Question

Crown-Rump Length. In the article "The Human Vomeronasal Organ. Part II: Prenatal Development" (hormonal of Anatomy, Vol. 197. Issue 3, pp. 421-436). T. Smith and $\mathrm{K}$. 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 the length of crown-rump (y), in millimeters. For part (g). predict the crown-rump length of a 19-week-old fetus.

X	10	10	13	13	18	19	19	23	25	28
Y	66	66	108	106	161	166	177	228	235	280

• a. fond 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.

Short answer

Ans:

part (a):$\hat{y}=-48.1284+11.6533x$

part (b): Graph representation

part (c): The length of the crown-rump increases as the age of fetuses increases.

part (d): The slope of the regression line is $11.6533$

part (e): The predictor variable is the age of fetuses and the response variable is the length of the crown rump.

part (f): All the points are near the straight line, so there are no outliers and influential observations.

Part (g). The predicted crown-rump length of a 19 -week-old status is $173.2843$millimeters.

Step-by-step solution

Step 1. GIven information:

Given Data points:

X	10	10	13	13	18	19	19	23	25	28
Y	66	66	108	106	161	166	177	228	235	280

Step 2. Solving part (a):

$\begin{array}{|cccc|}\hline x& y& xy& x^2\\ 10& 66& 660& 100\\ 10& 66& 660& 100\\ 13& 108& 1404& 169\\ 13& 106& 1378& 169\\ 18& 161& 2898& 324\\ 19& 166& 3154& 361\\ 19& 177& 3363& 361\\ 23& 228& 5244& 529\\ 25& 235& 5875& 625\\ 28& 280& 7840& 784\\ \sum x-178& \sum y=1593& \sum xy=32476& \sum x^2-3522\\ \hline\end{array}$

Step 3. Continue:

Number of observations, $n=10$

The slope of the regression line is,

$$\begin{gathered} b_1=\frac{\sum x_i{y}_i-\left(\sum x_i\right)\left(\sum y_i\right)/n}{\sum x_i^2-{\left(\sum x_i\right)}^2/n} \\ =\frac{32476-\left(178\right)\left(1593\right)/10}{3522-(178{)}^2/10} \\ =\frac{4120.6}{353.6} \\ =11.6533 \end{gathered}$$

The intercept is,

$$\begin{gathered} b_0=\frac{1}{n}\left(\sum y_i-b_1\sum x_i\right) \\ =\frac{1}{10}(1593-(11.65328\left)\right(178\left)\right) \\ =-48.1284 \end{gathered}$$

Therefore, the regression equation is $\hat{y}=-48.1284+11.6533x$

Step 4. Solving part (b):

Graph the regression equation and data points.

Using Excel, follow the steps below to obtain the data points and the regression line.

1) Enter the data into an excel sheet.

2) Select two columns, and insert, a scatter plot.

3) Right-click on the points and add a trend line.

4) Select the display trend line.

The graph representation for the equation:

Step 5. Solving part (c):

From the scatter plot obtained in part (b), it can be observed that the length of the crown-rump increases as the age of fetuses increases.

Step 6. Solving part (d):

The slope of the regression line is $11.6533$, it means that a unit change in the age of fetuses causes a change of $11.6533$millimeters increases in the length of the crown rump.

Step 7. Solving part (e):

The predictor variable is the age of fetuses and the response variable is the length of the crown rump.

Step 8. Solving part (f):

From the scatter plot obtained in part (b), it can be observed that all the points are near to the straight line, so there are no outliers and influential observations.

. Solving part (g):

The predicted crown-rump length of a 19-week-old fetus is,

$$\begin{gathered} \hat{y}=-48.1284+11.6533x \\ =-48.1284+11.6533\left(19\right) \\ =173.2843 \end{gathered}$$

I, therefore, the predicted crown-rump length of a 19 -week-old status is$173.2843$ millimeters.