Question

find the correlation r step - by - step. let x denote the sea surface temperature and y denote the coral growth. first, find the mean and standard deviation of each variable. give your answer to at least two decimal places but if you are calculating the values by hand, avoid rounding during intermediate steps. mean for the x data set, $\bar{x}=$ degrees mean for the y data set, $\bar{y}=$ cm standard deviation for the x data set, $s_x=$ degrees standard deviation for the y data set, $s_y=$ cm

Explanation:

Step1: Calculate the mean of x - data

The x - data is \(x = [0.85,0.85,0.79,0.86,0.89,0.92]\). The formula for the mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\), where \(n = 6\), \(\sum_{i=1}^{6}x_{i}=0.85 + 0.85+0.79+0.86+0.89+0.92=5.16\). So \(\bar{x}=\frac{5.16}{6}=0.86\) degrees.

Step2: Calculate the mean of y - data

The y - data is \(y=[26.7,26.6,26.6,26.5,26.3,26.1]\). Using the mean formula \(\bar{y}=\frac{\sum_{i = 1}^{n}y_{i}}{n}\), with \(n = 6\), \(\sum_{i = 1}^{6}y_{i}=26.7+26.6+26.6+26.5+26.3+26.1 = 158.8\). So \(\bar{y}=\frac{158.8}{6}\approx26.47\) cm.

Step3: Calculate the standard deviation of x - data

The formula for the sample - standard deviation \(s_{x}=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}\).
\((x_1-\bar{x})=(0.85 - 0.86)=-0.01\), \((x_2-\bar{x})=(0.85 - 0.86)=-0.01\), \((x_3-\bar{x})=(0.79 - 0.86)=-0.07\), \((x_4-\bar{x})=(0.86 - 0.86)=0\), \((x_5-\bar{x})=(0.89 - 0.86)=0.03\), \((x_6-\bar{x})=(0.92 - 0.86)=0.06\).
\(\sum_{i = 1}^{6}(x_{i}-\bar{x})^{2}=(-0.01)^{2}+(-0.01)^{2}+(-0.07)^{2}+0^{2}+0.03^{2}+0.06^{2}=0.0001 + 0.0001+0.0049+0+0.0009+0.0036 = 0.0096\).
\(s_{x}=\sqrt{\frac{0.0096}{5}}\approx0.044\) degrees.

Step4: Calculate the standard deviation of y - data

The formula for the sample - standard deviation \(s_{y}=\sqrt{\frac{\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}}{n - 1}}\).
\((y_1-\bar{y})=(26.7 - 26.47)=0.23\), \((y_2-\bar{y})=(26.6 - 26.47)=0.13\), \((y_3-\bar{y})=(26.6 - 26.47)=0.13\), \((y_4-\bar{y})=(26.5 - 26.47)=0.03\), \((y_5-\bar{y})=(26.3 - 26.47)=-0.17\), \((y_6-\bar{y})=(26.1 - 26.47)=-0.37\).
\(\sum_{i = 1}^{6}(y_{i}-\bar{y})^{2}=(0.23)^{2}+(0.13)^{2}+(0.13)^{2}+(0.03)^{2}+(-0.17)^{2}+(-0.37)^{2}=0.0529+0.0169+0.0169+0.0009+0.0289+0.1369 = 0.2534\).
\(s_{y}=\sqrt{\frac{0.2534}{5}}\approx0.23\) cm.

Answer:

\(\bar{x}=0.86\) degrees, \(\bar{y}\approx26.47\) cm, \(s_{x}\approx0.044\) degrees, \(s_{y}\approx0.23\) cm