Question

how strongly do physical characteristics of sisters and brothers correlate? the data in the table give the heights (in inches) of 12 adult pairs.
brother\tsister
71\t69
68\t64
68\t65
67\t63
70\t65
71\t62
70\t65
73\t64
72\t66
65\t59
66\t62
70\t64
to access the data, click the link for your preferred software format.
csv excel (xls) excel (xlsx) jmp mac-text minitab14-18 minitab18+ pc-text r spss ti crunchit!
assume damien is 70 inches tall. predict the height of his sister tonya. give your answer to one decimal place.
tonyas predicted height:
in

Explanation:

Step1: Find the regression equation

First, we need to calculate the regression equation for the height of brothers (x) and sisters (y). Let's list the data points:
Brother (x): 71, 68, 68, 67, 70, 71, 70, 73, 72, 65, 66, 70
Sister (y): 69, 64, 65, 63, 65, 62, 65, 64, 66, 59, 62, 64

First, calculate the mean of x ($\bar{x}$) and mean of y ($\bar{y}$):

$\bar{x}=\frac{71 + 68 + 68 + 67 + 70 + 71 + 70 + 73 + 72 + 65 + 66 + 70}{12}$

$=\frac{849}{12}=70.75$

$\bar{y}=\frac{69 + 64 + 65 + 63 + 65 + 62 + 65 + 64 + 66 + 59 + 62 + 64}{12}$

$=\frac{758}{12}\approx63.1667$

Next, calculate the slope (b) of the regression line:

$b=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})}{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}$

First, calculate $(x_{i}-\bar{x})(y_{i}-\bar{y})$ and $(x_{i}-\bar{x})^{2}$ for each data point:

For (71, 69):
$(71 - 70.75)(69 - 63.1667)=(0.25)(5.8333)=1.4583$
$(71 - 70.75)^{2}=0.0625$

For (68, 64):
$(68 - 70.75)(64 - 63.1667)=(-2.75)(0.8333)= - 2.2917$
$(68 - 70.75)^{2}=7.5625$

For (68, 65):
$(68 - 70.75)(65 - 63.1667)=(-2.75)(1.8333)= - 5.0417$
$(68 - 70.75)^{2}=7.5625$

For (67, 63):
$(67 - 70.75)(63 - 63.1667)=(-3.75)(-0.1667)=0.6251$
$(67 - 70.75)^{2}=14.0625$

For (70, 65):
$(70 - 70.75)(65 - 63.1667)=(-0.75)(1.8333)= - 1.3750$
$(70 - 70.75)^{2}=0.5625$

For (71, 62):
$(71 - 70.75)(62 - 63.1667)=(0.25)(-1.1667)= - 0.2917$
$(71 - 70.75)^{2}=0.0625$

For (70, 65):
$(70 - 70.75)(65 - 63.1667)=(-0.75)(1.8333)= - 1.3750$
$(70 - 70.75)^{2}=0.5625$

For (73, 64):
$(73 - 70.75)(64 - 63.1667)=(2.25)(0.8333)=1.8750$
$(73 - 70.75)^{2}=5.0625$

For (72, 66):
$(72 - 70.75)(66 - 63.1667)=(1.25)(2.8333)=3.5417$
$(72 - 70.75)^{2}=1.5625$

For (65, 59):
$(65 - 70.75)(59 - 63.1667)=(-5.75)(-4.1667)=23.9584$
$(65 - 70.75)^{2}=33.0625$

For (66, 62):
$(66 - 70.75)(62 - 63.1667)=(-4.75)(-1.1667)=5.5417$
$(66 - 70.75)^{2}=22.5625$

For (70, 64):
$(70 - 70.75)(64 - 63.1667)=(-0.75)(0.8333)= - 0.6250$
$(70 - 70.75)^{2}=0.5625$

Now, sum up the $(x_{i}-\bar{x})(y_{i}-\bar{y})$ terms:

$1.4583-2.2917 - 5.0417+0.6251-1.3750 - 0.2917 - 1.3750+1.8750+3.5417+23.9584+5.5417 - 0.6250$

Let's calculate step by step:

$1.4583-2.2917=-0.8334$

$-0.8334 - 5.0417=-5.8751$

$-5.8751+0.6251=-5.25$

$-5.25-1.3750=-6.625$

$-6.625 - 0.2917=-6.9167$

$-6.9167 - 1.3750=-8.2917$

$-8.2917+1.8750=-6.4167$

$-6.4167+3.5417=-2.875$

$-2.875+23.9584=21.0834$

$21.0834+5.5417=26.6251$

$26.6251 - 0.6250=26.0001\approx26$

Now sum up the $(x_{i}-\bar{x})^{2}$ terms:

$0.0625+7.5625+7.5625+14.0625+0.5625+0.0625+0.5625+5.0625+1.5625+33.0625+22.5625+0.5625$

$=0.0625\times4 + 7.5625\times2+14.0625+0.5625\times3+5.0625+1.5625+33.0625+22.5625$

$=0.25+15.125+14.0625+1.6875+5.0625+1.5625+33.0625+22.5625$

$=0.25+15.125=15.375$

$15.375+14.0625=29.4375$

$29.4375+1.6875=31.125$

$31.125+5.0625=36.1875$

$36.1875+1.5625=37.75$

$37.75+33.0625=70.8125$

$70.8125+22.5625=93.375$

So, $b=\frac{26}{93.375}\approx0.2784$

Now, calculate the intercept (a) of the regression line:

$a=\bar{y}-b\bar{x}$

$a = 63.1667-0.2784\times70.75$

$a = 63.1667 - 19.69\approx43.4767$

So the regression equation is $y = 43.4767+0.2784x$

Step2: Predict Tonya's height

Now, we are given that Damien's height (x) is 70 inches. We need to predict Tonya's height (y) using the regression equation:

$y = 43.4767+0.2784\times70$

$y = 43.4767+19.488$

$y\approx62.9647\approx63.0$ (rounded to one decimal place)

Answer:

63.0