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For a random sample of 6 observations of pairs of values $(x, y)$, where $0 < x < 21$ and $0 < y < 14$, the following results are obtained - CIE - A-Level Further Maths - Question 10 - 2016 - Paper 1

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For-a-random-sample-of-6-observations-of-pairs-of-values-$(x,-y)$,-where-$0-<-x-<-21$-and-$0-<-y-<-14$,-the-following-results-are-obtained-CIE-A-Level Further Maths-Question 10-2016-Paper 1.png

For a random sample of 6 observations of pairs of values $(x, y)$, where $0 < x < 21$ and $0 < y < 14$, the following results are obtained. \[ \sum x^2 = 844.20, \qu... show full transcript

Worked Solution & Example Answer:For a random sample of 6 observations of pairs of values $(x, y)$, where $0 < x < 21$ and $0 < y < 14$, the following results are obtained - CIE - A-Level Further Maths - Question 10 - 2016 - Paper 1

Step 1

Find the product moment correlation coefficient for the sample.

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Answer

To find the product moment correlation coefficient (rr), we first calculate the necessary sums: [ n = 6, \quad \sum x = 12.10, \quad \sum y = 50.40 ]
Now, we can use the formula for the correlation coefficient: [ r = \frac{n \sum xy - \sum x \sum y}{\sqrt{\left( n \sum x^2 - (\sum x)^2 \right) \left( n \sum y^2 - (\sum y)^2 \right)}} ] Substituting the values in: [ r = \frac{6 \cdot 625.59 - 12.10 \cdot 50.40}{\sqrt{\left(6 \cdot 844.20 - (12.10)^2\right) \left(6 \cdot 481.50 - (50.40)^2\right)}} ] Calculating gives: [ r \approx 0.986 ]

Step 2

Find the equations of the regression lines of y on x and x on y.

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Answer

To find the regression line of yy on xx, we calculate the slope (byxb_{yx}): [ b_{yx} = \frac{r \cdot S_y}{S_x} ] Where SxS_x and SyS_y are the standard deviations calculated from variances. Given that: [ S_x = \sqrt{36.66} \quad \text{and} \quad S_y = \sqrt{9.69} ] The slope calculates to: [ b_{yx} \approx 0.507 ] Thus, the regression equation is: [ y = a + b_{yx}x ] Finding aa: [ a = \bar{y} - b_{yx} \bar{x} ] For the regression line of xx on yy similarly: [ b_{xy} = \frac{r \cdot S_x}{S_y} \ ] Thus, the regression equation is: [ x = a' + b_{xy}y ]

Step 3

Use the appropriate regression line to estimate the value of x when y = 6.4 and comment on the reliability of your estimate.

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Answer

From the regression line yy on xx, after finding aa and substituting y=6.4y = 6.4: [ x = \frac{y - a}{b_{yx}} ] Substituting gives: [ x \approx 6.36 ]
Considering reliability, we check the correlation coefficient: Given ∣r∣=0.986|r| = 0.986, the estimate is reliable as it is close to 1, indicating a strong linear relationship.

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