Question

The table below shows the life expectancy for an individual born in the United States in certain years.

Year of Birth	Life Expectancy
1930	59.7
1940	62.9
1950	70.2
1965	69.7
1973	71.4
1982	74.5
1987	75
1992	75.7
2010	78.7

1. Find the estimated life expectancy for an individual born in 1973

2. Use the two points in part (e) to plot the least squares line on your graph from part (b).

3. Are there any outliers in the data?Yes, 1930 and 2010 are outliers.Yes, 1930 and 1950 are outliers.     Yes, 1950 is an outlier.No, there are no outliers

4. Using the least squares line, find the estimated life expectancy for an individual born in 1870. (Round your answer to one decimal place.)
Does the least squares line give an accurate estimate for that year? Explain why or why not. Yes, because the estimate is over 50 years.No, because 1870 is outside the domain of the least squares line.    

Answer 1

1) -

To find the estimated life expectancy for year 1973, we have to find equation of regression line-

Equation of regression line - Y = b₀ + b₁X

Where,

Observation table -

Year(x)	Life expectancy(y)	(x=x_bar)	(y-y_bar)	(x-x_bar)(y-y_bar)	(x-x_bar)^2	(y-y_bar)^2
1930	59.7	-39.889	-11.1667	445.4284963	1591.132321	124.6951889
1940	62.9	-29.889	-7.9667	238.1166963	893.352321	63.46830889
1950	70.2	-19.889	-0.6667	13.2599963	395.572321	0.44448889
1965	69.7	-4.889	-1.1667	5.7039963	23.902321	1.36118889
1973	71.4	3.111	0.5333	1.6590963	9.678321	0.28440889
1982	74.5	12.111	3.6333	44.0028963	146.676321	13.20086889
1987	75	17.111	4.1333	70.7248963	292.786321	17.08416889
1992	75.7	22.111	4.8333	106.8690963	488.896321	23.36078889
2010	78.7	40.111	7.8333	314.2014963	1608.892321	61.36058889
17729	637.8			1239.966667	5450.888889	305.26

Calculations -

  

Hence, the equation of regression line is -

Y = -377.283 + 0.2275X

Estimated value of life expectancy for year 1973 -

Y = -377.283 + 0.2275X = -377.283 + (0.2275)(1973) =  -377.283 + 448.8575 = 71.5745

Estimated life expectancy for year 1973 is 71.5745 years.

b) -

Least square regression line on the graph -

c) -

To find outliers, we have to find out the standardized residual for each year life expectancy. If any standardized residual is greater than 3, then it is an outlier.

Formula for standardized residual is -

Where ,

n = Total number of observation = 9

k = parametres estimated = 2

Observation table -

Year(x)	Life expectancy(y)	y_hat	ei = Y-y_hat	ei^2
1930	59.7	61.792	-2.092	4.376464
1940	62.9	64.067	-1.167	1.361889
1950	70.2	66.342	3.858	14.884164
1965	69.7	69.7545	-0.0545	0.00297025
1973	71.4	71.5745	-0.1745	0.03045025
1982	74.5	73.622	0.878	0.770884
1987	75	74.7595	0.2405	0.05784025
1992	75.7	75.897	-0.197	0.038809
2010	78.7	79.992	-1.292	1.669264
				23.19273475

Calculations -

Observation table for standardized residuals -

ei = Y-y_hat	Standardized residuals
-2.092	-1.18201523
-1.167	-0.659374653
3.858	2.179834971
-0.0545	-0.030793418
-0.1745	-0.098595439
0.878	0.496084786
0.2405	0.13588655
-0.197	-0.111308318
-1.292	-0.730001758

Here, we can see that no standardized residual is greater than 3, there are no outlier.

So, There are no outlier.

4) -

We have to estimate life expectancy for year 1870 -

Y = 0.2275X - 377.283 = (1870)(0.2275) - 377.283 = 425.425 - 377.283 = 48.142 48.1

Least square regression line does not give accurate value,because 1870 is outside the domain of the least square line.