91Ó°ÊÓ

Construct a scatterplot, and find the value of the linear correlation coefficient \(r\). Also find the P-value or the critical values of \(r\) from Table \(A-5 .\) Use a significance level of \(\alpha=0.05 .\) Determine whether there is sufficient evidence to support a claim of a linear correlation between the two variables. (Save your work because the same data sets will be used in Section \(10-2\) exercises.) Listed below are annual data for various years. The data are weights (metric tons) of lemons imported from Mexico and U.S. car crash fatality rates per 100,000 population [based on data from "The Trouble with QSAR (or How I Learned to Stop Worrying and Embrace Fallacy)," by Stephen Johnson, Journal of Chemical Information and Modeling, Vol. 48, No. 1]. Is there sufficient evidence to conclude that there is a linear correlation between weights of lemon imports from Mexico and U.S. car fatality rates? Do the results suggest that imported lemons cause car fatalities? $$\begin{array}{|l|c|c|c|c|c|} \hline \text { Lemon Imports } & 230 & 265 & 358 & 480 & 530 \\ \hline \text { Crash Fatality Rate } & 15.9 & 15.7 & 15.4 & 15.3 & 14.9 \\ \hline \end{array}$$

Step by step solution

01

Create a scatterplot

Plot the lemon imports (in metric tons) on the x-axis and the crash fatality rates on the y-axis. For this, plot the pairs (230, 15.9), (265, 15.7), (358, 15.4), (480, 15.3), and (530, 14.9).

02

Calculate the linear correlation coefficient (r)

Use the formula for the linear correlation coefficient:\[ r = \frac{n(\sum xy) - (\sum x)(\sum y)}{\sqrt{[n(\sum x^2) - (\sum x)^2][n(\sum y^2) - (\sum y)^2]}} \]Compute the necessary summations: \(\sum x\), \(\sum y\), \(\sum xy\), \(\sum x^2\), \(\sum y^2\) and substitute into the formula. Here, \(n=5\).

03

Step 2.1: Compute the necessary sums

\[\sum x = 230 + 265 + 358 + 480 + 530 = 1863\]\[\sum y = 15.9 + 15.7 + 15.4 + 15.3 + 14.9 = 77.2\]\[\sum xy = (230 \cdot 15.9) + (265 \cdot 15.7) + (358 \cdot 15.4) + (480 \cdot 15.3) + (530 \cdot 14.9) = 3661.2 + 4160.5 + 5513.2 + 7344 + 7897 = 28576.9\]\[\sum x^2 = 230^2 + 265^2 + 358^2 + 480^2 + 530^2 = 52900 + 70225 + 128164 + 230400 + 280900 = 762589\]\[\sum y^2 = 15.9^2 + 15.7^2 + 15.4^2 + 15.3^2 + 14.9^2 = 252.81 + 246.49 + 237.16 + 234.09 + 222.01 = 1192.56\]

04

Step 2.2: Substitute into the formula

\[ r = \frac{5 \times 28576.9 - 1863 \times 77.2}{\sqrt{[5 \times 762589 - 1863^2][5 \times 1192.56 - 77.2^2]}} \]First, simplify the terms inside the square root:\[ 5 \times 762589 = 3812945\]\[ 1863^2 = 3471969\]\[ 5 \times 1192.56 = 5962.8\]\[ 77.2^2 = 5959.84\]Then, compute:\[ r = \frac{5 \times 28576.9 - 1863 \times 77.2}{\sqrt{(3812945 - 3471969) (5962.8 - 5959.84)}} = \frac{142884.5 - 143877.6}{\sqrt{340976 \times 2.96}} = \frac{-993.1}{999.979} = -0.99 \]

05

Find the critical values from Table A-5

With \(\alpha = 0.05\) and degrees of freedom \( df = n - 2 = 5 - 2 = 3 \), the table value for \( r \) at a 0.05 significance level is \( \pm 0.878 \).

06

Compare the correlation coefficient with the critical values

Since the absolute value of the calculated \( r = -0.99 \) is greater than the critical value 0.878, this indicates significant linear correlation.

07

Draw conclusions

The evidence supports a linear correlation between lemon imports and car fatality rates. However, correlation does not imply causation, so it cannot be concluded that lemon imports cause car fatalities.

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Key Concepts

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

scatterplot

A scatterplot is a graph used to display and compare data with two variables. Each point in the scatterplot represents a pair of values (x, y), showing how one variable relates to the other. To construct a scatterplot:
1. Plot the independent variable (like lemon imports) on the x-axis.
2. Plot the dependent variable (like car crash fatality rates) on the y-axis.
3. Each pair of values (e.g., (230, 15.9)) corresponds to a point on the graph.
By examining the scatterplot, you can get a sense of the relationship between the two variables. For instance, you'll observe the trend of whether higher lemon imports seem to coincide with higher, lower, or stable car crash fatality rates.

linear correlation coefficient

The linear correlation coefficient, often denoted as 'r', measures the strength and direction of a linear relationship between two variables.

Values of 'r' range between -1 and 1:
1. If r = 1, there is a perfect positive linear correlation.
2. If r = -1, there is a perfect negative linear correlation.
3. If r = 0, there is no linear correlation.

The formula for 'r' is:

P-value

In statistical hypothesis testing, the P-value helps determine the significance of results. It's the probability of observing results as extreme as, or more extreme than, the ones observed, under the assumption the null hypothesis is true.
A low P-value (< 0.05) indicates strong evidence against the null hypothesis, suggesting the observed data is unlikely to have occurred by random chance.

significance level

The significance level, represented by α (alpha), is the threshold set by researchers to determine whether the null hypothesis should be rejected. Common significance levels are 0.05, 0.01, and 0.10. In this case, α = 0.05:
1. If the P-value < α, reject the null hypothesis.
2. If the P-value ≥ α, do not reject the null hypothesis.

Using this framework helps ensure that conclusions drawn from data analysis are statistically justified.

causation vs correlation

Understanding the difference between causation and correlation is crucial.

Correlation means there is a relationship between two variables, but it doesn't answer whether one variable causes the other to change. For instance, lemon imports and car fatality rates could be correlated, but it doesn't mean lemon imports cause car fatalities.

Causation indicates that one event is the result of the occurrence of the other event; for example, smoking causes lung cancer. Establishing causation usually requires controlled experiments and deeper analysis beyond correlation.