Respuesta :
Answer:
1) [tex]m=\frac{77}{10.5}=7.333[/tex]
2) [tex]b=\bar y -m \bar x=80.857-(7.333*3.5)=55.190[/tex]
3) [tex]\hat y=7.333(0)+55.190=55.190[/tex]
4) False. The values predited will fall on the same line since we are estimating the values with just one line.
5) [tex]\hat y=7.333(4.5)+55.190=88.190[/tex]
6) [tex]R^2 = (0.971^2) =0.943[/tex]
And that means that the linear model explains 94.29% of the variation.
Step-by-step explanation:
We assume that the data is this one:
x:1.5,2.5, 3, 3.5 , 4, 4.5 ,5.5
y: 67, 69, 80, 81, 86, 89, 94.
Step 1 of 6: Find the estimated slope. Round your answer to three decimal places.
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 = 1.5+2.5+3+3.5+4+4.5+5.5=24.5[/tex]
[tex]\sum_{i=1}^n y_i =67+ 69+ 80+ 81+ 86+ 89+ 94=566[/tex]
[tex]\sum_{i=1}^n x^2_i =1.5^2+2.5^2+3^2+3.5^2+4^2+4.5^2+5.5^2=96.25[/tex]
[tex]\sum_{i=1}^n y^2_i =67^2+69^2+80^2+81^2+86^2+89^2+94^2=46364[/tex]
[tex]\sum_{i=1}^n x_i y_i =1.5*67+2.5*69+3*80+3.5*81+4*86+4.5*89+5.5*94=2058[/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}=96.25-\frac{24.5^2}{7}=10.5[/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}=2058-\frac{24.5*566}{7}=77[/tex]
And the slope would be:
[tex]m=\frac{77}{10.5}=7.333[/tex]
Step 2 of 6: Find the estimated y-intercept. Round your answer to three decimal places.
Now we can find the means for x and y like this:
[tex]\bar x= \frac{\sum x_i}{n}=\frac{24.5}{7}=3.5[/tex]
[tex]\bar y= \frac{\sum y_i}{n}=\frac{566}{7}=80.857[/tex]
And we can find the intercept using this:
[tex]b=\bar y -m \bar x=80.857-(7.333*3.5)=55.190[/tex]
So the line would be given by:
[tex]\hat y=7.333 x +55.190[/tex]
Step 3 of 6: Determine the value of the dependent variable yˆ at x = 0.
[tex]\hat y=7.333(0)+55.190=55.190[/tex]
Step 4 of 6: Determine if the statement "Not all points predicted by the linear model fall on the same line" is true or false.
False. The values predited will fall on the same line since we are estimating the values with just one line.
Step 5 of 6: Find the estimated value of y when x = 4.5. Round your answer to three decimal places.
[tex]\hat y=7.333(4.5)+55.190=88.190[/tex]
Step 6 of 6: Find the value of the coefficient of determination. Round your answer to three decimal places
n=7 [tex] \sum x = 24.5, \sum y = 566, \sum xy =2058, \sum x^2 =96.25, \sum y^2 =46364[/tex]
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]
[tex]r=\frac{7(2058)-(24.5)(566)}{\sqrt{[7(96.25) -(24.5)^2][7(46364) -(566)^2]}}=0.971[/tex]
And the determination coeffcient is just the square of the correlation coefficient given by:
[tex]R^2 = (0.971^2) =0.943[/tex]
And that means that the linear model explains 94.3% of the variation.
