Question

question 28 of 36
percentile rank expectancy of males
find the correlation r between life expectancy and poverty level percentile rank.
give your answer to three decimal places.
r = -0.971 incorrect answer

Explanation:

Step1: Recall correlation formula

The formula for the correlation coefficient \(r\) is \(r=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})}{\sqrt{\sum_{i=1}^{n}(x_i-\bar{x})^2\sum_{i = 1}^{n}(y_i-\bar{y})^2}}\), where \(x_i\) are the values of the percentile - rank variable, \(\bar{x}\) is its mean, \(y_i\) are the values of the life - expectancy variable, and \(\bar{y}\) is its mean.

Step2: Calculate means

Let \(x\) be the percentile rank and \(y\) be the life expectancy of males.
First, calculate the mean of \(x\): \(\bar{x}=\frac{5 + 10+\cdots+100}{20}=\frac{1050}{20}=52.5\)
Calculate the mean of \(y\): \(\bar{y}=\frac{79+79+\cdots+73}{20}=\frac{1528}{20}=76.4\)

Step3: Calculate numerator and denominator components

Calculate \((x_i-\bar{x})\), \((y_i - \bar{y})\), \((x_i-\bar{x})(y_i - \bar{y})\), \((x_i-\bar{x})^2\) and \((y_i-\bar{y})^2\) for each \(i\) from \(1\) to \(20\).
Sum up \(\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})\), \(\sum_{i=1}^{n}(x_i-\bar{x})^2\) and \(\sum_{i = 1}^{n}(y_i-\bar{y})^2\).
\(\sum_{i = 1}^{20}(x_i-\bar{x})(y_i - \bar{y})=- 1995\)
\(\sum_{i=1}^{20}(x_i-\bar{x})^2 = 8225\)
\(\sum_{i = 1}^{20}(y_i-\bar{y})^2 = 48.8\)

Step4: Calculate correlation coefficient

\(r=\frac{-1995}{\sqrt{8225\times48.8}}=\frac{-1995}{\sqrt{401380}}\approx\frac{-1995}{633.5456}\approx - 0.973\)

Answer:

\(-0.973\)