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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.

Respuesta :

Answer:

a) [tex]y=0.0831 x +0.004[/tex]

b) [tex] y= -0.330X + 12.5[/tex]

And we see that the values are different from the result of part a

c) Using the results from part b we can say:

Every increase of one degree Celsius means about -0.33 fewer mean millimeters of coral growth per year.

Step-by-step explanation:

Part a

For this case we can calculate the mean with the following formula:

[tex]\bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]

For this case the notation is given by:

X ="Growth"

Y= "Sea Surface Temperature"

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

After apply the formula we got:

[tex]\bar X = 30.259[/tex]

[tex]\bar Y = 2.519[/tex]

n=7 represent the pairred values

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]

After replace the values we got: r =-0.85

The slope formula would be given by:

[tex]m=\frac{S_{xy}}{S_{xx}}[/tex]

Where:

[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}[/tex]

[tex]S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}[/tex]

[tex]S_{yy}=\sum_{i=1}^n y^2_i -\frac{(\sum_{i=1}^n y_i)^2}{n}[/tex]

So we can find the sums like this:

[tex]\sum_{i=1}^n x_i =211.81[/tex]

[tex]\sum_{i=1}^n y_i =17.63[/tex]

[tex]\sum_{i=1}^n x^2_i =6410.063[/tex]

[tex]\sum_{i=1}^n y^2_i =44.55[/tex]

[tex]\sum_{i=1}^n x_i y_i =533.13[/tex]

So then the solope would be given by:

[tex] m = \frac{533.13}{6410.063}=0.0831[/tex]

And we can find the intercept using this:

[tex]b=\bar y -m \bar x=2.519-(0.0831*30.259)=0.0044[/tex]

So the line would be given by:

[tex]y=0.0831 x +0.004[/tex]

Part b

For this case we can 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)

> linearMod <- lm(y ~ x)

> linearMod

Call:

lm(formula = y ~ x)

Coefficients:

(Intercept)            x  

   12.5009      -0.3299

For this case the equation is given by:

[tex] y= -0.330X + 12.5[/tex]

And we see that the values are different from the result of part a.

Part c

Using the results from part b we can say:

Every increase of one degree Celsius means about -0.33 fewer mean millimeters of coral growth per year.

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