated range of values which is likely to include the true population parameter \(\beta\). If this interval includes zero, it suggests that \(\beta\) is not significantly different from zero. Therefore, it implies that there is no statistically significant relationship between the variables. If zero is not in the interval, it suggests that \(\beta\) has been precisely estimated and there is a significant relationship between the variables.

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Key Concepts

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

Standard Error of the Slope
When performing regression analysis, understanding the standard error of the slope, abbreviated as SEb, is critical. SEb measures the amount of variability in the estimate of the slope coefficient (\beta) in the regression equation. In simpler terms, it tells us how much the slope you calculated from your sample data might vary if you were to take different samples from the same population.

To calculate SEb, you take the square root of the sum of squared residuals (SSResid), divided by the degrees of freedom (n - 2), which then is divided by the sum of the squared differences of the predictor variable (x) from its mean (\bar{x}). In mathematical notation, the formula is:
\[ SEb = \sqrt{\frac{SSResid}{(n-2)}}/\sqrt{\sum(x-\bar{x})^{2}} \]
The smaller the SEb, the more precise is the estimate of the slope. A large SEb implies that the slope estimate is less reliable. Knowing the standard error helps determine the margin of error and construct confidence intervals around the estimated slope, which are essential to make inferences about the population from sample data.
Confidence Interval
The confidence interval is a range constructed around the slope estimate from a sample that likely contains the true slope of the population regression line. It's a way to measure the precision of your estimate and is useful for making predictions and testing hypotheses. The 95% confidence interval means we're 95% confident that the interval includes the actual value of the slope in the population.

To compute a confidence interval for the slope, we use the following formula:
\[ b \pm t_{(n-2)} * SEb \]
Here, b is the estimated slope from the sample, and t_{(n-2)} is the critical value based on the t-distribution for the chosen confidence level, in this case, 95%, with n - 2 degrees of freedom. The SEb is the standard error of the slope we previously discussed. This interval provides bounds within which we expect the population slope to fall, giving us a way to gauge the estimate's reliability.
Population Regression Line
The population regression line is what we aim to estimate when we perform regression analysis on a sample. It represents the relationship between the predictor (x) and response (y) variable in the entire population. The equation of the true population regression line looks like:
\[ y = \beta_0 + \beta_1x \]
where \(\beta_0\) is the true intercept and \(\beta_1\) is the true slope of the line. As these are the true population parameters, we typically don't know them and estimate them using sample data. The estimated regression line we compute from a sample is thus an approximation meant to come as close as possible to this ideal line. It's important to note that every sample might give us a slightly different regression equation. Therefore, the concept of a confidence interval for the slope, as discussed above, is vital for understanding how well our sample estimates might match the true population parameters.
T-distribution
The t-distribution plays a fundamental role in estimating the population parameters from a sample. It's especially important when the sample size is small, which is commonly the case in practical scenarios. The t-distribution resembles the normal distribution but has heavier tails, which means it is more prone to producing values that fall far from its mean. This is useful when you are dealing with uncertainty and small sample sizes because it accounts for that variability.

In the context of constructing confidence intervals, the t-distribution is used to provide the critical value (t-value) needed to calculate the margin of error. This critical value varies based on the chosen confidence level (e.g., 95%) and the degrees of freedom, which is typically the sample size minus two for simple linear regression. With a larger sample size, the t-distribution becomes similar to the normal distribution. The t-distribution ensures that the confidence interval accurately reflects the increased uncertainty in the estimate when we have fewer data points.

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Most popular questions from this chapter

A simple linear regression model was used to describe the relationship between sales revenue \(y\) (in thousands of dollars) and advertising expenditure \(x\) (also in thousands of dollars) for fast-food outlets during a 3 -month period. A sample of 15 outlets yielded the accompanying summary quantities. $$ \begin{array}{l} \sum x=14.10 \sum y=1438.50 \quad \sum x^{2}=13.92 \\ \sum y^{2}=140,354 \quad \sum x y=1387.20 \\ \sum(y-\vec{y})^{2}=2401.85 \quad \sum(y-\hat{y})^{2}=561.46 \end{array} $$ a. What proportion of observed variation in sales revenue can be attributed to the linear relationship between revenue and advertising expenditure? b. Calculate \(s_{e}\) and \(s_{b}\). c. Obtain a \(90 \%\) confidence interval for \(\beta,\) the average change in revenue associated with a \(\$ 1000\) (that is, 1-unit) increase in advertising expenditure.

Exercise 13.16 described a regression analysis in which \(y=\) sales revenue and \(x=\) advertising expenditure. Summary quantities given there yield \(n=15 \quad b=52.27 \quad s_{b}=8.05\) a. Test the hypothesis \(H_{0}: \beta=0\) versus \(H_{x}: \beta \neq 0\) using a significance level of .05. What does your conclusion say about the nature of the relationship between \(x\) and \(y\) ? b. Consider the hypothesis \(H_{0}: \beta=40\) versus \(H_{A} \cdot \beta>\) 40\. The null hypothesis states that the average change in sales revenue associated with a 1 -unit increase in advertising expenditure is (at most) \(\$ 40,000\). Carry out a test using significance level .01 .

The article "Performance Test Conducted for a Gas Air-Conditioning System" (American Society of Heating, Refrigerating. and Alr Conditioning Engineering [1969]: 54\()\) reported the following data on maximum outdoor temperature \((x)\) and hours of chiller operation per day \((y)\) for a 3 -ton residential gas air- conditioning system: \(\begin{array}{rrrrrrr}x & 72 & 78 & 80 & 86 & 88 & 92 \\ y & 4.8 & 7.2 & 9.5 & 14.5 & 15.7 & 17.9\end{array}\) Suppose that the system is actually a prototype model, and the manufacturer does not wish to produce this model unless the data strongly indicate that when maximum outdoor temperature is \(82^{\circ} \mathrm{F}\), the true average number of hours of chiller operation is less than 12 . The appropriate hypotheses are then \(H_{0}: \alpha+\beta(82)=12 \quad\) versus \(\quad H_{d}: \alpha+\beta(82)<12\) Use the statistic \(t=\frac{a+b(82)-12}{s_{a+b(82)}}\) which has a \(t\) distribution based on \((n-2)\) df when \(H_{0}\) is true, to test the hypotheses at significance level .01 .

A sample of \(n=61\) penguin burrows was selected, and values of both \(y=\) trail length \((\mathrm{m})\) and \(x=\) soil hardness (force required to penetrate the substrate to a depth of \(12 \mathrm{~cm}\) with a certain gauge, in \(\mathrm{kg}\) ) were determined for each one ("Effects of Substrate on the Distribution of Magellanic Penguin Burrows," The Auk [1991]: \(923-933\) ). The equation of the least-squares line was \(\hat{y}=11.607-1.4187 x,\) and \(r^{2}=.386 .\) a. Does the relationship between soil hardness and trail length appear to be linear, with shorter trails associated with harder soil (as the article asserted)? Carry out an appropriate test of hypotheses. b. Using \(s_{\mathrm{e}}=2.35, \bar{x}=4.5,\) and \(\sum(x-\bar{x})^{2}=250,\) predict trail length when soil hardness is 6.0 in a way that conveys information about the reliability and precision of the prediction. c. Would you use the simple linear regression model to predict trail length when hardness is \(10.0 ?\) Explain your reasoning

A random sample of \(n=347\) students was selected, and each one was asked to complete several questionnaires, from which a Coping Humor Scale value \(x\) and a Depression Scale value \(y\) were determined ("Depression and Sense of Humor." (Psychological Reports \([1994]: 1473-1474)\). The resulting value of the sample correlation coefficient was -.18 . a. The investigators reported that \(P\) -value \(<.05 .\) Do you agree? b. Is the sign of \(r\) consistent with your intuition? Explain. (Higher scale values correspond to more developed sense of humor and greater extent of depression.) c. Would the simple linear regression model give accurate predictions? Why or why not?

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