Question

1. below, results of linear regression analysis illustrate the correlation (dashed line) between latitudes and the average temperature from 17 weather stations. in the equation, t represents mean temperature and l represents latitude range.

a. is this an example of one-variable or two-variable data?

b. what type of relationship exists between temperature and latitude (positive, negative, or none)?

c. determine the value of the correlation coefficient (r-value).

d. based on your answer from c), is the correlation between temperature and latitude, moderate or weak/none?

e. use the equation from the linear regression to predict the temperature for a latitude of \\(50^{\circ}\text{n}\\).

Explanation:

Identify data variables

The problem asks whether the data is one-variable or two-variable. The graph plots Latitude (\(L\)) on the horizontal axis and Temperature (\(T\)) on the vertical axis. Because we are analyzing the relationship between two distinct numerical characteristics measured from the same weather stations, this is an example of two-variable (bivariate) data.

Determine relationship direction

We look at the trend of the data points and the slope of the regression line. The equation of the line is given as:

$$T = -0.395L + 34.426$$

The slope is \(-0.395\), which is negative. Visually, as the latitude increases, the temperature decreases. Therefore, a negative relationship exists between temperature and latitude.

Calculate correlation coefficient

We are given the coefficient of determination:

$$R^2 = 0.956$$

The correlation coefficient \(r\) is the square root of \(R^2\). Since the relationship is negative (as determined by the negative slope of the regression line), \(r\) must be negative:

$$r = -\sqrt{R^2} = -\sqrt{0.956} \approx -0.978$$

Evaluate correlation strength

The value of \(r \approx -0.978\) is extremely close to \(-1\). In statistics, an absolute value of \(|r|\) greater than \(0.8\) indicates a strong correlation. Therefore, the correlation between temperature and latitude is strong.

Predict temperature at latitude

We use the given linear regression equation:

$$T = -0.395L + 34.426$$

Substitute \(L = 50\) into the equation:

$$T = -0.395(50) + 34.426$$

$$T = -19.75 + 34.426 = 14.676^\circ\text{C}$$

Answer:

Question 3a

Two-variable data

Question 3b

Negative

Question 3c

\(r \approx -0.978\)

Question 3d

Strong

Question 3e

\(14.676^\circ\text{C}\) (or approximately \(14.7^\circ\text{C}\))