Question

Marriage Prospects Data released by the Census Bureau in 1986 indicated the likelihood that never-married women would eventually marry. The data indicated that the older the woman, the less the likelihood of marriage. Specifically, two statistics indicated that women who were 45 and never-married had an 18 percent chance of marriage and women 25 years old had a 78 percent chance of marriage. Assume that a linear fit to these two data points provides a reasonable approximation for the function p=f(a), where p equals the probability of marriage and a equals the age of a never- married woman.

(a) Determine the linear function p=f(a).
(b) Interpret the slope and p intercept.
(c) Do the values in part b seem reasonable?
(d) If the restricted domain on this function is 20 sa s 50, determine f(20), f(30), f(40), and f(50).

Answer 1

a.
`f(a) = -0.03a +1.53`

b. See Explanation

c. The slope is reasonable but the p intercept is not

d.
`f(20) = 93\%`
`f(30) = 63\%`
`f(40) = 33\%`
`f(50) = 3\%`

Explanation:

Given

`a = age`

`p = probability\ of\ marriage`

`a = 45` when
`p = 18\%`

`a = 25` when
`p = 78\%`

Solving (a): The linear function

We start by calculating the slope, m

`m = (p_2 - p_1)/(a_2 - a_1)`

`m = (78\% - 18\%)/(25- 45)`

`m = (60\%)/(-20)`

`m = -3\%`

`m = -0.03`

The function is then calculated as follows

`p - p_1 = m(a - a_1)`

This gives:

`p - 18\% = -0.03(a - 45)`

`p - 0.18 = -0.03(a - 45)`

`p - 0.18 = -0.03a +1.35`

Solve for p

`p= -0.03a +1.35+0.18`

`p= -0.03a +1.53`

Hence,

`f(a) = -0.03a +1.53`

Solving (b): Interpret the slope and the p intercept

The slope is calculated as:

`m = -0.03`

And it implies that, there is a 3% reduction in change of getting older as women get older

The p intercept implies that, there is a 1.53 chance for 0 years old female child to get married.

Solving (c): Is (b) reasonable

The slope is reasonable.

However, the p intercept is not because of the age of the woman

Solving (d): Determine f(20), f(30), f(40), f(50)

We have that:

`f(a) = -0.03a +1.53`

`f(20) = -0.03 * 20 + 1.53`

`f(20) = -0.6 + 1.53`

`f(20) = 0.93`

`f(20) = 93\%`

`f(30) = -0.03 * 30 + 1.53`

`f(30) = -0.9 + 1.53`

`f(30) = 0.63`

`f(30) = 63\%`

`f(40) = -0.03 * 40 + 1.53`

`f(40) = -1.2 + 1.53`

`f(40) = 0.33`

`f(40) = 33\%`

`f(50) = -0.03 * 50 + 1.53`

`f(50) = -1.5 + 1.53`

`f(50) = 0.03`

`f(50) = 3\%`