14.96 Crown-Rump Length. Following are the data on age of fetuses and length of crown-rump from Exercise 14.26.

x	10	10	13	13	18	19	19	23	25	28
y	66	66	108	106	161	166	177	288	235	280

a. Determine a point estimate for the mean crown-rump length of all $19$-week-old fetuses.
b. Find a $90\%$ confidence interval for the mean crown-rump length of all 19-week-old fetuses.
c. Find the predicted crown-rump length of a $19$-week-old fetus.

d. Determine a $90\%$prediction interval for the crown-rump length of a $19$ -week-old fetus.

To determine a point estimate for the mean crown-rump length of all $19$-week-old fetuses.

The table is calculated as:

Let, $S_{xy}=\sum x_i{y}_i-\left(\sum x_i\right)\left(\sum y_i\right)/n$
$=32476-\left(178\right)\left(1593\right)/10$

$=32476-283554/10$

$S_n=\sum x_i^2-{\left(\sum x_i\right)}^2/n$
$=3522-(178{)}^2/10$
$=3522-31684/10$
$=3522-3168.4$
$=353.6$

The total $SST$ sum of squares is calculated as follows:

$S_y=\sum y^2-{\left(\sum y_i\right)}^2/n$
$=302027-(1593{)}^2/10$
$=302027-2537649/10$
$=302027-253764.9$
$=48262.1$
The $SSR$regression sum of squares is calculated as follows:

$SSR=\frac{S_{xy}^2}{S_{xx}}$
$=\frac{(4120.6{)}^2}{353.6}$
$=\frac{16979344.36}{353.6}$
$=48018.50781$

Since,

$SSE=SST-SSR$

$=48262.1-48018.50781$

$=243.5921946$

Determine the standard error of the estimate as follows:
$s_e=\sqrt{\frac{SSE}{n-2}}$
$=\sqrt{\frac{243.5921946}{10-2}}$
$=5.518063458$
Determine the slope of the regression line as follows:
$b_1=\frac{S_w}{S_w}$
$=\frac{4120.6}{353.6}$
$=11.65328054$

Determine the value of $y$-intercept as follows:
$b_0=\frac{1}{n}\left(\sum y_i-b_i\sum x_i\right)$
$=\frac{1}{10}(1593-11.6532(178\left)\right)$
$=\frac{1}{10}(-481.2696)$
$\approx -48.127$

As a result, the regression equation is ${\hat{y}}_p=-48.127+11.6532x_p$

Substituting the value of $x_p=19$into the regression equation yields the method for determining the value of the point estimate.

${\hat{y}}_p=-48.127+11.6532x_p$

$=-48.127+11.6532\left(19\right)$

$=173.2838$

As a result, the point estimate is ${\hat{y}}_p=173.28$.

To find a $90\%$ confidence interval for the mean crown-rump length of all $19-$week-old fetuses.

Let, $\alpha =0.10$for a $90\%$confidence interval.

Since, $n=10$,
$df=n-2$
$=10-2$
$=8$
According to calculation:

$t_{a2}=t_{0.10/2}$

$=t_{005}$

Computing the confidence interval end points for the conditional mean of the response variable is as follows:
${\hat{y}}_p\pm t_{\alpha 2}{s}_e\sqrt{\frac{1}{n}+\frac{{\left(x_p-\sum x_i/n\right)}^2}{S_x}}$
Note that: $x_p=19$
${\hat{y}}_p=173.28$
$s_e=5.52$
$S_{xx}=353.6$
Hence,

$173.28\pm 1.860(5.52)\sqrt{\frac{1}{10}+\frac{(19-178/10{)}^2}{353.6}}$
$173.28\pm 10.2672\sqrt{0.104072398}$
Also,

$173.28\pm 3.312224789$,
Or $169.97$to $176.59$
Asa result,$90\%$ confident that the mean crown-rump length of all $19$-week-old fetuses is somewhere between $169.97$and $176.59$

To find the predicted crown-rump length of a $19$-week-old fetus.

Since, the regression equation is ${\hat{y}}_p=-48.127+11.6532x_p$
By substituting the value of $x_p=19$ in the regression equation, the projected value is obtained.
${\hat{y}}_p=-48.127+11.6532x_p$
$=-48.127+11.6532\left(19\right)$
$=173.2838$
Asa result, the point estimate is $\hat{y}=173.28$.

To determine a $90\%$ prediction interval for the crown-rump length of a $19$ -week-old fetus.

Let, $\alpha =0.10$for a $90\%$confidence interval.

Since, $n=10$
$df=n-2$
$=10-2$
$=8$
According to calculation:

$t_{a2}=t_{0.10/2}$

$=t_{005}$

$=1.860$

Determining the end points of the prediction interval for the response variable's value is as follows:
${\hat{y}}_p\pm t_{\alpha 2}{s}_e\sqrt{1+\frac{1}{n}+\frac{{\left(x_p-\sum x_i/n\right)}^2}{S_{xx}}}$
Note that: $x_p=19$
${\hat{y}}_p=173.28$
$s_e=5.52$
$S_{xx}=353.6$
Hence,

$173.28\pm 1.860(5.52)\sqrt{1+\frac{1}{10}+\frac{(19-178/10{)}^2}{353.6}}$
$173.28\pm 10.2672\sqrt{1+0.104072398}$
Also,

$173.28\pm 10.78824494$Or $162.50$to $184.07$
As a result, $90\%$certain that the crown-rump length of a $19$-week-old fetuses will be between $162.50$and $184.07$