Question

A study is given in which scientists examined data on mean sea surface temperatures (in degrees Celsius) and mean coral growth (in millimeters per year) over a several-year period at different locations. Here are the data:
Sea Surface Temperature 29.67 29.87 30.15 30.21 30.47 30.64 30.80
Growth 2.64 2.59 2.69 2.60 2.48 2.38 2.25
(a) Use your calculator to find the mean and standard deviation of both sea surface temperature x and growth y and the correlation r between x and y. Use these basic measures to find the equation of the least-squares line for predicting y from x. (Round your answers to three decimal places.)
y = + x

(b) Enter the data into your software or calculator, and use the regression function to find the least-squares line. The result should agree with your work in part (a) up to roundoff error. (Round your answers to three decimal places.)
y = + x

(c) Say in words what the numerical value of the slope tells you. (Round your answer to three decimal places.)

Every increase of one degree Celsius means about ___________ fewer mean millimeters of coral growth per year.

Answer 1

Answer:

a) $\bar x= \frac{\sum x_i}{n}=\frac{211.81}{7}=30.26$

$\bar y= \frac{\sum y_i}{n}=\frac{17.63}{7}=2.52$

$r=\frac{7(533.1304)-(211.81)(17.63)}{\sqrt{[7(6410.063) -(211.81)^2][7(44.5511) -(17.63)^2]}}=-0.85$  

$b = \frac{S_{xy}}{S_{xx}}$

Where:

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=6410.063-\frac{211.81^2}{7}=0.9950$  

$S_{yy}=\sum_{i=1}^n y^2_i -\frac{(\sum_{i=1}^n y_i)^2}{n}=44.5511-\frac{17.63^2}{7}=0.1487$  

$b = \frac{S_{xy}}{S_{xx}}$

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i){n}}=533.1304-\frac{211.81*17.63}{7}=-0.3282$  

And the slope would be:  

$m=\frac{-0.3282}{0.9950}=-0.330$  

And we can find the intercept using this:  

$b=\bar y -m \bar x=2.52-(-0.3299*30.26)=12.502$  

So then the linear model would be:

$y =-0.330 x + 12.502$

b) > x<-c(29.67, 29.87, 30.15, 30.21, 30.47, 30.64, 30.80)

> y<-c(2.64,2.59, 2.69, 2.60, 2.48, 2.38, 2.25)

> lm<- lm(y~x)

> lm

Call:

lm(formula = y ~ x)

Coefficients:

(Intercept)            x  

   12.5009      -0.3299  

And the model is given by:

$y= -0.330 x +12.501$

c) Every increase of one degree Celsius means about -0.330 fewer mean millimeters of coral growth per year.

Step-by-step explanation:

For this case we have the following data:

X: 29.67 29.87 30.15 30.21 30.47 30.64 30.80

Y: 2.64 2.59 2.69 2.60 2.48 2.38 2.25

Part a

$\bar x= \frac{\sum x_i}{n}=\frac{211.81}{7}=30.26$

$\bar y= \frac{\sum y_i}{n}=\frac{17.63}{7}=2.52$

The correlation coefficient is a "statistical measure that calculates the strength of the relationship between the relative movements of two variables". It's denoted by r and its always between -1 and 1.

And in order to calculate the correlation coefficient we can use this formula:  

$r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}$  

For our case we have this:

n=7 $\sum x = 211.81, \sum y = 17.63, \sum xy = 533.1304, \sum x^2 =6410.063, \sum y^2 =44.5511$  

$r=\frac{7(533.1304)-(211.81)(17.63)}{\sqrt{[7(6410.063) -(211.81)^2][7(44.5511) -(17.63)^2]}}=-0.85$  

We can calculate the slope for the regression model with this formula:

$b = \frac{S_{xy}}{S_{xx}}$

Where:

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=6410.063-\frac{211.81^2}{7}=0.9950$  

$S_{yy}=\sum_{i=1}^n y^2_i -\frac{(\sum_{i=1}^n y_i)^2}{n}=44.5511-\frac{17.63^2}{7}=0.1487$  

And if we replace we got:

$b = \frac{S_{xy}}{S_{xx}}$

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i){n}}=533.1304-\frac{211.81*17.63}{7}=-0.3282$  

And the slope would be:  

$m=\frac{-0.3282}{0.9950}=-0.330$  

And we can find the intercept using this:  

$b=\bar y -m \bar x=2.52-(-0.3299*30.26)=12.502$  

So then the linear model would be:

$y =-0.330 x + 12.502$

Part b

For this case we use the following R code:

> x<-c(29.67, 29.87, 30.15, 30.21, 30.47, 30.64, 30.80)

> y<-c(2.64,2.59, 2.69, 2.60, 2.48, 2.38, 2.25)

> lm<- lm(y~x)

> lm

Call:

lm(formula = y ~ x)

Coefficients:

(Intercept)            x  

   12.5009      -0.3299  

And the model is given by:

$y= -0.330 x +12.501$

Part c

For this case the best interpretation is:

Every increase of one degree Celsius means about -0.330 fewer mean millimeters of coral growth per year.