Math / Data & Statistics

QuestionThe table lists the average tuition and fees at private colleges and universities for selected years. \begin{tabular}{|c|c|c|c|c|c|} \hline Year & 1985 & 1990 & 1995 & 2000 & 2008 \\ \hline \begin{tabular}{c} Tuition and \\ Fees \\ (in dollars) \end{tabular} & 5328 & 9369 & 12,336 & 16,154 & 25,111 \\ \hline \end{tabular} (a) Find the equation of the least-squares regression line that models the data. $y \approx$ $\square$ (Type the slope as a decimal rounded to three decimal places. Round the $y$ -intercept to the nearest integer.)

Studdy Solution

STEP 1

1. The independent variable $x$ represents the year.
2. The dependent variable $y$ represents the tuition and fees in dollars.
3. We will use the least-squares method to find the regression line.
4. The years will be transformed into a form suitable for regression analysis, typically by setting the first year as $x = 0$ .

STEP 2

1. Transform the year data for regression analysis.
2. Calculate the necessary sums for the least-squares formulas.
3. Compute the slope $m$ of the regression line.
4. Compute the y-intercept $b$ of the regression line.
5. Write the equation of the regression line.

STEP 3

Transform the year data by setting 1985 as $x = 0$ . The transformed years are:
\begin{align} 1985 & \rightarrow 0 \\ 1990 & \rightarrow 5 \\ 1995 & \rightarrow 10 \\ 2000 & \rightarrow 15 \\ 2008 & \rightarrow 23 \\ \end{align}

STEP 4

Calculate the necessary sums:
\begin{align} \sum x &= 0 + 5 + 10 + 15 + 23 = 53 \\ \sum y &= 5328 + 9369 + 12336 + 16154 + 25111 = 68298 \\ \sum xy &= (0 \times 5328) + (5 \times 9369) + (10 \times 12336) + (15 \times 16154) + (23 \times 25111) = 100,7853 \\ \sum x^2 &= 0^2 + 5^2 + 10^2 + 15^2 + 23^2 = 899 \\ \end{align}

STEP 5

Compute the slope $m$ using the formula:
$m = \frac{n(\sum xy) - (\sum x)(\sum y)}{n(\sum x^2) - (\sum x)^2}$
where $n = 5$ .
$m = \frac{5(1007853) - (53)(68298)}{5(899) - (53)^2} = \frac{5039265 - 3629794}{4495 - 2809} = \frac{1409471}{1686} \approx 835.993$

STEP 6

Compute the y-intercept $b$ using the formula:
$b = \frac{\sum y - m(\sum x)}{n}$
$b = \frac{68298 - 835.993 \times 53}{5} = \frac{68298 - 44207.629}{5} \approx 4818.874$
Round $b$ to the nearest integer: $b \approx 4819$ .

STEP 7

Write the equation of the regression line:
$y \approx 835.993x + 4819$
The equation of the least-squares regression line is:
$y \approx 835.993x + 4819$

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Studdy Solution

STEP 1

STEP 2

1. Transform the year data for regression analysis.
2. Calculate the necessary sums for the least-squares formulas.
3. Compute the slope $m$ of the regression line.
4. Compute the y-intercept $b$ of the regression line.
5. Write the equation of the regression line.

STEP 3

Transform the year data by setting 1985 as $x = 0$ . The transformed years are:
\begin{align} 1985 & \rightarrow 0 \\ 1990 & \rightarrow 5 \\ 1995 & \rightarrow 10 \\ 2000 & \rightarrow 15 \\ 2008 & \rightarrow 23 \\ \end{align}

STEP 4

STEP 5

Compute the slope $m$ using the formula:
$m = \frac{n(\sum xy) - (\sum x)(\sum y)}{n(\sum x^2) - (\sum x)^2}$
where $n = 5$ .
$m = \frac{5(1007853) - (53)(68298)}{5(899) - (53)^2} = \frac{5039265 - 3629794}{4495 - 2809} = \frac{1409471}{1686} \approx 835.993$

STEP 6

Compute the y-intercept $b$ using the formula:
$b = \frac{\sum y - m(\sum x)}{n}$
$b = \frac{68298 - 835.993 \times 53}{5} = \frac{68298 - 44207.629}{5} \approx 4818.874$
Round $b$ to the nearest integer: $b \approx 4819$ .

STEP 7

Write the equation of the regression line:
$y \approx 835.993x + 4819$
The equation of the least-squares regression line is:
$y \approx 835.993x + 4819$

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Studdy Solution

STEP 1

STEP 2

1. Transform the year data for regression analysis.2. Calculate the necessary sums for the least-squares formulas.3. Compute the slope m m m of the regression line.4. Compute the y-intercept b b b of the regression line.5. Write the equation of the regression line.

STEP 3

Transform the year data by setting 1985 as x=0 x = 0 x=0. The transformed years are:\begin{align*} 1985 & \rightarrow 0 \\ 1990 & \rightarrow 5 \\ 1995 & \rightarrow 10 \\ 2000 & \rightarrow 15 \\ 2008 & \rightarrow 23 \\ \end{align*}

STEP 4

STEP 5

STEP 6

STEP 7

Write the equation of the regression line:y≈835.993x+4819y \approx 835.993x + 4819y≈835.993x+4819 The equation of the least-squares regression line is:y≈835.993x+4819 y \approx 835.993x + 4819 y≈835.993x+4819

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1. Transform the year data for regression analysis.
2. Calculate the necessary sums for the least-squares formulas.
3. Compute the slope $m$ of the regression line.
4. Compute the y-intercept $b$ of the regression line.
5. Write the equation of the regression line.

Transform the year data by setting 1985 as $x = 0$ . The transformed years are:
\begin{align} 1985 & \rightarrow 0 \\ 1990 & \rightarrow 5 \\ 1995 & \rightarrow 10 \\ 2000 & \rightarrow 15 \\ 2008 & \rightarrow 23 \\ \end{align}

Write the equation of the regression line:
$y \approx 835.993x + 4819$
The equation of the least-squares regression line is:
$y \approx 835.993x + 4819$