Question

The table below shows the average annual cost of health insurance for a single individual, from 1999 to 2019, according to the Kaiser Family Foundation.YearCost1999$2,1962000$2,4712001$2,6892002$3,0832003$3,3832004$3,6952005$4,0242006$4,2422007$4,4792008$4,7042009$4,8242010$5,0492011$5,4292012$5,6152013$5,8842014$6,0252015$6,2512016$6,1962017$6,4352017$6,8962019$7,186(a) Using only the data from the first and last years, build a linear model to describe the cost of individual health insurance from 1999 onward. Use t to represent years after 1999 (treating 1999 as year 0).Pt = (b) Using this linear model, predict the cost of insurance in 2030.$ (c) = According to this model, when do you expect the cost of individual insurance to reach $12,000? Give your answer as a calendar year (ex: 2020).During the year (d) Using a calculator or spreadsheet program, build a linear regression model to describe the cost of individual insurance from 1999 onward. Use t to represent years after 1999 (treating 1999 as year 0), and round the values of P0 and d to the nearest dollar.Pt= (e) Using the regression model, predict the cost of insurance in 2030.$ (f) According to the regression model, when do you expect the cost of individual insurance to reach $12,000? Give your answer as a calendar year (ex: 2020).During the year

Answer 1

Part (a) Using only the data from the first and last years, build a linear model to describe the cost of individual health insurance from 1999 onward. Use t to represent years after 1999 (treating 1999 as year 0).

we have the ordered pairs

(1999, 2,196) -------> (0,2,196)

(2019,7,186) -------> (20,7,196)

Find out the slope

where

t -----> is the number of years since 1999

P ----> the cost

m=(7,196-2,196)/(20-0)

m=5,000/20

m=250

Find the equation of the linear model in slope-intercept form

P=mt+b

we have

m=250

point (0,2,196)

substitute and solve for b

2,196=250(0)+b

b=2,196

therefore

P=250t+2,196

Part b

Using this linear model, predict the cost of insurance in 2030

For t=2030=2030-1999=31 years

substitute

P=250(31)+2,196

P=$9,946

Part c

According to this model, when do you expect the cost of individual insurance to reach $12,000? Give your answer as a calendar year (ex: 2020).

For P=$12,000

substitute in the linear model

12,000=250t+2,196

250t=12,000-2,196

250t=9,804

t=39 years

therefore

1999+39=year 2038

Part d

Using a calculator or spreadsheet program, build a linear regression model to describe the cost of individual insurance from 1999 onward. Use t to represent years after 1999 (treating 1999 as year 0), and round the values of

P0 and d to the nearest dollar

using a regression calculator

the equation is

ŷ = 239.15065X + 2406.39827

y=239x+2,406

Part e

Using the regression model, predict the cost of insurance in 2030

For t=2030-1999=31 years

P=239(31)+2,406

P=$9,815

Part f

According to the regression model, when do you expect the cost of individual insurance to reach $12,000? Give your answer as a calendar year (ex: 2020).

For P=$12,000

substitute

12,000=239x+2,406

239x=12,000-2,406

239x=9,594

t=40 years

year=1999+40=2039