Question

presidents and first ladies: the following table presents the ages of the last 10 u.s. presidents and their wives on the first day of their presidencies. use a ti - 84 calculator to compute the following.
name her age his age
donald and melania trump 46 70
barack and michelle obama 45 47
george w. and laura bush 54 54
bill and hillary clinton 45 46
george and barbara bush 63 64
ronald and nancy reagan 59 69
jimmy and rosalynn carter 49 52
gerald and betty ford 56 61
richard and pat nixon 56 56
lyndon and lady bird johnson 50 55
send data to excel
part: 0 / 4
part 1 of 4
(a) compute the least - square regression line for predicting the presidents age from the first ladys age. round the slope and y - intercept values to at least four decimal places.
\\( \hat{y}=\\)

Explanation:

Step1: Calculate necessary sums

Let $x$ be the first - lady's age and $y$ be the president's age.
We need to calculate $\sum x$, $\sum y$, $\sum x^{2}$, $\sum xy$, and $n$ (where $n = 10$).
For the given data:
$\sum x=46 + 45+54 + 45+63+59+49+56+56+50=523$
$\sum y=70 + 47+54 + 46+64+69+52+61+56+55=574$
$\sum x^{2}=46^{2}+45^{2}+54^{2}+45^{2}+63^{2}+59^{2}+49^{2}+56^{2}+56^{2}+50^{2}$
$=2116+2025+2916+2025+3969+3481+2401+3136+3136+2500 = 27605$
$\sum xy=46\times70 + 45\times47+54\times54+45\times46+63\times64+59\times69+49\times52+56\times61+56\times56+50\times55$
$=3220+2115+2916+2070+4032+4071+2548+3416+3136+2750 = 29274$

Step2: Calculate the slope $b_1$

The formula for the slope $b_1$ of the least - squares regression line $\hat{y}=b_0 + b_1x$ is $b_1=\frac{n\sum xy-\sum x\sum y}{n\sum x^{2}-(\sum x)^{2}}$
Substitute $n = 10$, $\sum x = 523$, $\sum y = 574$, $\sum xy = 29274$, and $\sum x^{2}=27605$ into the formula:
$n\sum xy-\sum x\sum y=10\times29274-523\times574$
$=292740 - 299102=-6362$
$n\sum x^{2}-(\sum x)^{2}=10\times27605-(523)^{2}$
$=276050-273529 = 2521$
$b_1=\frac{-6362}{2521}\approx - 2.5236$

Step3: Calculate the y - intercept $b_0$

The formula for the y - intercept $b_0$ is $b_0=\bar{y}-b_1\bar{x}$, where $\bar{x}=\frac{\sum x}{n}$ and $\bar{y}=\frac{\sum y}{n}$
$\bar{x}=\frac{523}{10}=52.3$
$\bar{y}=\frac{574}{10}=57.4$
$b_0=57.4-(-2.5236)\times52.3$
$=57.4 + 132.0843=189.4843$

Answer:

$\hat{y}=189.4843-2.5236x$