Question

The table below gives the age and bone density for five randomly selected women. Using this data, consider the equation of the regression line, y-+bf predicting a woman's bone density based on her age. Keep in mind, the correlation coefficient may or may not be statistically significant for the data gven Remember, i practice, it would not be appropriate to use the regression line to make a prediction if the correlation coefficient is not statistically significant Age 36 45 48 57 65 Bone Density 351 339 335 325 320 Table Copy Daa

i need help ASAP with step 1,2,3,4,5 and 6 thank you

Answer 1

necessary calculation table:-

age(x)	density(y)	x^2	y^2	xy
36	351	1296	123201	12636
45	339	2025	114921	15255
48	335	2304	112225	16080
57	325	3249	105625	18525
65	320	4225	102400	20800
sum=251	sum=1670	sum=13099	sum=558372	sum=83296

1).regression slope be:-

$\beta_{1}=\frac{n\sum xy-\sum x\sum y}{n\sum x^2-(\sum x)^2}$

  $\beta_{1}=\frac{(5*83296)-(251*1670)}{(5*13099)-(251)^2 }\approx -1.079$

2).estimated y-intercept:-

$\beta_{0}=\bar{y}-(\beta_{1}*\bar{x})$

$=\frac{1670}{5}-(-1.079*\frac{251}{5})=388.1658\approx 388.169$

3).this statement that " not all points predicted by the regression model fall on the same line" is false.

[ explanation:-

table containing the predicted value of y based on the regression equation:-

density(y)	predicted value of y
351	349.316
339	339.6087
335	336.3729
325	326.6656
320	318.0369

predicted value of y vs age(x) predicted value of y 35 40 45 50 age(x) 55 60 65

4). the estimated regression model is:-

bone density = 388.169 -1.079 age

i.e $\hat{y}=388.169-1.079 x$

according to this model if the independent variable increases by one unit then the dependent variable will decrease by -1.709 unit.

5).at x= 0 the dependent variable $\hat{y}$ be:- 388.169 ($\beta_{0}$)

6).here $\bar{y}$ = (1670 / 5 ) = 334

table for calculation:-

predicted value of y ($\hat{y}$)	$(y-\bar{y})^2$	$(\hat{y}-\bar{y})^2$
349.316	289	234.5786
339.6087	25	31.45708
336.3729	1	5.63063
326.6656	81	53.79346
318.0369	196	254.8209
sum	592	580.281

so, coefficient of determination:-

$R^2=\frac{SSR}{SST}=\frac{\sum (\hat{y}-\bar{y})^2}{\sum (y-\bar{y})^2}=\frac{580.281}{592}\approx 0.980$

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