To find and interpret a confidence interval, at the specified confidence level $90\%$for the slope of the population regression line that relates the response variables to the predictor variable.

The given data is

Regression $t-$interval

For a $90\%$confidence interval,

$\alpha =0.10$

$n=10$

$df=n-2$

$$\begin{gathered} =10-2 \\ =8 \end{gathered}$$

$t_{\alpha /2}=t_{0.10/2}$

$$\begin{gathered} =t_{0.05} \\ =1.860 \end{gathered}$$

The formula for calculating the endpoints of the confidence interval for ${\beta }_1$are

$b_1\pm t_{\alpha /2}\frac{s_e}{\sqrt{S_{xx}}}$

Computation table

$S_{xy}=\sum x_i{y}_i-\left(\sum x_i\right)\left(\sum y_i\right)/n$

$=32476-\left(178\right)\left(1593\right)/10$

$=32476-283554/10$

$$\begin{gathered} =32476-28355.4 \\ =4120.6 \end{gathered}$$

$S_{xx}=\sum x_i^2-{\left(\sum x_i\right)}^2/n$

$=3522-(178{)}^2/10$

$=3522-31684/10$

$$\begin{gathered} =3522-3168.4 \\ =353.6 \end{gathered}$$

The total sum of square SST is

$SSR=\frac{S_{xy}^2}{S_{xx}}$

$=\frac{(4120.6{)}^2}{353.6}$

$=\frac{16979344.36}{353.6}$

$=48018.50781$

$SSE=SST-SSR$

$$\begin{gathered} =48262.1-48018.50781 \\ =243.5921946 \end{gathered}$$

The slope of the regression line is

$b_1=\frac{S_{xy}}{S_{xx}}$

$=\frac{4120.6}{353.6}$

$=11.65328054$

The standard error of the estimate is

$s_e=\sqrt{\frac{SSE}{n-2}}$

$=\sqrt{\frac{243.5921946}{10-2}}$

$$\begin{gathered} =5.518063458 \\ \approx 5.52 \end{gathered}$$

Substituting the values

$11.65328054\pm 1.860\times \frac{5.52}{\sqrt{353.6}}$

$11.65328054\pm 0.545812537,Or11.107to12.199$

We can be $90\%$confident that, for the Volatile emission, the increase in the mean length of Crown-Rump per 1 -week increase in age of foetuses is somewhere between $11.10$and $12.2$millimetres.