Question

The average annual cost (including tuition, room, board, books and fees) to attend a public college takes nearly a third of the annual income of a typical family with college-age children (Money, April 2012). At private colleges, the average annual cost is equal to about 60% of the typical family's income. The following random samples show the annual cost of attending private and public colleges. Data are in thousands of dollars.

Private Colleges
52.8 30.6 43.2 45.8 33.3
33.3 50.5 44.0 42.0 37.8

Public Colleges
20.3 22.8 28.2 18.5 15.6 25.6
24.1 14.4 28.5 21.8 22.0 25.8

Required:
a. Compute the sample mean and sample standard deviation for private and public colleges.
b. What is the point estimate of the difference between the two population means?
c. Develop a 95% confidence interval of the difference between the annual cost of attending private and pubic colleges.

Answer 1

Answer:

(a)

$\bar x_1 =41.33$     $\sigma_1 = 7.48$

$\bar x_2 =22.3$       $\sigma_2 = 4.53$

(b)

$d = 19.03$

(c)

$CI = 16.33$ to $21.73$

Step-by-step explanation:

Given

$n_1 = 10\\ n_2 = 12$

And the accompanying data for private (1) and public (2) colleges

Solving (a): The sample mean and sample standard deviations

Private College

Calculating sample mean

This is calculated as:

$\bar x_i = \frac{\sum x_i}{n_i}$

$\bar x_1 =\frac{52.8+30.6+43.2 +45.8 +33.3+33.3 +50.5 +44.0 +42.0 +37.8}{10}$

$\bar x_1 =\frac{413.3}{10}$

$\bar x_1 =41.33$

Calculating sample standard deviation

This is calculated as:

$\sigma_i = \sqrt\frac{\sum(x_i - \bar x_i)}{n_i-1}$

$\sigma_1 = \sqrt\frac{(52.8-41.33)^2+(30.6-41.33)^2+.........+(37.8-41.33)^2}{10-1}$

$\sigma_1 = \sqrt\frac{503.261}{9}$

$\sigma_1 = \sqrt{55.9178888889$

$\sigma_1 = 7.4778264816$

$\sigma_1 = 7.48$ -- approximated

Public College

Calculating sample mean

$\bar x_2 =\frac{20.3 +22.8 +28.2 +18.5 +15.6 +25.6+ 24.1 +14.4 +28.5 +21.8 +22.0 +25.8}{12}$

$\bar x_2 =\frac{267.6}{12}$

$\bar x_2 =22.3$

Calculating sample standard deviation

$\sigma_2 = \sqrt\frac{(20.3 -22.3)^2+(22.8 -22.3)^2+(28.2 -22.3)^2+(18.5 -22.3)^2+(15.6 -22.3)^2+................+(25.8-22.3)^2}{12-1}$$\sigma_2 = \sqrt\frac{225.96}{11}$

$\sigma_2 = \sqrt{20.5418181818$

$\sigma_2 = 4.532308262$

$\sigma_2 = 4.53$ -- approximated

Solving (b): Point estimate of the difference between the two population means

This is calculated as:

$d = \bar x_1 - \bar x_2$

$d = 41.33 - 22.3$

$d = 19.03$

Solving (c): 95% confidence interval

First, calculate the degrees of freedom:

$df = \frac{(\sigma_1^2/n_1 + \sigma_2^2/n_2)^2}{\frac{(\sigma_1^2/n_1)^2}{n_1 - 1} + \frac{(\sigma_2^2/n_2)^2}{n_2 - 1}}$

$df = \frac{(7.48^2/10 + 4.53^2/12)^2}{\frac{(7.48^2/10)^2}{10 - 1} + \frac{(4.53^2/12)^2}{12 - 1}}$

$df = \frac{53.3647}{31.3045/9 + 2.9244/11}$

$df = \frac{53.3647}{3.7441}$

$df = 14.2530$

$df = 14$

Calculate the critical value (t)

At 95% confidence interval and df = 14;

$df = 14$

The confidence interval is then calculated as:

$CI = (\bar x_1 - \bar x_2) \± \sqrt{\sigma_1^2/n_1 + \sigma_2^2/n_2}$

$CI = (41.33 - 22.3) \± \sqrt{7.48^2/10 + 4.53^2/12}$

$CI = 19.03 \± \sqrt{7.305115$

$CI = 19.03 \± 2.70$

Split

$CI = 19.03 - 2.70$ to $19.03 + 2.70$

$CI = 16.33$ to $21.73$

This implies that:

Private colleges have population mean annual cost of $16.33 to $21.73 more expensive than public colleges