#3.) Below are the average heights for American boys. Consider “birth” to be 0 y
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Question
#3.) Below are the average heights for American boys. Consider “birth” to be 0 years old.
a) Calculate the least squares (best-fit) line. Put the equation in the form of: ??? = a + bx. (Round your answers to two decimal places.)
b) If you plot these points on a graph (using your TI’s) does it appear from inspection that there is a relationship between the variables? Why or why not? (NOTE you do not need to draw a graph of your plot)
(c) Find the correlation coefficient r. (Round your answer to four decimal places.) Is it significant at the 0.01 level? If so why?
(d) Find the estimated height in cm for a 11 year old boy. (Use your equation from part a. Round your answer to one decimal place.)
(e) If you saw a boy who was 93.5 cm tall but you didn’t know how old he was, what age would you guess to equate to 93.5 cm? (Round your answer to the nearest year)
f) What percent of the variation in Height is explained by the variation in Age? (Round your answer to the nearest whole percent.)
Age (years) Height (cm) birth 50.8 2 83.8 3 91.4 5 106.6 7 119.3 10 137.1 14 157.5Explanation / Answer
(a) Let 'x' be Age (in years) and 'y' be height( in cm)
a = y-bar - b * x-bar
b = cov( x , y)/ var(x)
y-bar = mean(y) = 106.643
x-bar = mean(x) = 5.857
cov(x, y) = mean(xy) - mean (x) * mean(y) = 144.79
b = 7.095
a = 106.645 - 5.857 * 7.095 = 65.0895
So, the equation will be y = 65.09 + 7.09 x
(b) Yes ,if we plot these points on a graph it appear from inspection that there is a relationship between the variables because as the Age increases the height also increases. Hence, there is a positive relationship between Age and height.
(c) Correlation coefficient r = Cov(x, y) / ?( var(x) * var(y) ) = 0.9760
(d) we are given x = 11
So , y = 65.0895 + 11 * 7.095 = 143.1 cm
(e) we are given y = 93.5
So , x = (93.5 - 65.0895 )/ 7.095 = 4 years
(f) Percent of the variation in Height is explained by the variation in Age = (Correlation coefficient )2 = (0.976)2 = 0.9526 = 95%
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