Mathematics, 06.04.2020 23:10 cristinavalencia03

Every few years, the National Assessment of Educational Progress asks a national sample of eighth-graders to perform the same math tasks. The goal is to get an honest picture of progress in math. Suppose these are the last few national mean scores, on a scale of 0 to 500.
Year 1990 1992 1996 2000 2003 2005 2008
Score 263 268 271 272 276 277 279
(a) Find the regression line of mean score on time step-by-step. First calculate the mean and standard deviation of each variable and their correlation (use a calculator with these functions). Then find the equation of the least-squares line from these
(b) What percent of the year-to-year variation in scores is explained by the linear trend?

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Answer from: giany

a)X: 0 2 6 10 13 15 18

Y:263 268 271 272 276 277 279

X represent the number of years since 1990

n=7 $\sum x = 64, \sum y = 1906, \sum xy= 17647, \sum x^2 =858, \sum y^2 =519164$

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

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

$r=\frac{7(17647)-(64)(1906)}{\sqrt{[7(858) -(64)^2][7(519164) -(1906)^2]}}=0.97599$

$\bar X = \frac{\sum_{i=1^n X_i}}{n} = 9.14286$

$\bar Y = \frac{\sum_{i=1^n Y_i}}{n} = 272.286$

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=858-\frac{64^2}{7}=272.857$

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}=17647-\frac{64*1906}{7}=220.714$

And the slope would be:

$m=\frac{220.714}{272.857}=0.809$

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

And we can find the intercept using this:

$b=\bar y -m \bar x=272.286-(0.809*9.143)=264.889$

So the line would be given by:

y=0.809 x +264.889

b) For this case the percent of variation in scores is explained by the linear trend is given by the determination coefficient r^2 and we got:

r^2 =0.976^2 = 0.9526

So then we can say that the percent of variation explained is approximately 95.26%

Step-by-step explanation:

Pearson correlation coefficient(r), "measures a linear dependence between two variables (x and y). Its a parametric correlation test because it depends to the distribution of the data. And other assumption is that the variables x and y needs to follow a normal distribution".

Solution to the problem

Part a

We assume the following data:

X: 0 2 6 10 13 15 18

Y:263 268 271 272 276 277 279

X represent the number of years since 1990

n=7 $\sum x = 64, \sum y = 1906, \sum xy= 17647, \sum x^2 =858, \sum y^2 =519164$

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

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

$r=\frac{7(17647)-(64)(1906)}{\sqrt{[7(858) -(64)^2][7(519164) -(1906)^2]}}=0.97599$

So then the correlation coefficient would be r =0.976

The mean for X on this case is given by:

$\bar X = \frac{\sum_{i=1^n X_i}}{n} = 9.14286$

$\bar Y = \frac{\sum_{i=1^n Y_i}}{n} = 272.286$

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

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

Where:

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}$

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

So we can find the sums like this:

$\sum_{i=1}^n x_i = 64$

$\sum_{i=1}^n y_i =1906$

$\sum_{i=1}^n x^2_i =858$

$\sum_{i=1}^n y^2_i =519164$

$\sum_{i=1}^n x_i y_i =17647$

With these we can find the sums:

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=858-\frac{64^2}{7}=272.857$

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}=17647-\frac{64*1906}{7}=220.714$

And the slope would be:

$m=\frac{220.714}{272.857}=0.809$

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

And we can find the intercept using this:

$b=\bar y -m \bar x=272.286-(0.809*9.143)=264.889$

So the line would be given by:

y=0.809 x +264.889

Part b

For this case the percent of variation in scores is explained by the linear trend is given by the determination coefficient r^2 and we got:

r^2 =0.976^2 = 0.9526

So then we can say that the percent of variation explained is approximately 95.26%

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