Problem 12 - Calculus and its applications

Question

Consider the following data showing the average life expectancy of men in various years. $$ \begin{array}{|c|c|} \hline \begin{array}{c} \text { NUMBER OF YEARS, } x, \\ \text { SINCE } 1990 \end{array} & \begin{array}{c} \text { LIFE EXPECTANCY } \\ \text { OF MEN, } \boldsymbol{y} \text { (years) } \end{array} \\ \hline 0 & 71.8 \\ 10 & 74.1 \\ 13 & 74.8 \\ 17 & 75.4 \\ 21 & 76.3 \\ \hline \end{array} $$ a) Find the regression line, \(y=m x+b\). b) Use the regression line to predict the life expectancy of men in 2020 and \(2025 .\)

Step-by-Step Solution

Answer

The regression line is \(y = 3.66x + 29.828\). In 2020, life expectancy is predicted to be 109.7 years, and in 2025, it is 157.0 years.

Step 1: Understand the Data

We are given pairs of data \(x, y\): \(x\) is the number of years since 1990, and \(y\) is the life expectancy of men in those years. Our task is to find a linear relation between \(x\) and \(y\) to predict future values.

Step 2: Calculate Slope (m) of the Line

First, calculate the slope \(m\) using the formula \(m = \frac{n(\sum xy) - (\sum x)(\sum y)}{n(\sum x^2) - (\sum x)^2}\). Use the given data to compute:- \sum x = 0 + 10 + 13 + 17 + 21 = 61\,- \sum y = 71.8 + 74.1 + 74.8 + 75.4 + 76.3 = 372.4\,- \sum xy = (0)(71.8) + (10)(74.1) + (13)(74.8) + (17)(75.4) + (21)(76.3) = 5036.3\,- \sum x^2 = 0^2 + 10^2 + 13^2 + 17^2 + 21^2 = 879\, and- n = 5 (the number of data points).Now calculate \(m\): \(m = \frac{5 \times 5036.3 - 61 \times 372.4}{5 \times 879 - 61^2} = \frac{25181.5 - 22713.4}{4395 - 3721} = \frac{2468.1}{674} \approx 3.66\).

Step 3: Calculate y-intercept (b) of the Line

The y-intercept \(b\) is found using the formula \(b = \frac{\sum y - m(\sum x)}{n}\). Substitute the known values: \(b = \frac{372.4 - 3.66 \times 61}{5} = \frac{372.4 - 223.26}{5} = \frac{149.14}{5} = 29.828\).

Step 4: Write the Regression Line Equation

With \(m = 3.66\) and \(b = 29.828\), our regression line is \( y = 3.66x + 29.828 \). This equation models the life expectancy of men in terms of years since 1990.

Step 5: Predict Life Expectancy for 2020 and 2025

To find the life expectancy in 2020, calculate \(x = 2020 - 1990 = 30\) and substitute into the regression equation: \(y_{2020} = 3.66 \times 30 + 29.828 = 109.728\).For 2025, calculate \(x = 2025 - 1990 = 35\): \(y_{2025} = 3.66 \times 35 + 29.828 = 157.028\).

Key Concepts

Life Expectancy

Life expectancy is a statistical measure that estimates the average number of years a person can expect to live. It's an essential indicator in understanding the overall health of a population. The data we have tracks the life expectancy of men in years since 1990. Here, life expectancy is represented as the dependent variable, denoted by \(y\), and the number of years since 1990 is the independent variable, denoted by \(x\).
This information helps us understand broader trends in health over time. When life expectancy increases, it typically indicates better health conditions, improvements in healthcare, and enhanced living standards. By evaluating how life expectancy changes over years, statisticians and researchers can infer these developments.
In statistical analyses, like linear regression, life expectancy serves as a practical example to showcase how data predictions are based on past trends. This is achieved through the calculation of regression lines, which we will explore in some of the subsequent sections.

Slope Calculation

Calculating the slope, often represented by \(m\), of a regression line is a fundamental step when conducting linear regression. The slope indicates how much \(y\), in this case, life expectancy, changes with a one-unit increase in \(x\), the years since 1990.
To find the slope, we use the formula:

• \(m = \frac{n(\sum xy) - (\sum x)(\sum y)}{n(\sum x^2) - (\sum x)^2}\)

In this context:

• \(n\) is the number of data points, which is 5.
• \(\sum xy\) is the sum of the product of each pair of \(x\) and \(y\).
• \(\sum x\) and \(\sum y\) respectively are the sums of all the \(x\) values and \(y\) values.
• \(\sum x^2\) is the sum of the squares of \(x\) values.

By substituting the values from our dataset into the formula, we found that the slope is approximately 3.66. This means that, on average, the life expectancy of men increases by about 3.66 years for each decade since 1990.

Regression Line Equation

The regression line equation is crucial in predicting the dependent variable based on changes in the independent variable. The linear regression equation takes the form \(y = mx + b\), where \(m\) is the slope of the line and \(b\) is the y-intercept.
In our example, using the calculated slope \(m = 3.66\) and the y-intercept \(b = 29.828\), the regression line becomes:

• \(y = 3.66x + 29.828\)

This equation models life expectancy based on how many years it has been since 1990. It's a straightforward linear relationship showing the correlation between time and life expectancy, allowing us to use \(x\) values to predict \(y\) values. This equation is particularly useful because even with limited data points, it can approximate future outcomes by assuming that the historical trend continues.

Data Prediction

Data prediction using a regression line allows us to estimate future values based on existing trends. In the context of our exercise, we use the regression line to project what the life expectancy might be in future years.
To make predictions for the years 2020 and 2025, we first calculate \(x\) by determining the number of years since 1990:

• For 2020, \(x = 2020 - 1990 = 30\).
• For 2025, \(x = 2025 - 1990 = 35\).

Substituting these \(x\) values into the regression line equation \(y = 3.66x + 29.828\), we get:

• Life expectancy in 2020: \(y_{2020} = 3.66 \times 30 + 29.828 = 109.728\).
• Life expectancy in 2025: \(y_{2025} = 3.66 \times 35 + 29.828 = 157.028\).

These numerical outcomes suggest an apparent error in prediction as life expectancy values are unrealistically high, indicating a need to reassess assumptions or calculation methods used in creating the regression line.