91Ó°ÊÓ

Given table is

and$\mathrm{Î\pm }=0.10$we have to find out whether the data provide sufficient evidence to conclude that the predictor variable is useful for providing the response variable.

STEP-1 The null and alternative hypothesis are:

$H_0:{\mathrm{β}}_1=0\&H_{\mathrm{Î\pm }}:{\mathrm{β}}_1≠0$

STEP-2 determine significance level

The hypothesis test should be run at a significance level of $10$percent, or $\mathrm{Î\pm }=0.10$

Table of computation

$$\begin{gathered} S_{xy}=∑x_i{y}_i-\left(∑x_i\right)\left(∑y_i\right)/n \\ =32476-\left(178\right)\left(1593\right)/10 \\ =4120.6 \end{gathered}$$

$$\begin{gathered} S_{xx}=∑x_i^2-{\left(∑x_i\right)}^2/n \\ =3522-(178{)}^2/10 \\ =353.6 \end{gathered}$$

SST is given by

$$\begin{gathered} S_{yy}=∑y_i^2-{\left(∑y_i\right)}^2/n \\ =302027-(1593{)}^2/10 \\ =48262.1 \end{gathered}$$

SSR is given by

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

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

$b_1=11.65328054$

$s_e≈5.52$

STEP-3 The value of test statistic can be calculated as

STEP-4

$$\begin{gathered} df=n-1 \\ =10-2 \end{gathered}$$

We discovered that critical values are $Â\pm t_{\mathrm{Î\pm }/2}=Â\pm t_{0.05}=Â\pm 1.860$using technology.

Or

STEP 4: The test statistic's value is $t=39.71$as of Step 3. The P-value is the chance of seeing a value of $tof39.71$or more in magnitude if the null hypothesis is true, because the test is two-tailed. We get $p=1.78×{10}^{-10}$through using technology.

STEP 5: Because the test statistic's value exceeds the critical value. Our null hypothesis$H_0$is rejected. i.e. $t=39.71<t_{0.05,8}=1.86$

STEP 5: Because $P=1.78×{10}^{-10}<\mathrm{Î\pm }=0.10$, Our null hypothesis $H_0$is rejected.

STEP 6: The data give adequate evidence to infer $t$the variable, age $x$is beneficial as a predictor of crown-rump $y$length at the $10\%$significant level.