Question

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.

Short answer

(a) A point estimate for the mean crown-rump length is ${\hat{y}}_p=173.28$.

(b) A $90\%$confident that the mean crown-rump length of all $19$-week-old fetuses is between $169.97$and $176.59$.

(c) The predicted crown-rump length is $\hat{y}=173.28$.

(d) A $90\%$prediction interval for the crown-rump length is between $162.50$ to $184.07$

Step-by-step solution

Part (a) Step 1: Given information

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

Part (a) Step 2: Explanation

The table is calculated as:

x	y	$xy$	$x^2$	$y^2$
10	66	660	100	4356
10	66	660	100	4356
13	108	1404	169	11664
13	106	1378	169	11236
18	161	2898	324	25921
19	166	3154	361	27556
19	177	3363	361	31329
23	228	5244	529	51984
25	235	5875	625	55225
28	280	7840	784	78400
$\sum x_i=178$	$\sum y_i=1593$	$\sum x_i{y}_i=32476$	$\sum {x_i}^2=3522$	$\sum {y_i}^2=302027$

Part (a) Step 3: Explanation

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

$=32476-283554/10$

$=32476-283554/10$
$=32476-28355.4$
$=4120.6$
Then,

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

Part (a) Step 4: Explanation

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,

$SSE=SST-SSR$

$=48262.1-48018.50781$

$=243.5921946$

Part (a) Step 5: Explanation

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:
$b_1=\frac{S_w}{S_w}$
$=\frac{4120.6}{353.6}$
$=11.65328054$

Part (a) Step 6: Explanation

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 ${\hat{y}}_p=173.28$.

Part (b) Step 1: Given information

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

Part (b) Step 2: Explanation

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

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

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

$=t_{005}$

Part (b) Step 3: Explanation

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 $176.59$

Part (c) Step 1: Given information

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

Part (c) Step 2: Explanation

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 $\hat{y}=173.28$.

Part (d) Step 1: Given information

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

Part (d) Step 2: Explanation

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

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

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

$=t_{005}$

$=1.860$

Part (d) Step 2: Explanation

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 $184.07$