/*! 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 64 The accompanying data on \(x=\ma... [FREE SOLUTION] | 91Ó°ÊÓ

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The accompanying data on \(x=\mathrm{UV}\) transparency index and \(y=\) maximum prevalence of infection was read from a graph in the article "Solar Radiation Decreases Parasitism in Daphnia" (Ecology Letters, 2012: 47-54): $$ \begin{array}{l|ccccccccc} x & 1.3 & 1.4 & 1.5 & 2.0 & 2.2 & 2.7 & 2.7 & 2.7 & 2.8 \\ \hline y & 16 & 3 & 32 & 1 & 13 & 0 & 8 & 16 & 2 \\ x & 2.9 & 3.0 & 3.6 & 3.8 & 3.8 & 4.6 & 5.1 & 5.7 \\ \hline y & 1 & 7 & 36 & 25 & 10 & 35 & 58 & 56 \end{array} $$ Summary quantities include \(S_{x x}=25.5224, S_{y y}=\) \(5593.0588\), and \(S_{x y}=264.4882 .\) a. Calculate and interpret the value of the sample correlation coefficient. b. If you decided to fit the simple linear regression model to this data, what proportion of observed variation in maximum prevalence could be explained by the model relationship? c. If you decided to regress UV transparency index on maximum prevalence (i.e., interchange the roles of \(x\) and \(y\) ), what proportion of observed variation could be attributed to the model relationship? d. Carry out a test of \(H_{0}: \rho=.5\) versus \(H_{\mathrm{a}}: \rho>.5\) using a significance level of .05.

Short Answer

Expert verified
a) r = 0.7004; b) 49.06%; c) 49.06%; d) Reject \( H_0 \).

Step by step solution

01

Calculate the sample correlation coefficient (r)

The sample correlation coefficient is calculated using the formula:\[ r = \frac{S_{xy}}{\sqrt{S_{xx} \cdot S_{yy}}} \]Substitute the given values: \( S_{xy} = 264.4882 \), \( S_{xx} = 25.5224 \), and \( S_{yy} = 5593.0588 \). Calculate the denominator first:\[ \sqrt{25.5224 \times 5593.0588} = 377.5472 \]Then, calculate \( r \):\[ r = \frac{264.4882}{377.5472} \approx 0.7004 \]This indicates a positive, moderately strong linear relationship between the UV transparency index and the maximum prevalence of infection.
02

Determine the proportion of variation explained by the regression model

The proportion of variation explained by the model, also called the coefficient of determination, is \( R^2 \) and can be calculated as:\[ R^2 = r^2 \]Substitute the value of \( r \):\[ R^2 = (0.7004)^2 \approx 0.4906 \]About 49.06% of the observed variation in maximum prevalence can be explained by the regression model with the UV transparency index as the predictor.
03

Calculate the proportion of variation for regressing UV transparency on prevalence

When swapping the roles of \( x \) and \( y \), the correlation coefficient remains the same. Hence, the proportion of explained variation \( R^2 \) is unchanged:\[ R^2 = 0.7004^2 = 0.4906 \]Thus, about 49.06% of the observed variation in the UV transparency index can be attributed to the model when regressing on maximum prevalence.
04

Hypothesis test for the correlation coefficient

To test \( H_0: \rho = 0.5 \) versus \( H_a: \rho > 0.5 \), use the test statistic for the correlation coefficient:\[ t = \frac{r \sqrt{n-2}}{\sqrt{1-r^2}} \]Here, \( n = 17 \) (as there are 17 data points), and \( r = 0.7004 \).Calculate \( t \):\[ t = \frac{0.7004 \sqrt{17-2}}{\sqrt{1-(0.7004)^2}} = \frac{0.7004 \times 3.872}{\sqrt{1-0.4906}} \approx 3.872 \times 1.395 = 5.398 \]With 15 degrees of freedom, compare the calculated \( t \) value to the critical value for a one-tailed test at \( \alpha = 0.05 \). The critical value from a t-distribution table is approximately 1.753. Since 5.398 > 1.753, we reject \( H_0 \). There is sufficient evidence to conclude that \( \rho > 0.5 \).

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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 method used to model the relationship between a dependent variable and one or more independent variables. When working with linear regression, the idea is to find the best-fitting line through your data points, which minimizes the differences between the predicted values and the actual data values. This line is known as the regression line.

In the context of this data involving the UV transparency index as the independent variable (\( x \)) and the maximum prevalence of infection as the dependent variable (\( y \)), linear regression is being used to predict how changes in UV transparency can influence the prevalence of infection. The slope or gradient of the line reveals the direction and strength of the relationship. If it's a positive slope, as found here, it means that as the UV transparency index increases, the maximum prevalence of infection also tends to increase, although not necessarily at the same rate.

The calculation of the line's parameters includes determining the slope (\( b \)) and the y-intercept (\( a \)), which define the equation of the line:\[y = a + bx\]The formula helps us make predictions or analyze the correlation strength between the variables.
Hypothesis Testing
In hypothesis testing, you start with a null hypothesis and an alternative hypothesis. The goal is to determine which of these two hypotheses best fits the data. For this exercise, the null hypothesis (\( H_0 \)) asserts that the population correlation coefficient is 0.5, and the alternative hypothesis (\( H_a \)) suggests it is greater than 0.5.

To test these hypotheses, we compute a test statistic (in this case a t-statistic) and compare it with a critical value from statistical tables given a specific significance level. This test assesses whether the sample correlation (\( r \) = 0.7004) significantly exceeds 0.5. By using this approach, we determine if the observed correlation can be considered statistically significant or if it might have occurred by random chance.

For the data provided, the calculated t-value was 5.398, which surpassed the critical t-value for a one-tailed test at the given 0.05 significance level. Therefore, the null hypothesis is rejected, suggesting the evidence supports \( H_a \): the true correlation is indeed greater than 0.5.
Coefficient of Determination
The coefficient of determination, denoted as \( R^2 \), is a key statistic in linear regression that explains how much of the variance in the dependent variable can be predicted from the independent variable. It essentially tells us the percentage of data points that fall within the line of best fit.

In simpler terms, \( R^2 \) measures the strength and utility of the model. A higher \( R^2 \) value means a better fit for the line through the data points. For this specific exercise involving the UV transparency index and the maximum prevalence of infection, \( R^2 \) was calculated to be approximately 0.4906 or 49.06%.

This indicates that nearly half of the variability in the maximum prevalence of infection can be explained by changes in the UV transparency index. However, it also implies that there is still over 50% of the variation due to other factors not incorporated into the model, highlighting the complexity of factors influencing infection prevalence.

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Most popular questions from this chapter

The article "Quantitative Estimation of Clay Mineralogy in Fine-Grained Soils" (J. of Geotechnical and Geoenvironmental Engr., 2011: 997-1008) reported on various chemical properties of natural and artificial soils. Here are observations on \(x=\) cation exchange capacity (CEC, in meq/100 g) and \(y=\) specific surface area (SSA, in \(\mathrm{m}^{2} / \mathrm{g}\) ) of 20 natural soils. $$ \begin{array}{c|cccccccccc} x & 66 & 121 & 134 & 101 & 77 & 89 & 63 & 57 & 117 & 118 \\ \hline y & 175 & 324 & 460 & 288 & 205 & 210 & 295 & 161 & 314 & 265 \\ x & 76 & 125 & 75 & 71 & 133 & 104 & 76 & 96 & 58 & 109 \\ \hline y & 236 & 355 & 240 & 133 & 431 & 306 & 132 & 269 & 158 & 303 \end{array} $$ Minitab gave the following output in response to a request for \(r\) : Normal probability plots of \(x\) and \(y\) are quite straight. a. Carry out a test of hypotheses to see if there is a positive linear association in the population from which the sample data was selected. b. With \(n=20\), how small would the value of \(r\) have to be in order for the null hypothesis in the test of (a) to not be rejected at significance level .01? c. Calculate a confidence interval for \(\rho\) using a \(95 \%\) confidence level.

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