Answer to Question #218371 in Statistics and Probability for Leah

Question

The table below depicts information on the retirement age and age at which one dies. Use the table to answer the questions that follow.
Retirement age(x)| 55 | 61| 52| 65| 70 |45 |60 |
Death age(y)| 60 |70 |88 |90 |62 | 50 | 77|
i) Calculate the Pearson correlation coefficient and interpret
ii) From a regression equation of the form y=α+βX+ε, where y is the retirement age, X is age at which one dies and ε, estimate the values for the constant and slope of the regression and interpret.
iii) Predict the age at which one will die if the person retires at age 78.
iv) Calculate the coefficient of determination and interpret.

Answer 1

$i) \; bar{x}= \frac{\sum^n_{i=1} x_i}{n} \\ = \frac{408}{7}= 58.285 \\ \bar{y} = \frac{\sum^n_{i=1} y_i}{n} \\ = \frac{497}{7}=71 \\ SS_{xx}= \sum^n_{i=1} x^2_i - \frac{( \sum^n_{i=1} x_i )^2}{n} \\ = 24200 - \frac{408^2}{7} \\ = 419.429 \\ SS_{yy}= \sum^n_{i=1} y^2_i - \frac{( \sum^n_{i=1} y_i )^2}{n} \\ = 36617 - \frac{497^2}{7} \\ = 1330 \\ SS_{xy}= \sum^n_{i=1} x_iy_i - \frac{( \sum^n_{i=1} x_i )( \sum^n_{i=1} y_i )}{n} \\ = 29206 - \frac{408 \times 497}{7} \\ = 238$

The correction coefficient is:

$r = \frac{SS_{xy}}{\sqrt{ SS_{xx} \times SS_{yy} }} \\ = \frac{238}{\sqrt{419.429 \times 1330}} \\ = 0.319$

r=0.319 so retirement age and death age are low positive correlated.

ii) In b part one is mistakes they say that where y is the retirement age, X is the age at which one dies and, is a logical mistake.

As per data we consider Retirement age is(x) and Death age is (y)"

The regression coefficient (the slope m, and the y-intercept n) are:

$m = \frac{SS_{xy}}{SS_{xx}} \\ = \frac{238}{419.429} \\ = 0.5674 \\ n = \bar{y} - \bar{x} \times m \\ = 71 -58.285 \times 0.5674 \\ = 37.9264$

The regression equation is:

$y = 37.9264 +0.5674x \\ iii) \; y(78) = 37.9264 + 0.5674 \times 78 \\ = 82.18 = 82$

iv) The coefficient of determination

$r^2 = 0.319^2 = 0.101$

So, 0.1 R-square means that your model explains 10% of the variation within the data.