contestada

Global warming. Have average global temperatures been increasing in recent years?
Here are annual average global temperatures for the last 20 year’s in degrees Celsius.

A. Find the least-square regression line for predicting average global temperature from year.

Temperature = _+ (0.01) ? + (_+ 0.0001) ?XYear

B. An environmentalist uses the line, based on the 20 years, to predict average global temperatures in 2050. What is the prediction?

Temperature (_+ 0.01) = ? degrees Celsius

Respuesta :

Answer:

a) [tex]y=0.0106 x +14.34[/tex]

We are assuming that the starting year 1994 represent the value x=0.

b) [tex]y(56)=0.0106(56) +14.34 =14.93[/tex]

Step-by-step explanation:

We assume that the data is this one:

x: 0, 1, 2, 3, 4, 5, 6,7, 8,9,10,11,12,13,14,15,16,17,18,19

y: 14.23, 14.35, 14.22, 14.42, 14.54, 14.36, 14.33, 14.45, 14.51, 14.52, 14.48, 14.55, 14.50, 14.49, 14.41, 14.50, 14.56, 14.43, 14.48, 14.52

We are assuming that the starting year 1994 represent the value x=0.

Part a

For this case we need to calculate the slope with the following formula:

[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]

So we can find the sums like this:

[tex]\sum_{i=1}^n x_i = 0+1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19=190[/tex]

[tex]\sum_{i=1}^n y_i =14.23+ 14.35+ 14.22+ 14.42+ 14.54+ 14.36+ 14.33+ 14.45+ 14.51+ 14.52+ 14.48+ 14.55+ 14.50+ 14.49+ 14.41+ 14.50+ 14.56+ 14.43+ 14.48+ 14.52=288.85[/tex]

[tex]\sum_{i=1}^n x^2_i =0^2+1^2+2^2+3^2+4^2+5^2+6^2+7^2+8^2+9^2+10^2+11^2+12^2+13^2+14^2+15^2+16^2+17^2+18^2+19^2=2470[/tex]

[tex]\sum_{i=1}^n y^2_i =14.23^2+ 14.35^2+ 14.22^2+ 14.42^2+ 14.54^2+ 14.36^2+ 14.33^2+ 14.45^2+ 14.51^2+ 14.52^2+ 14.48^2+ 14.55^2+ 14.50^2+ 14.49^2+ 14.41^2+ 14.50^2+ 14.56^2+ 14.43^2+ 14.48^2+ 14.52^2=4171.904[/tex]

[tex]\sum_{i=1}^n x_i y_i =0*14.23+ 1*14.35+ 2*14.22+ 3*14.42+ 4*14.54+ 5*14.36+ 6*14.33+ 7*14.45+ 8*14.51+ 9*14.52+ 10*14.48+ 11*14.55+ 12*14.50+ 13*14.49+ 14*14.41+ 15*14.50+ 16*14.56+ 17*14.43+ 18*14.48+ 19*14.52=2751.15[/tex]

With these we can find the sums:

[tex]S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=2470-\frac{190^2}{20}=665[/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}=2751.15-\frac{190*288.15}{20}=7.075[/tex]

And the slope would be:

[tex]m=\frac{7.075}{665}=0.0106[/tex]

NoW we we can find the means for x and y like this:

[tex]\bar x= \frac{\sum x_i}{n}=\frac{190}{20}=9.5[/tex]

[tex]\bar y= \frac{\sum y_i}{n}=\frac{288.85}{20}=14.4425[/tex]

And we can find the intercept using this:

[tex]b=\bar y -m \bar x=14.4425-(0.0106*9.5)=14.34[/tex]

So the line would be given by:

[tex]y=0.0106 x +14.34[/tex]

Part b

For this case we just need to find how many years are between 1994 and 2050 and we got 2050-1994= 56.

And replacinf x=56 into our model we got:

[tex]y(56)=0.0106(56) +14.34 =14.93[/tex]