91Ó°ÊÓ

STEP 1: $\mathrm{Î\pm }=0.10$for a $90\%$confidence interval. Given that $\mathrm{n}=10$

df $=n-2$

$=10-2$

$=8$

From technology, $t_{\mathrm{Î\pm }/2}=t_{0.10/2}=t_{0.05}=1.860$

STEP 2:

For ${\mathrm{β}}_1$, the formula for finding the end points of the confidence interval is

$b_1Â\pm t_{\mathrm{Î\pm }2}\frac{s_e}{\sqrt{S_{xx}}}$

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

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

$=32476-283554/10$

$=32476-28355.4$

$=4120.6$

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

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

$=3522-31684/10$

$=3522-3168.4$

$=353.6$

The entire amount of square SST is calculated as follows:

$S_{yy}=∑y_i^2-{\left(∑y_i\right)}^2/n$

$=302027-(1593{)}^2/10$

$=302027-2537649/10$

$=302027-253764.9$

$=48262.1$

The regression sum of squares $mathrmSSR$ is calculated as follows:

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

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

$=\frac{16979344.36}{353.6}$

$=48018.50781$

$\mathrm{SSE}=\mathrm{SST}-\mathrm{SSR}$

$=48262.1-48018.50781$

$=243.5921946$

The slope of the regression line is calculated using the formula,

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

$=\frac{4120.6}{353.6}$

$=11.65328054$

The standard error of the estimate is calculated using the formula:

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

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

$=5.518063458$

$≈5.52$

We have $b\_\left\{1\right\}=11.65328054,$

$s\_\left\{e\right\}=5.52$

$S\_\{xx\}=353.6$

So, $11.65328054Â\pm 1.860×\frac{5.52}{\sqrt{353.6}}$

Or $11.65328054Â\pm 0.545812537$, or $11.107$to $12.199$

STEP 3:

Interpretation: We can be $90\%$confident that the increase in the mean length of Crown-Rump per 1-week rise in foetus age is between $11.10$and$12.2$ millimetres for the Vomeronasal Organ.