Respuesta :
Answer:
a) [tex]\bar x= \frac{\sum x_i}{n}=\frac{211.81}{7}=30.26[/tex]
[tex]\bar y= \frac{\sum y_i}{n}=\frac{17.63}{7}=2.52[/tex]
[tex]r=\frac{7(533.1304)-(211.81)(17.63)}{\sqrt{[7(6410.063) -(211.81)^2][7(44.5511) -(17.63)^2]}}=-0.85[/tex]
[tex] b = \frac{S_{xy}}{S_{xx}}[/tex]
Where:
[tex]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[/tex]
[tex]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[/tex]
[tex] b = \frac{S_{xy}}{S_{xx}}[/tex]
[tex]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[/tex]
And the slope would be:
[tex]m=\frac{-0.3282}{0.9950}=-0.330[/tex]
And we can find the intercept using this:
[tex]b=\bar y -m \bar x=2.52-(-0.3299*30.26)=12.502[/tex]
So then the linear model would be:
[tex] y =-0.330 x + 12.502[/tex]
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:
[tex] y= -0.330 x +12.501[/tex]
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
[tex]\bar x= \frac{\sum x_i}{n}=\frac{211.81}{7}=30.26[/tex]
[tex]\bar y= \frac{\sum y_i}{n}=\frac{17.63}{7}=2.52[/tex]
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:
[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}[/tex]
For our case we have this:
n=7 [tex] \sum x = 211.81, \sum y = 17.63, \sum xy = 533.1304, \sum x^2 =6410.063, \sum y^2 =44.5511[/tex]
[tex]r=\frac{7(533.1304)-(211.81)(17.63)}{\sqrt{[7(6410.063) -(211.81)^2][7(44.5511) -(17.63)^2]}}=-0.85[/tex]
We can calculate the slope for the regression model with this formula:
[tex] b = \frac{S_{xy}}{S_{xx}}[/tex]
Where:
[tex]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[/tex]
[tex]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[/tex]
And if we replace we got:
[tex] b = \frac{S_{xy}}{S_{xx}}[/tex]
[tex]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[/tex]
And the slope would be:
[tex]m=\frac{-0.3282}{0.9950}=-0.330[/tex]
And we can find the intercept using this:
[tex]b=\bar y -m \bar x=2.52-(-0.3299*30.26)=12.502[/tex]
So then the linear model would be:
[tex] y =-0.330 x + 12.502[/tex]
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:
[tex] y= -0.330 x +12.501[/tex]
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.
