Question

how sensitive to changes in water temperature are coral reefs? to find out, scientists examined data on sea surface temperatures and coral growth per year at locations in the tropical western atlantic. the table shows the data for a subset of the study locations.
sea surface temperature growth
26.7 0.85
26.6 0.85
26.6 0.79
26.5 0.86
26.3 0.89
26.1 0.92
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csv excel jmp mac - text minitab14 - 18 minitab18+ pc - text r spss ti crunchit!
(c) enter these data into your calculator or software and use the correlation function to find r. check that you get the same result as the correlation result that you obtained in the step - by - step calculations, up to roundoff error. round your answer to three decimal places.

Explanation:

Step1: Recall correlation formula

The Pearson - correlation coefficient formula for a sample of size $n$ with paired data $(x_i,y_i)$ is $r=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})}{\sqrt{\sum_{i=1}^{n}(x_i-\bar{x})^2\sum_{i = 1}^{n}(y_i-\bar{y})^2}}$, where $\bar{x}=\frac{1}{n}\sum_{i = 1}^{n}x_i$ and $\bar{y}=\frac{1}{n}\sum_{i = 1}^{n}y_i$. But we can also use a calculator or software. Let $x$ be the sea - surface temperature and $y$ be the growth.

Step2: Use statistical software or calculator

Assume we use a scientific calculator with a correlation function. Enter the values of sea - surface temperature ($x$: 26.7, 26.6, 26.6, 26.5, 26.3, 26.1) and growth ($y$: 0.85, 0.85, 0.79, 0.86, 0.89, 0.92) into the calculator.

Step3: Calculate and round

After performing the correlation calculation on the calculator, we get the correlation coefficient $r$. Round the result to three decimal places.

Using a calculator or software (e.g., Excel's CORREL function), we find that:
Let $x=\{26.7,26.6,26.6,26.5,26.3,26.1\}$ and $y = \{0.85,0.85,0.79,0.86,0.89,0.92\}$
$\bar{x}=\frac{26.7 + 26.6+26.6+26.5+26.3+26.1}{6}=\frac{158.8}{6}\approx26.467$
$\bar{y}=\frac{0.85 + 0.85+0.79+0.86+0.89+0.92}{6}=\frac{5.16}{6}=0.86$

$\sum_{i = 1}^{6}(x_i-\bar{x})(y_i-\bar{y})=(26.7 - 26.467)(0.85 - 0.86)+(26.6 - 26.467)(0.85 - 0.86)+(26.6 - 26.467)(0.79 - 0.86)+(26.5 - 26.467)(0.86 - 0.86)+(26.3 - 26.467)(0.89 - 0.86)+(26.1 - 26.467)(0.92 - 0.86)$
$=(0.233)\times(- 0.01)+(0.133)\times(-0.01)+(0.133)\times(-0.07)+(0.033)\times0+( - 0.167)\times0.03+( - 0.367)\times0.06$
$=-0.00233-0.00133 - 0.00931+0 - 0.00501-0.02202=-0.03999$

$\sum_{i = 1}^{6}(x_i-\bar{x})^2=(26.7 - 26.467)^2+(26.6 - 26.467)^2+(26.6 - 26.467)^2+(26.5 - 26.467)^2+(26.3 - 26.467)^2+(26.1 - 26.467)^2$
$=0.233^2+0.133^2+0.133^2+0.033^2+( - 0.167)^2+( - 0.367)^2$
$=0.054289+0.017689+0.017689+0.001089+0.027889+0.134689 = 0.253334$

$\sum_{i = 1}^{6}(y_i-\bar{y})^2=(0.85 - 0.86)^2+(0.85 - 0.86)^2+(0.79 - 0.86)^2+(0.86 - 0.86)^2+(0.89 - 0.86)^2+(0.92 - 0.86)^2$
$=(-0.01)^2+(-0.01)^2+(-0.07)^2+0^2+0.03^2+0.06^2$
$=0.0001+0.0001 + 0.0049+0+0.0009+0.0036=0.0096$

$r=\frac{\sum_{i = 1}^{6}(x_i-\bar{x})(y_i-\bar{y})}{\sqrt{\sum_{i = 1}^{6}(x_i-\bar{x})^2\sum_{i = 1}^{6}(y_i-\bar{y})^2}}=\frac{-0.03999}{\sqrt{0.253334\times0.0096}}=\frac{-0.03999}{\sqrt{0.0024319064}}\approx - 0.805$

Answer:

$-0.805$