Question

chapter 10 homework
score: 60/190 answered: 7/19
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question 7
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here is a bivariate data set.

x	y
64.2	-9.5
61.4	-44.5
71.8	22.5
58.8	-42.6
107.6	69.6

find the correlation coefficient and report it accurate to three decimal places.
r = 0.976 ×

Explanation:

Step1: Recall the formula for correlation coefficient \( r \)

The formula for the Pearson correlation coefficient \( r \) is:
\[ r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^{2}-(\sum x)^{2}][n\sum y^{2}-(\sum y)^{2}]}} \]
where \( n \) is the number of data points. Here, \( n = 5 \).

Step2: Calculate \( \sum x \), \( \sum y \), \( \sum xy \), \( \sum x^{2} \), \( \sum y^{2} \)

- \( \sum x=64.2 + 61.4+71.8 + 58.8+107.6=363.8 \)
- \( \sum y=- 9.5-44.5 + 22.5-42.6 + 69.6=15.5 \)
- \( \sum xy=(64.2\times(-9.5))+(61.4\times(-44.5))+(71.8\times22.5)+(58.8\times(-42.6))+(107.6\times69.6) \)
- \( 64.2\times(-9.5)=-609.9 \)
- \( 61.4\times(-44.5)=-2732.3 \)
- \( 71.8\times22.5 = 1615.5 \)
- \( 58.8\times(-42.6)=-2494.88 \)
- \( 107.6\times69.6 = 7490.96 \)
- \( \sum xy=-609.9-2732.3 + 1615.5-2494.88 + 7490.96=3269.38 \)
- \( \sum x^{2}=64.2^{2}+61.4^{2}+71.8^{2}+58.8^{2}+107.6^{2} \)
- \( 64.2^{2}=4121.64 \)
- \( 61.4^{2}=3769.96 \)
- \( 71.8^{2}=5155.24 \)
- \( 58.8^{2}=3457.44 \)
- \( 107.6^{2}=11577.76 \)
- \( \sum x^{2}=4121.64 + 3769.96+5155.24 + 3457.44+11577.76=28082.04 \)
- \( \sum y^{2}=(-9.5)^{2}+(-44.5)^{2}+22.5^{2}+(-42.6)^{2}+69.6^{2} \)
- \( (-9.5)^{2}=90.25 \)
- \( (-44.5)^{2}=1980.25 \)
- \( 22.5^{2}=506.25 \)
- \( (-42.6)^{2}=1814.76 \)
- \( 69.6^{2}=4844.16 \)
- \( \sum y^{2}=90.25+1980.25 + 506.25+1814.76+4844.16=9235.67 \)

Step3: Substitute values into the formula

- Numerator: \( n(\sum xy)-(\sum x)(\sum y)=5\times3269.38-363.8\times15.5 \)
- \( 5\times3269.38 = 16346.9 \)
- \( 363.8\times15.5=5638.9 \)
- Numerator: \( 16346.9 - 5638.9=10708 \)
- Denominator part 1: \( n\sum x^{2}-(\sum x)^{2}=5\times28082.04-(363.8)^{2} \)
- \( 5\times28082.04 = 140410.2 \)
- \( 363.8^{2}=132350.44 \)
- Denominator part 1: \( 140410.2-132350.44 = 8059.76 \)
- Denominator part 2: \( n\sum y^{2}-(\sum y)^{2}=5\times9235.67-(15.5)^{2} \)
- \( 5\times9235.67 = 46178.35 \)
- \( 15.5^{2}=240.25 \)
- Denominator part 2: \( 46178.35 - 240.25=45938.1 \)
- Denominator: \( \sqrt{8059.76\times45938.1}=\sqrt{37024444.5616}\approx6084.8 \) (approximate calculation)
- \( r=\frac{10708}{6084.8}\approx1.759 \)? Wait, no, I must have made a mistake in calculation. Let's recalculate \( \sum xy \) and other sums.

Wait, let's recalculate \( \sum x \): \( 64.2+61.4 = 125.6; 125.6 + 71.8=197.4; 197.4+58.8 = 256.2; 256.2+107.6 = 363.8 \) (correct). \( \sum y \): \( -9.5-44.5=-54; -54 + 22.5=-31.5; -31.5-42.6=-74.1; -74.1 + 69.6=-4.5 \). Oh! I made a mistake in calculating \( \sum y \) earlier. \( \sum y=-9.5-44.5 + 22.5-42.6 + 69.6=(-9.5-44.5)+(22.5-42.6)+69.6=(-54)+(-20.1)+69.6= - 4.5 \).

Now recalculate \( \sum xy \):
- \( 64.2\times(-9.5)=-609.9 \)
- \( 61.4\times(-44.5)=-2732.3 \)
- \( 71.8\times22.5 = 1615.5 \)
- \( 58.8\times(-42.6)=-2494.88 \)
- \( 107.6\times69.6 = 7490.96 \)
\( \sum xy=-609.9-2732.3 + 1615.5-2494.88 + 7490.96=(-609.9-2732.3-2494.88)+(1615.5 + 7490.96)=(-5837.08)+(9106.46)=3269.38 \) (correct).

Now recalculate numerator: \( n(\sum xy)-(\sum x)(\sum y)=5\times3269.38-363.8\times(-4.5) \)
- \( 5\times3269.38 = 16346.9 \)
- \( 363.8\times(-4.5)=-1637.1 \)
- Numerator: \( 16346.9+1637.1 = 17984 \)

Denominator part 1: \( n\sum x^{2}-(\sum x)^{2}=5\times28082.04-(363.8)^{2} \)
- \( 5\times28082.04 = 140410.2 \)
- \( 363.8^{2}=132350.44 \)
- Denominator part 1: \( 140410.2 - 132350.44=8059.76 \)

Denominator part 2: \( n\sum y^{2}-(\sum y)^{2}=5\times9235.67-(-4.5)^{2} \)
- \( 5\times9235.67 = 46178.35 \)
- \( (-4.5)^{2}=20.25 \)
- Denominator part 2: \( 46178.35-20.25 = 46158.1 \)

Denominator: \( \sqrt{8059.76\times46158.1}=\sqrt{8059.76\times46158.1}\approx\sqrt{372000000}\approx19287.3 \) (Wait, no, 8059.76*46158.1: 8000*46000=368000000, 8059.76*46158.1≈8059.76*46000 + 8059.76*158.1≈370748960+1274158≈372023118. So square root of 372023118≈19287.9)

Now \( r=\frac{17984}{19287.9}\approx0.932 \)

Wait, let's use a more accurate way. Let's use the formula step by step with correct \( \sum y=-4.5 \)

Alternative approach: Use the formula for \( r \) using deviations.

Let \( \bar{x}=\frac{\sum x}{n}=\frac{363.8}{5}=72.76 \)

\( \bar{y}=\frac{\sum y}{n}=\frac{-4.5}{5}=-0.9 \)

\( SS_{x}=\sum(x - \bar{x})^{2}=\sum x^{2}-n\bar{x}^{2}=28082.04-5\times(72.76)^{2} \)

\( 72.76^{2}=5294.0176 \), \( 5\times5294.0176 = 26470.088 \)

\( SS_{x}=28082.04 - 26470.088 = 1611.952 \)

\( SS_{y}=\sum(y - \bar{y})^{2}=\sum y^{2}-n\bar{y}^{2}=9235.67-5\times(-0.9)^{2}=9235.67 - 5\times0.81=9235.67 - 4.05 = 9231.62 \)

\( SP=\sum(x - \bar{x})(y - \bar{y})=\sum xy - n\bar{x}\bar{y}=3269.38-5\times72.76\times(-0.9) \)

\( 5\times72.76\times(-0.9)=-327.42 \)

\( SP=3269.38 + 327.42=3596.8 \)

Then \( r=\frac{SP}{\sqrt{SS_{x}\times SS_{y}}}=\frac{3596.8}{\sqrt{1611.952\times9231.62}} \)

\( 1611.952\times9231.62\approx1611.952\times9000=14507568, 1611.952\times231.62\approx373300, total\approx14880868 \)

\( \sqrt{14880868}\approx3857.57 \)

\( r=\frac{3596.8}{3857.57}\approx0.932 \)

So the correct correlation coefficient is approximately \( 0.932 \) (after correcting the sum of \( y \) values).

Answer:

\( 0.932 \) (rounded to three decimal places)