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Take 200 and 150 respectively as the assumed mean for [tex]\(X\)[/tex] and [tex]\(Y\)[/tex] series of 11 values. Then

[tex]\[
d_x = X - 200, \quad d_y = Y - 150
\][/tex]

Given data:

[tex]\[
\sum d_x = 13, \quad \sum d_x^2 = 2667, \quad \sum d_y = 42, \quad \sum d_y^2 = 6964, \quad \sum d_x d_y = 3943
\][/tex]

The value of [tex]\(r\)[/tex] is:

Sagot :

GinnyAnsweridn
To find the value of the correlation coefficient [tex]\( r \)[/tex], follow these detailed steps:

1. Calculate the Means of [tex]$dx$[/tex] and [tex]$dy$[/tex]:

We have the sums [tex]\(\sum dx\)[/tex] and [tex]\(\sum dy\)[/tex], and we know the number of values, [tex]\( n = 11 \)[/tex].

[tex]\[ \text{Mean of } dx = \frac{\sum dx}{n} = \frac{13}{11} \approx 1.18182 \][/tex]

[tex]\[ \text{Mean of } dy = \frac{\sum dy}{n} = \frac{42}{11} \approx 3.81818 \][/tex]

2. Calculate the Standard Deviations of [tex]$dx$[/tex] and [tex]$dy$[/tex]:

We use the sums of squares [tex]\(\sum dx^2\)[/tex] and [tex]\(\sum dy^2\)[/tex] to find the standard deviations.

[tex]\[ \text{Variance of } dx = \frac{\sum dx^2}{n} - (\text{Mean of } dx)^2 = \frac{2667}{11} - (1.18182)^2 \][/tex]

Simplifying this:

[tex]\[ \text{Variance of } dx \approx \frac{2667}{11} - 1.39669 \approx 242.4545 \][/tex]

[tex]\[ \text{Standard deviation of } dx = \sqrt{242.4545} \approx 15.52604 \][/tex]

Similarly, for [tex]\( dy \)[/tex]:

[tex]\[ \text{Variance of } dy = \frac{\sum dy^2}{n} - (\text{Mean of } dy)^2 = \frac{6964}{11} - (3.81818)^2 \][/tex]

Simplifying this:

[tex]\[ \text{Variance of } dy \approx \frac{6964}{11} - 14.5818 \approx 618.4287 \][/tex]

[tex]\[ \text{Standard deviation of } dy = \sqrt{618.4287} \approx 24.86991 \][/tex]

3. Calculate the Covariance between [tex]$dx$[/tex] and [tex]$dy$[/tex]:

We use [tex]\(\sum dx \, dy\)[/tex]:

[tex]\[ \text{Covariance} = \frac{\sum dx \, dy}{n} - (\text{Mean of } dx \cdot \text{Mean of } dy) = \frac{3943}{11} - (1.18182 \cdot 3.81818) \][/tex]

Simplifying this:

[tex]\[ \text{Covariance} \approx 358.4545 - 4.51235 \approx 353.94215 \][/tex]

4. Calculate the Correlation Coefficient [tex]\( r \)[/tex]:

Now we use the standard deviations and the covariance:

[tex]\[ r = \frac{\text{Covariance}(dx, dy)}{\text{Standard deviation of } dx \cdot \text{Standard deviation of } dy} = \frac{353.94215}{15.52604 \cdot 24.86991} \][/tex]

Simplifying this:

[tex]\[ r \approx \frac{353.94215}{386.0093} \approx 0.91664 \][/tex]

Therefore, the value of the correlation coefficient [tex]\( r \)[/tex] is approximately [tex]\( 0.91664 \)[/tex].
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