Question

the table lists the average tuition and fees at private colleges and universities for selected years.

year	1985	1990	1995	2000	2008

(a) find the equation of the least-squares regression line that models the data.
y ≈
(type the slope as a decimal rounded to three decimal places. round the y-intercept to the nearest integer.)

Explanation:

Step1: Define Variables

Let \( x \) be the number of years since 1985. So for 1985, \( x = 0 \); 1990, \( x = 5 \); 1995, \( x = 10 \); 2000, \( x = 15 \); 2008, \( x = 23 \). The \( y \)-values are the tuition and fees: \( y_1 = 5344 \), \( y_2 = 9345 \), \( y_3 = 12370 \), \( y_4 = 16150 \), \( y_5 = 25144 \).

Step2: Calculate Necessary Sums

- \( \sum x = 0 + 5 + 10 + 15 + 23 = 53 \)
- \( \sum y = 5344 + 9345 + 12370 + 16150 + 25144 = 68353 \)
- \( \sum xy = (0\times5344)+(5\times9345)+(10\times12370)+(15\times16150)+(23\times25144) = 0 + 46725 + 123700 + 242250 + 578312 = 990987 \)
- \( \sum x^2 = 0^2 + 5^2 + 10^2 + 15^2 + 23^2 = 0 + 25 + 100 + 225 + 529 = 879 \)

Step3: Calculate Slope (\( m \))

The formula for the slope of the least - squares regression line is \( m=\frac{n\sum xy-\sum x\sum y}{n\sum x^2 - (\sum x)^2} \), where \( n = 5 \).

Substitute the values:
\[ \begin{align*} m&=\frac{5\times990987-53\times68353}{5\times879-(53)^2}\\ &=\frac{4954935 - 3622709}{4395 - 2809}\\ &=\frac{1332226}{1586}\\ &\approx 840.0 \end{align*} \]
Wait, that can't be right. Wait, recalculate \( \sum xy \) and \( \sum x^2 \) and \( \sum x \) and \( \sum y \) again.

Wait, 23*25144: 25144*20 = 502880, 25144*3 = 75432, so 502880+75432 = 578312. Correct. 15*16150 = 242250, correct. 10*12370 = 123700, correct. 5*9345 = 46725, correct. So \( \sum xy = 46725+123700 = 170425 + 242250 = 412675 + 578312 = 990987 \). Correct.

\( \sum x = 0 + 5+10 + 15+23 = 53 \), correct. \( \sum y = 5344+9345 = 14689 + 12370 = 27059 + 16150 = 43209 + 25144 = 68353 \), correct.

\( \sum x^2 = 0 + 25+100 + 225+529 = 879 \), correct. \( n = 5 \).

So numerator: \( 5\times990987=4954935 \), \( 53\times68353 = 53\times(68000 + 353)=53\times68000+53\times353 = 3604000+18709 = 3622709 \). Then \( 4954935 - 3622709 = 1332226 \).

Denominator: \( 5\times879 = 4395 \), \( 53^2 = 2809 \), \( 4395 - 2809 = 1586 \).

\( m=\frac{1332226}{1586}\approx 840.0 \)? Wait, no, 1332226 ÷ 1586: 1586×840 = 1586×800+1586×40 = 1,268,800+63,440 = 1,332,240. Oh! So 1332226 is 14 less than 1586×840. So \( m\approx 839.991 \approx 840.0 \) (rounded to three decimal places).

Step4: Calculate Y - Intercept (\( b \))

The formula for the \( y \)-intercept is \( b=\frac{\sum y - m\sum x}{n} \).

Substitute the values:
\[ \begin{align*} b&=\frac{68353-839.991\times53}{5}\\ &=\frac{68353 - 44519.523}{5}\\ &=\frac{23833.477}{5}\\ &\approx 4767 \end{align*} \]

Wait, let's check with a more accurate calculation of \( m \). Let's do the division 1332226 ÷ 1586 more accurately. 1586×839 = 1586×(800 + 39)=1586×800+1586×39 = 1,268,800+61,854 = 1,330,654. Then 1332226 - 1,330,654 = 1,572. 1,572÷1586≈0.991. So \( m = 839.991 \).

Then \( b=\frac{\sum y - m\sum x}{n}=\frac{68353-(839.991\times53)}{5} \). 839.991×53 = (840 - 0.009)×53 = 840×53-0.009×53 = 44520 - 0.477 = 44519.523. Then 68353 - 44519.523 = 23833.477. 23833.477÷5 = 4766.6954≈4767.

So the equation of the least - squares regression line is \( y\approx 839.991x + 4767 \).

Wait, but let's verify with a calculator - like approach. Maybe we made a mistake in defining \( x \). Let's re - define \( x \) as the actual year minus 1985, so 1985: \( x = 0 \), 1990: \( x = 5 \), 1995: \( x = 10 \), 2000: \( x = 15 \), 2008: \( x = 23 \) (correct).

Alternatively, use the formula for linear regression:

The general form is \( y = mx + b \), where

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

and

\( b=\frac{\sum y - m\sum x}{n} \)

We can also use a calculator - like method. Let's use the following table:

| \( x \) | \( y \) | \( xy \) | \( x^2 \) |
|---|---|---|---|
| 0 | 5344 | 0 | 0 |
| 5 | 9345 | 46725 | 25 |
| 10 | 12370 | 123700 | 100 |
| 15 | 16150 | 242250 | 225 |
| 23 | 25144 | 578312 | 529 |

Sums:

\( \sum x = 53 \), \( \sum y = 68353 \), \( \sum xy = 990987 \), \( \sum x^2 = 879 \), \( n = 5 \)

Calculating \( m \):

\[ m=\frac{5\times990987 - 53\times68353}{5\times879 - 53^2}=\frac{4954935-3622709}{4395 - 2809}=\frac{1332226}{1586}\approx 839.991 \]

Calculating \( b \):

\[ b=\frac{68353-839.991\times53}{5}=\frac{68353 - 44519.523}{5}=\frac{23833.477}{5}\approx 4767 \]

Answer:

\( y\approx 839.991x + 4767 \)