A withdrawal of $225

Mathematics

1 answer:

igor_vitrenko [27]3 years ago

4 0

Answer:

ok bro

Step-by-step explanation:

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Consider this scenario: The population of a city increased steadily over a ten-year span. The following ordered pairs show the p

vova2212 [387]

Answer:

Regression function: $y=1986.406+0.0059x$

The function predicts that population will reach 14,000 in year 2068.

Step-by-step explanation:

We have to determine a function $y=b_0+b_1x_1$ by applying linear regression. The data we have is 5 pair of points which relates population to year.

According to the simple regression model (one independent variable), if we minimize the error between the model (the linear function) and the points given, the parameters are:

$b_0=\bar{y}+b_1\bar{x}\\\\b_1=\frac{\sum\limits^5_{i=1} {(x_i-\bar x)(y_i-\bar y)}}{\sum\limits^5_{i=1} {(x_i-\bar x)^2}}$

We start calculating the average of x and y

$\bar x=\frac{2500+2650+3000+3500+4200}{5}=\frac{15850}{5}=3170\\\\ \bar y=\frac{2001+2002+2004+2007+2011}{5}=\frac{10025}{5}=2005$

The sample covariance can be calculated as

$\sum\limits^5_{i=1} {(x_i-\bar x)(y_i-\bar y)}=(2500-3170)(2001-2005)+(2650-3170)(2002-2005)+(3000-3170)(2004-2005)+(3500-3170)(2007-2005)+(4200-3170)(2011-2005)\\\\\sum\limits^5_{i=1} {(x_i-\bar x)(y_i-\bar y)}=2680+1560+170+660+6180\\\\ \sum\limits^5_{i=1} {(x_i-\bar x)(y_i-\bar y)}=11250$

The variance of x can be calculated as

$\sum\limits^5_{i=1} {(x_i-\bar x)^2}=(2500-3170)^2+(2650-3170)^2+(3000-3170)^2+(3500-3170)^2+(4200-3170)^2\\\\\sum\limits^5_{i=1} {(x_i-\bar x)^2}=448900+270400+28900+108900+1060900\\\\\sum\limits^5_{i=1} {(x_i-\bar x)^2}=1918000$

Now we can calculate the parameters of the regression model

$b_1=\frac{\sum\limits^5_{i=1} {(x_i-\bar x)(y_i-\bar y)}}{\sum\limits^5_{i=1} {(x_i-\bar x)^2}}=\frac{11250}{1918000}=0.005865485 \\\\ b_0=\bar{y}+b_1\bar{x}=2005-0.005865485*3170=1986.406413$