Question

How sensitive to changes in water temperature are coral reefs? To find out, scientists examined data on sea surface temperatures, in degrees Celsius, and mean coral growth, in centimeters per year, over a several‑year period at locations in the Gulf of Mexico and the Caribbean Sea. The table shows the data for the Gulf of Mexico.

Sea surface temperature 26.726.7 26.626.6 26.626.6 26.526.5 26.326.3 26.126.1
Growth 0.850.85 0.850.85 0.790.79 0.860.86 0.890.89 0.920.92
Find the correlation rr step by step. Round off to two decimals places in each step. First, find the mean and standard deviation of each variable. Then, find the six standardized values for each variable. Finally, use the formula for rr . Round your answer to three decimal places.

Enter these data into your calculator or software, and use the correlation function to find rr . Check that you get the same results as using the formula, up to round off error. Round your answer to three decimal places.

Answer 1

Answer:

2153.21574

Step-by-step explanation:

Answer 2

Correlation simply describes the association between two variables.

• The calculated correlation is -0.8110
• The software result is -0.8111

Given that:

$x:\{26.7, 26.7, 26.6, 26.6, 26.6, 26.6, 26.5, 26.5, 26.3,26.3, 26.1,26.1\}$

$y = \{0.85, 0.85, 0.85, 0.85, 0.79, 0.79, 0.86, 0.86, 0.89, 0.89, 0.92, 0.92\}$

(a) Calculate the correlation (r), manually

Start by calculating the mean of x and y

$\bar x = \frac{\sum x}{n}$

$\bar x = \frac{26.7+26.7+26.6+26.6+26.6+26.6+26.5+26.5+26.3+26.3+26.1+26.1}{12}$

$\bar x = \frac{317.6}{12}$

$\bar x = 26.47$

$\bar y = \frac{\sum y}{n}$

$\bar y = \frac{0.85+ 0.85+0.85+ 0.85+ 0.79+0.79+ 0.86+ 0.86+ 0.89+ 0.89+ 0.92+ 0.92}{12}$

$\bar y = \frac{10.32}{12}$

$\bar y = 0.86$

Calculate square of x deviation $SS_x$

$SS_x = \sum(x - \bar x)^2$

$SS_x = (26.7 - 26.47)^2+.....+(26.1 - 26.47)^2+(26.1 - 26.47)^2$

$SS_x = 0.5068$

Calculate square of y deviation $SS_y$

$SS_y = \sum(y - \bar y)^2$

$SS_y = (0.85 - 0.86)^2 + (0.85- 0.86)^2 +........+ (0.92- 0.86)^2$

$SS_y = 0.0192$

Calculate the sum of product of x and y deviation $\sum(x - \bar x) \times \sum(y - \bar y)$

$\sum(x - \bar x) \times \sum(y - \bar y) = (26.7-26.47) \times (0.85 - 0.86) + (26.7-26.47) \times (0.85- 0.86) + ......+(26.1-26.47) \times (0.92- 0.86)$

$\sum(x - \bar x) \times \sum(y - \bar y) = -0.08$

The coefficient (r) is then calculated as follows:

$r = \frac{\sum(x - \bar x) \times \sum(y - \bar y)}{\sqrt{SS_x \times SS_y}}$

$r = \frac{-0.08}{\sqrt{0.5068 \times 0.0192}}$

$r = \frac{-0.08}{\sqrt{0.00973056}}$

$r = \frac{-0.08}{0.0986436009}$

$r = -0.8110$

(b) Calculate the correlation (r), using a software

Using a software,

$r = -0.8111$

See attachment for the results

Hence, the correlation is -0.8110

Read more about correlation at:

https://brainly.com/question/20804169