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Math  /  Data & Statistics

Question2.2 The following are measurements of the total dissolved salts (TDS) and hardness index of 10 samples of water: \begin{tabular}{cc|cccccccccc} TDS & X\boldsymbol{X} & 202 & 327 & 112 & 463 & 582 & 926 & 681 & 290 & 776 & 378 \\ \hline \begin{tabular}{c} Hardness \\ index \end{tabular} & Y\boldsymbol{Y} & 14 & 23 & 9 & 35 & 44 & 68 & 50 & 21 & 57 & 28 \end{tabular}
Given that SX=262.435S_{X}=262.435 and SY=19.393S_{Y}=19.393, compute the correlation coefficient.

Studdy Solution

STEP 1

1. The data consists of paired measurements for TDS (XX) and Hardness index (YY).
2. The standard deviations for XX and YY are given as SX=262.435S_X = 262.435 and SY=19.393S_Y = 19.393.
3. The correlation coefficient rr is calculated using the formula:
r=∑(Xi−Xˉ)(Yi−Yˉ)(n−1)SXSYr = \frac{\sum (X_i - \bar{X})(Y_i - \bar{Y})}{(n-1)S_X S_Y}
where nn is the number of samples, Xˉ\bar{X} and Yˉ\bar{Y} are the means of XX and YY respectively.

STEP 2

1. Calculate the means Xˉ\bar{X} and Yˉ\bar{Y}.
2. Compute the covariance ∑(Xi−Xˉ)(Yi−Yˉ)\sum (X_i - \bar{X})(Y_i - \bar{Y}).
3. Calculate the correlation coefficient rr.

STEP 3

Calculate the mean of XX:
Xˉ=202+327+112+463+582+926+681+290+776+37810=473710=473.7\bar{X} = \frac{202 + 327 + 112 + 463 + 582 + 926 + 681 + 290 + 776 + 378}{10} = \frac{4737}{10} = 473.7

STEP 4

Calculate the mean of YY:
Yˉ=14+23+9+35+44+68+50+21+57+2810=34910=34.9\bar{Y} = \frac{14 + 23 + 9 + 35 + 44 + 68 + 50 + 21 + 57 + 28}{10} = \frac{349}{10} = 34.9

STEP 5

Compute the covariance ∑(Xi−Xˉ)(Yi−Yˉ)\sum (X_i - \bar{X})(Y_i - \bar{Y}):
\begin{align*} \sum (X_i - \bar{X})(Y_i - \bar{Y}) = & (202 - 473.7)(14 - 34.9) + (327 - 473.7)(23 - 34.9) \\ & + (112 - 473.7)(9 - 34.9) + (463 - 473.7)(35 - 34.9) \\ & + (582 - 473.7)(44 - 34.9) + (926 - 473.7)(68 - 34.9) \\ & + (681 - 473.7)(50 - 34.9) + (290 - 473.7)(21 - 34.9) \\ & + (776 - 473.7)(57 - 34.9) + (378 - 473.7)(28 - 34.9) \end{align*}
Calculate each term and sum them up.

STEP 6

Calculate each term:
\begin{align*} (202 - 473.7)(14 - 34.9) &= -271.7 \times -20.9 = 5682.53 \\ (327 - 473.7)(23 - 34.9) &= -146.7 \times -11.9 = 1745.73 \\ (112 - 473.7)(9 - 34.9) &= -361.7 \times -25.9 = 9378.03 \\ (463 - 473.7)(35 - 34.9) &= -10.7 \times 0.1 = -1.07 \\ (582 - 473.7)(44 - 34.9) &= 108.3 \times 9.1 = 985.53 \\ (926 - 473.7)(68 - 34.9) &= 452.3 \times 33.1 = 14972.13 \\ (681 - 473.7)(50 - 34.9) &= 207.3 \times 15.1 = 3130.23 \\ (290 - 473.7)(21 - 34.9) &= -183.7 \times -13.9 = 2553.43 \\ (776 - 473.7)(57 - 34.9) &= 302.3 \times 22.1 = 6680.83 \\ (378 - 473.7)(28 - 34.9) &= -95.7 \times -6.9 = 660.33 \end{align*}
Sum these values:
5682.53+1745.73+9378.03−1.07+985.53+14972.13+3130.23+2553.43+6680.83+660.33=45187.975682.53 + 1745.73 + 9378.03 - 1.07 + 985.53 + 14972.13 + 3130.23 + 2553.43 + 6680.83 + 660.33 = 45187.97

STEP 7

Calculate the correlation coefficient rr:
r=45187.97(10−1)×262.435×19.393r = \frac{45187.97}{(10-1) \times 262.435 \times 19.393}
r=45187.979×262.435×19.393r = \frac{45187.97}{9 \times 262.435 \times 19.393}
r=45187.9745906.61≈0.984r = \frac{45187.97}{45906.61} \approx 0.984
The correlation coefficient is:
0.984 \boxed{0.984}

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