Question

according to data collected by the world bank, the number of internet users in the united states and china has varied over the past decade as follows:

a) find a linear - regression modeling the number (in millions) of internet users u in the united states t years from 1998. round your slope and vertical intercept to three decimal places.
the number of internet users in the united states t years from 1998 is u(t) =

b) find a linear - regression modeling the number (in millions) of internet users c in china t years from 1998. round your slope and vertical intercept to three decimal places.
the number of internet users in china t years from 1998 is c(t) =

c) according to your models when will china have more internet users than the united states? round to the nearest year.
after the year, china will have more internet users.

Explanation:

Step1: Recall linear - regression formula

The linear - regression equation is of the form $y = mx + b$, where $m$ is the slope and $b$ is the y - intercept. For the US data points $(t_i,U_i)$ where $t$ is the number of years since 1998 and $U$ is the number of internet users (in millions).
Let's assume we have data points $(t_1,U_1),(t_2,U_2),\cdots,(t_n,U_n)$. The slope $m=\frac{n\sum_{i = 1}^{n}t_iU_i-\sum_{i = 1}^{n}t_i\sum_{i = 1}^{n}U_i}{n\sum_{i = 1}^{n}t_i^{2}-(\sum_{i = 1}^{n}t_i)^{2}}$ and the y - intercept $b=\frac{\sum_{i = 1}^{n}U_i - m\sum_{i = 1}^{n}t_i}{n}$.
For the US data: Let $t = 0$ for 1998, $t = 2$ for 2000, $t = 4$ for 2002, $t = 6$ for 2004, $t = 8$ for 2006, $t = 10$ for 2008. The data points for the US are $(0,41.31),(2,81.0),(4,131.17),(6,178.1),(8,208.5),(10,223.4)$.
$n = 6$, $\sum_{i = 1}^{6}t_i=0 + 2+4 + 6+8 + 10=30$, $\sum_{i = 1}^{6}U_i=41.31+81.0+131.17+178.1+208.5+223.4 = 863.48$, $\sum_{i = 1}^{6}t_iU_i=0\times41.31+2\times81.0+4\times131.17+6\times178.1+8\times208.5+10\times223.4=0 + 162+524.68+1068.6+1668+2234=5657.28$, $\sum_{i = 1}^{6}t_i^{2}=0^{2}+2^{2}+4^{2}+6^{2}+8^{2}+10^{2}=0 + 4+16+36+64+100 = 220$.
$m_{U}=\frac{6\times5657.28-30\times863.48}{6\times220 - 30^{2}}=\frac{33943.68-25904.4}{1320 - 900}=\frac{8039.28}{420}\approx19.141$.
$b_{U}=\frac{863.48-19.141\times30}{6}=\frac{863.48 - 574.23}{6}=\frac{289.25}{6}\approx48.208$.
So, $U(t)=19.141t + 48.208$.

Step2: Calculate linear - regression for China

For the China data points $(t_i,C_i)$. The data points for China are $(0,1.31),(2,22.5),(4,59.1),(6,94.04),(8,137.18),(10,221.68)$.
$n = 6$, $\sum_{i = 1}^{6}t_i = 30$, $\sum_{i = 1}^{6}C_i=1.31+22.5+59.1+94.04+137.18+221.68=535.81$, $\sum_{i = 1}^{6}t_iC_i=0\times1.31+2\times22.5+4\times59.1+6\times94.04+8\times137.18+10\times221.68=0 + 45+236.4+564.24+1097.44+2216.8=4160.88$, $\sum_{i = 1}^{6}t_i^{2}=220$.
$m_{C}=\frac{6\times4160.88-30\times535.81}{6\times220 - 30^{2}}=\frac{24965.28-16074.3}{420}=\frac{8890.98}{420}\approx21.170$.
$b_{C}=\frac{535.81-21.170\times30}{6}=\frac{535.81 - 635.1}{6}=\frac{- 99.29}{6}\approx - 16.548$.
So, $C(t)=21.170t-16.548$.

Step3: Find when China has more users than the US

We want to find $t$ such that $C(t)>U(t)$.
$21.170t-16.548>19.141t + 48.208$.
$21.170t-19.141t>48.208 + 16.548$.
$2.029t>64.756$.
$t=\frac{64.756}{2.029}\approx32$.
Since $t$ is the number of years since 1998, the year is $1998 + 32=2030$.

Answer:

a) $U(t)=19.141t + 48.208$
b) $C(t)=21.170t-16.548$
c) After the year 2030, China will have more internet users.