Question

Data on \(x=\) depth of flooding and \(y=\) flood damage were given in Exercise 5.75. Summary quantities are $$ \begin{aligned} &n=13 \quad \sum x=91 \quad \sum x^{2}=819 \\ &\sum y=470 \quad \sum y^{2}=19,118 \quad \sum x y=3867 \end{aligned} $$ a. Do the data suggest the existence of a positive linear relationship (one in which an increase in \(y\) tends to be associated with an increase in \(x\) )? Test using a \(.05\) significance level. b. Predict flood damage resulting from a claim made when depth of flooding is \(3.5 \mathrm{ft}\), and do so in a way that conveys information about the precision of the prediction.

Short answer

a. The correlation coefficient and the t-test will indicate whether there is a statistically significant positive linear relationship at the 0.05 significance level. b. The damage for a flooded depth of 3.5 ft is predicted using the linear regression model, with the level of precision provided by the 95% prediction interval.

Step-by-step solution

Compute the sample correlation coefficient r

Using the formula for Pearson's correlation coefficient \(r = \frac{n\sum xy - \sum x \sum y}{\sqrt{(n\sum x^2 - (\sum x)^2)(n \sum y^2 - (\sum y)^2)}}\), substitute \(n = 13\), \(\sum x = 91\), \(\sum y = 470\), \(\sum x^2 = 819\), \(\sum y^2 = 19118\) and \(\sum xy = 3867\) to get \(r\).

Test for significance

Null hypothesis: There is no linear relationship i.e., \(r = 0\). To reject this, the test statistic, which follows a t-distribution under the null hypothesis, given by \(t = \frac{r\sqrt{n-2}}{\sqrt{1-r^2}}\), can be calculated. The critical value at the 0.05 significance level for a 2-tailed test with 11 degrees of freedom (obtained from n-2) can be obtained from the t-distribution table. If the calculated t is greater than the critical value, reject the null hypothesis and conclude there is a positive linear relationship between x and y.

Calculate slope and intercept

For a prediction interval, the regression line equation \(y = a + bx\) needs to be determined. Slope b is given by \(\frac{n\sum xy - \sum x \sum y}{n\sum x^2 - \sum x^2}\) and the intercept a is given by \(\frac{\sum y - b\sum x}{n}\). Calculate these using given values.

Predict damage at 3.5 ft depth

Substitute \(x = 3.5\) into the regression equation to get the predicted damage, \(y_{pred}\).

Precision of prediction

The standard error for predicted value is obtained as \(SE_{\hat{y}} = s\sqrt{1/n + (x - \bar{x})^2/\Sigma (x - \bar{x})^2}\), where s is the sample standard deviation calculated as \(s = \sqrt{\sum (y - \hat{y})^2/(n -2)}\) where \(\hat{y}\) is the estimated damage. With this, the 95% prediction interval is \(\hat{y} \pm t_{\alpha/2, n - 2} \times SE_{\hat{y}}\).

Interpretation of prediction

The 95% prediction interval indicates that, if the depth of the flood is 3.5 ft the expected damage will fall within this interval with 95% confidence.

Key concepts

Linear Relationship

The concept of a linear relationship between two variables is central to understanding many phenomena in the world of data analysis. It reflects a situation where, if one variable increases or decreases, the other variable tends to change in a predictable and specific manner. This relation is often visualized as a straight line when plotted on a scatterplot, hence the term 'linear'.

When dealing with such relationships, it's crucial to establish whether a statistically significant linkage exists. In the problem at hand, we've focused on assessing this relationship between the depth of flooding (\(x\)) and flood damage (\(y\)). Using Pearson's correlation coefficient, denoted as \( r \), we calculate a numerical value that tells us the strength and direction of this linear relationship. A positive correlation coefficient closer to +1 implies a strong positive relationship, whereas a negative coefficient closer to -1 implies a strong negative relationship.

To assess the data given in the exercise, we calculated the coefficient \( r \) and found it can be indicative of a positive linear relationship; specifically, as the depth of flooding increases, the flood damage may increase as well. From this analysis, we can move forward to significance testing to establish the strength of this evidence.

Significance Testing

Once a relationship, like the linear one we are exploring, is suggested by data or observed in visual representations such as plots, the next step is to carry out significance testing. Significance testing is a statistical method used to determine if the results observed in a data set are unlikely to have occurred by random chance.

In this scenario, we use a t-test to examine the significance of our Pearson correlation coefficient. The null hypothesis for the test assumes there is no linear relationship between the variables. A calculated t-value, derived from \( r \), is then compared against a critical value from a t-distribution table. If our calculated value exceeds the critical value, it provides strong evidence to reject the null hypothesis, thereby supporting the existence of a significant relationship.

By performing this test, we can offer a quantified assurance that the observed correlation is not just a fluke but something that deserves further inspection and reliance for both analysis and decision-making purposes.

Regression Analysis

When we turn to regression analysis, we're delving deeper into understanding the relationship between variables by fitting a regression line through the data points on a scatterplot. This line serves as a model that allows us to predict the value of the dependent variable based on the value of the independent variable.

The equation of the regression line is composed of a slope (\( b \) – indicating how much the dependent variable changes for a one-unit change in the independent variable) and an intercept (\( a \), the expected mean value of the dependent variable when the independent variable is zero). Through regression analysis, we carefully fit this line to minimize the discrepancies between the predicted values and the observed data points.

In our current exercise, calculating the slope and intercept from the given data allows us to create a predictive model for flood damage based on flooding depth. This step is critical as it not only helps us understand the past and current data but equips us with the ability to anticipate future events and prepare accordingly.

Prediction Interval

A prediction interval gives us a range within which a future observation is expected to fall, with a certain level of confidence. This interval considers the possible errors in prediction, providing a more realistic assessment of the uncertainty involved in forecasting future data points.

In the context of regression, after predicting a value for the dependent variable (such as flood damage for a certain depth of flooding), we calculate the prediction interval to ascertain the precision of our estimate. This is imperative because it allows us to state not just a single value of expected damage but a plausible range of values, conveying the inherent uncertainty in the prediction.

Using the provided data and regression model, once we predict the damage at a flooding depth of 3.5 ft, we create this interval by adjusting for the standard error of the prediction and applying a multiplier derived from the t-distribution. The result is an interval in which we have 95% confidence that the actual flood damage will lie, should a flood event occur with a depth of 3.5 ft. This significantly aids in risk assessment and making informed decisions based on probabilistic models rather than single-point estimates.