## Related questions with answers

a. find the regression equation for the data points. b. graph the regression equation and the data points. c. describe the apparent relationship between the two variables under consideration. d. interpret the slope of the regression line. e. identify the predictor and response variables. f. identify outliers and potential influential observations. g. predict the values of the response variable for the specified values of the predictor variable, and interpret your results. In the article “The Human Vomeronasal Organ. Part II: Prenatal Development” (Journal of Anatomy, Vol. 197, Issue 3, pp. 421-436), T. Smith and K. Bhatnagar examined the controversial issue of the human vomeronasal organ, regarding its structure, function, and identity. The following table shows the age of fetuses (x), in weeks, and length of crown-rump (y), in millimeters. For part (g), predict the crown-rump length of a 19-week-old fetus.

$\begin{array}{l|llllllllll} \hline x & 10 & 10 & 13 & 13 & 18 & 19 & 19 & 23 & 25 & 28 \\ \hline y & 66 & 66 & 108 & 106 & 161 & 166 & 177 & 228 & 235 & 280 \\ \hline \end{array}$

Solution

VerifiedGiven:

$\begin{align*} n&=\text{Sample size}=10 \end{align*}$

(a) Let us first determine the necessary sums:

$\begin{align*} \sum x_i&=10+10+...+28=178 \\ \sum x_i^2&=10^2+10^2+...+28^2=3522 \\ \sum y_i&=66+66+...+280=1593 \\ \sum y_i^2&=66^2+66^2+...+280^2=302027 \\ \sum x_iy_i&=10\cdot 66+10\cdot 66+...+28\cdot 280=32476 \end{align*}$

Next, we can determine $S_{xx}$ and $S_{xy}$

$\begin{align*} S_{xx}&=\sum x_i^2-\dfrac{(\sum x_i)^2}{n}=3522-\dfrac{178^2}{10}=353.6 \\ S_{xy}&=\sum x_i y_i-\dfrac{(\sum x_i)(\sum y_i)}{n}=32476-\dfrac{178\cdot 1593}{10}=4120.6 \end{align*}$

The estimate $b$ of the slope $\beta$ is the ratio of $S_{xy}$ and $S_{xx}$:

$b=\dfrac{S_{xy}}{S_{xx}}=\dfrac{4120.6}{353.6}=11.6533$

The mean is the sum of all values divided by the number of values:

$\begin{align*} \overline{x}&=\dfrac{\sum x_i}{n}=\dfrac{178}{10}=17.8 \\ \overline{y}&=\dfrac{\sum y_i}{n}=\dfrac{1593}{10}=159.3 \end{align*}$

The estimate $a$ of the intercept $\alpha$ is the average of $y$ decreased by the product of the estimate of the slope and the average of $x$.

$a=\overline{y}-b\overline{x}=159.3-11.6533\cdot 17.8=-48.1284$

General least-squares equation: $\hat{y}=\alpha+\beta x$. Replace $\alpha$ by $a=-48.1284$ and $\beta$ by $b=11.6533$ in the general least-squares equation:

$y=a+bx=-48.1284+11.6533x$

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