QUESTION IMAGE
Question
find the correlation coefficient, r, of the data described below. researchers at a pharmaceutical company are testing a new drug that regulates blood sugar. in one test, subjects were prescribed a random and safe dose of the drug. once the drugs were administered, the researchers measured each subjects blood sugar levels before and after a meal. for each subject, the company recorded the given dose (in milligrams), x, and the rise in blood sugar (in milligrams per deciliter), y.
| dosage | rise in blood sugar level |
|---|---|
| 53 | 29 |
| 75 | 17 |
| 77 | 28 |
| 81 | 20 |
round your answer to the nearest thousandth.
Step1: Calculate the means
Let $x$ be the dosage and $y$ be the rise in blood - sugar level.
$n = 5$
$\bar{x}=\frac{46 + 53+75+77+81}{5}=\frac{332}{5}=66.4$
$\bar{y}=\frac{30 + 29+17+28+20}{5}=\frac{124}{5}=24.8$
Step2: Calculate the numerator and denominators
The formula for the correlation coefficient $r$ is $r=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})}{\sqrt{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}}}$
Calculate $(x_{i}-\bar{x})(y_{i}-\bar{y})$, $(x_{i}-\bar{x})^{2}$ and $(y_{i}-\bar{y})^{2}$ for each $i$:
| $x_{i}$ | $y_{i}$ | $x_{i}-\bar{x}$ | $y_{i}-\bar{y}$ | $(x_{i}-\bar{x})(y_{i}-\bar{y})$ | $(x_{i}-\bar{x})^{2}$ | $(y_{i}-\bar{y})^{2}$ |
|---|---|---|---|---|---|---|
| 53 | 29 | $53 - 66.4=-13.4$ | $29 - 24.8 = 4.2$ | $-13.4\times4.2=-56.28$ | $(-13.4)^{2}=179.56$ | $4.2^{2}=17.64$ |
| 75 | 17 | $75 - 66.4 = 8.6$ | $17 - 24.8=-7.8$ | $8.6\times(-7.8)=-67.08$ | $8.6^{2}=73.96$ | $(-7.8)^{2}=60.84$ |
| 77 | 28 | $77 - 66.4 = 10.6$ | $28 - 24.8 = 3.2$ | $10.6\times3.2 = 33.92$ | $10.6^{2}=112.36$ | $3.2^{2}=10.24$ |
| 81 | 20 | $81 - 66.4 = 14.6$ | $20 - 24.8=-4.8$ | $14.6\times(-4.8)=-70.08$ | $14.6^{2}=213.16$ | $(-4.8)^{2}=23.04$ |
$\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})=-106.08-56.28 - 67.08+33.92-70.08=-265.6$
$\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}=416.16+179.56+73.96+112.36+213.16=995.2$
$\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}=27.04+17.64+60.84+10.24+23.04=138.8$
Step3: Calculate the correlation coefficient
$r=\frac{-265.6}{\sqrt{995.2\times138.8}}=\frac{-265.6}{\sqrt{138133.76}}=\frac{-265.6}{371.663}$
$r\approx - 0.715$
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$-0.715$