Question

government funding: the following table presents the budget (in millions of dollars) for selected organizations that received u.s. government funding for arts and culture at two different points in time. use the ti - 84 plus calculator as needed.
organization\tfive years ago\tlast year
national endowment for the arts\t460\t447
kennedy center for the performing arts\t247\t231
national gallery of art\t142\t153
corporation for public broadcasting\t124\t172
institute of museum and library services\t95\t159
commission of fine arts\t18\t42
national endowment for the humanities\t2\t5

part 1 of 3
(a) compute the least - squares regression line for predicting last years budget from the five - years - ago budget. round the slope and y - intercept to at least four decimal places.
the equation for the least - squares regression line is (hat{y}=)

part 2 of 3
(b) if two institutions had budgets that differed by 12 million dollars five years ago, by how much would you predict their budgets to differ last year? round the answer to at least two decimal places.
last year, their budgets would differ by (square) million dollars.

part 3 of 3
(c) predict last years budget for an organization whose five - years - ago budget was 80 million dollars. round the answer to at least two decimal places.
last years budget is predicted to be (square) million dollars.

Explanation:

Step1: Recall least - squares regression formula

The least - squares regression line is of the form $\hat{y}=b_0 + b_1x$, where $b_1=\frac{n\sum_{i = 1}^{n}x_iy_i-\sum_{i = 1}^{n}x_i\sum_{i = 1}^{n}y_i}{n\sum_{i = 1}^{n}x_i^{2}-(\sum_{i = 1}^{n}x_i)^{2}}$ and $b_0=\bar{y}-b_1\bar{x}$, with $n$ being the number of data points, $x_i$ the five - years - ago budget, and $y_i$ the last year budget.
Let $x$ be the budget five years ago and $y$ be the budget last year. We have $n = 7$ data points.
Calculate $\sum_{i=1}^{7}x_i=460 + 247+142 + 124+95+18+2=1088$
$\sum_{i=1}^{7}y_i=447+231+153+172+159+42+5 = 1209$
$\sum_{i=1}^{7}x_iy_i=460\times447+247\times231+142\times153+124\times172+95\times159+18\times42+2\times5$
$=205620+57057+21726+21328+15105+756+10 = 321602$
$\sum_{i=1}^{7}x_i^{2}=460^{2}+247^{2}+142^{2}+124^{2}+95^{2}+18^{2}+2^{2}$
$=211600+61009+20164+15376+9025+324+4 = 317502$
$b_1=\frac{7\times321602-1088\times1209}{7\times317502 - 1088^{2}}$
$=\frac{2251214-1315392}{2222514-1183744}=\frac{935822}{1038770}\approx0.8912$
$\bar{x}=\frac{1088}{7}\approx155.4286$
$\bar{y}=\frac{1209}{7}\approx172.7143$
$b_0=172.7143-0.8912\times155.4286$
$=172.7143 - 138.4377=34.2766$

Step2: Write the regression equation

The least - squares regression line is $\hat{y}=34.2766 + 0.8912x$

Step3: Solve part (b)

If the difference in $x$ values ($\Delta x$) is 12 million dollars, we use the slope of the regression line.
The relationship between the change in $x$ and the change in $\hat{y}$ is given by $\Delta\hat{y}=b_1\Delta x$.
Since $b_1 = 0.8912$ and $\Delta x=12$, then $\Delta\hat{y}=0.8912\times12 = 10.6944\approx10.69$ million dollars.

Step4: Solve part (c)

When $x = 80$, substitute into the regression equation $\hat{y}=34.2766+0.8912\times80$
$\hat{y}=34.2766 + 71.296=105.5726\approx105.57$ million dollars

Answer:

(a) $\hat{y}=34.2766 + 0.8912x$
(b) $10.69$
(c) $105.57$