/*! 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 46 The article reported the followi... [FREE SOLUTION] | 91Ó°ÊÓ

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The article 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 $$ H_{0}: \alpha+\beta(82)=12 \text { versus } H_{a}: \alpha+\beta(82)<12 $$ 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 \(.01\).

Short Answer

Expert verified
To provide a short answer without having the actual calculations to hand, carry out the hypothesis test as outlined in the steps above. If the calculated t-statistic under the null hypothesis is less than the negative of the critical t value (which depends on (n-2) degrees of freedom and the 0.01 significance level), the null hypothesis is rejected, indicating that the chiller operates less than 12 hours on average when the outdoor temperature is 82 F.

Step by step solution

01

- Calculate the Coefficients of Linear Regression

Using the method of least squares, calculate the coefficients \(\alpha\) (intercept) and \(\beta\) (slope) for the linear regression model, where \(\alpha = \frac{n(\Sigma xy) - (\Sigma x)(\Sigma y)}{n \Sigma (x^2) - (\Sigma x)^2}\) and \(\beta = \frac{n \Sigma x^2 - (\Sigma x)^2}{n \Sigma (x^2) - (\Sigma x)^2}\). Use the given x and y values in these formulas to find the values of \(\alpha\) and \(\beta\).
02

- Calculate the t statistic

Insert the values of \(\alpha\) and \(\beta\) into the t statistic formula provided, \(t = \frac{{\alpha+\beta(82)-12}}{{s_{\alpha+\beta(82)}}}\), and compute the t value. \(s_{\alpha+\beta(82)}\) is the estimated standard error and can be found using the formula \(s_{\alpha+\beta(82)} = \sqrt{\frac{{\Sigma y^2 - \beta (\Sigma xy) - \alpha (\Sigma y)}}{{n - 2}}}\).
03

- Compare with the t critical value

The critical value for the t-distribution with \(n-2\) degrees of freedom at a significance level of 0.01 can be determined from a t-distribution table. If the calculated t-statistic is less than the negative critical t value, then reject the null hypothesis due to sufficient evidence that the system will operate less than 12 hours on average when the maximum temperature is 82 F.

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Key Concepts

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

Linear Regression
Linear regression is a statistical method used to model the relationship between a dependent variable and one or more independent variables by fitting a linear equation to observed data. This equation predicts the dependent variable value based on the independent variable(s).

The linear regression equation is given by \( y = \alpha + \beta x + \epsilon \) where \( y \) is the dependent variable, \( x \) is the independent variable, \( \alpha \) is the y-intercept, \( \beta \) is the slope of the line, and \( \epsilon \) is the error term. The slope, \( \beta \), indicates the change in \( y \) for a one-unit change in \( x \).

In the exercise, a simple linear regression is used to assess the relationship between outdoor temperature (independent variable) and hours of chiller operation (dependent variable) for a residential air-conditioning system.

To ensure clarity and ease of understanding, it's important to visualize the relationship by plotting the data points on a graph, with the vertical axis representing the hours of chiller operation and the horizontal axis for the temperature. Then, a line is drawn that best fits these points, which is the essence of the linear regression model.
t Distribution
The t distribution, also known as Student’s t distribution, is a probability distribution that arises when estimating the mean of a normally distributed population in situations where the sample size is small and population standard deviation is unknown.

It is symmetrical and bell-shaped, similar to the normal distribution, but with heavier tails, meaning it has a greater tendency to produce values that fall far from its mean. This property makes the t distribution particularly useful for hypothesis testing with small sample sizes.

When performing hypothesis testing, such as determining if the true average number of hours of chiller operation is less than 12, we use the t distribution to find the critical value associated with our significance level. If our calculated test statistic falls within the critical region (beyond the critical value), we have enough evidence to reject our null hypothesis. In the context of our exercise, we're testing at a significance level of 0.01, which means we're looking for strong evidence before we make a conclusion about the average chiller operation hours.
Least Squares Method
The least squares method is a statistical technique used to find the best-fitting curve to a given set of points by minimizing the sum of the squares of the offsets (the distances between the points and the curve).

This method is central to linear regression and helps in determining the line that best fits the observed data. By minimizing these offsets or residuals, we are essentially minimizing the error between the predicted values and the actual values.

The beauty of the least squares method is its ability to provide the most accurate estimate of the parameters of the linear regression model, which are the intercept (\( \alpha \)) and the slope (\( \beta \)). The formulas given in the step by step solution directly apply the least squares method to determine these coefficients.

A practical improvement when explaining least squares in an educational context is incorporating graphical interpretations such as residual plots. This helps students visually assess the fit of the model and understand the concept more intuitively. Additionally, discussing the characteristics of 'good' vs. 'poor' fits and explaining why we minimize the squares of the residuals rather than their absolute values can also enhance comprehension.

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Most popular questions from this chapter

The accompanying data were read from a plot (and are a subset of the complete data set) given in the article . The data represent the mean response times for a group of individuals with closed-head injury (CHI) and a matched control group without head injury on 10 different tasks. Each observation was based on a different study, and used different subjects, so it is reasonable to assume that the observations are independent. \begin{tabular}{ccc} & \multicolumn{2}{l} { Mean Response Time } \\ \cline { 2 - 3 } Study & Control & CHI \\ \hline 1 & 250 & 303 \\ 2 & 360 & 491 \\ 3 & 475 & 659 \\ 4 & 525 & 683 \\ 5 & 610 & 922 \\ 6 & 740 & 1044 \\ 7 & 880 & 1421 \\ 8 & 920 & 1329 \\ 9 & 1010 & 1481 \\ 10 & 1200 & 1815 \\ \hline \end{tabular} a. Fit a linear regression model that would allow you to predict the mean response time for those suffering a closed-head injury from the mean response time on the same task for individuals with no head injury. b. Do the sample data support the hypothesis that there is a useful linear relationship between the mean response time for individuals with no head injury and the mean response time for individuals with CHI? Test the appropriate hypotheses using \(\alpha=.05\). c. It is also possible to test hypotheses about the \(y\) intercept in a linear regression model. For these data, the null hypothesis \(H_{0}: \alpha=0\) cannot be rejected at the \(.05\) significance level, suggesting that a model with a \(y\) intercept of 0 might be an appropriate model. Fitting such a model results in an estimated regression equation of \(\mathrm{CHI}=1.48(\) Control \()\) Interpret the estimated slope of \(1.48\).

A sample of \(n=10,000(x, y)\) pairs resulted in \(r=.022\). Test \(H_{0}: \rho=0\) versus \(H_{a^{\circ}} \rho \neq 0\) at significance level .05. Is the result statistically significant? Comment on the practical significance of your analysis.

\(13.26\) In anthropological studies, an important characteristic of fossils is cranial capacity. Frequently skulls are at least partially decomposed, so it is necessary to use other characteristics to obtain information about capacity. One such measure that has been used is the length of the lambda- opisthion chord. The article reported the accompanying data for \(n=7\) Homo erectus fossils. \(\begin{array}{llllllll}x \text { (chord } & 78 & 75 & 78 & 81 & 84 & 86 & 87\end{array}\) length in \(\mathrm{mm}\) ) \(\begin{array}{llllllll}\text { (capacity } & 850 & 775 & 750 & 975 & 915 & 1015 & 1030\end{array}\) in \(\mathrm{cm}^{3}\) ) Suppose that from previous evidence, anthropologists had believed that for each \(1-\mathrm{mm}\) increase in chord length, cranial capacity would be expected to increase by \(20 \mathrm{~cm}^{3}\). Do these new experimental data strongly contradict prior belief?

Explain the difference between \(r\) and \(\rho\).

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. 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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