Question

Following are the data on age of fetuses and length of crown-rump.use$\alpha =0.10$presuming that the assumption for regression inference are met, decide at the specified significance level whether the data provide sufficient evidence to conclude that the predictor variable is useful for providing the response variable.

Short answer

The data support the conclusion that age $\left(x\right)$ is a useful predictor of crown-rump length $\left(y\right)$.

Step-by-step solution

Given Information

Given table is

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

Explanation

STEP-1 The null and alternative hypothesis are:

$H_0:{\beta }_1=0\&H_{\alpha }:{\beta }_1\ne 0$

STEP-2 determine significance level

The hypothesis test should be run at a significance level of $10$percent, or $\alpha =0.10$

Table of computation

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

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

SST is given by

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

SSR is given by

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

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

$b_1=11.65328054$

$s_e\approx 5.52$

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 $\pm t_{\alpha /2}=\pm t_{0.05}=\pm 1.860$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 $p=1.78\times {10}^{-10}$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 $P=1.78\times {10}^{-10}<\alpha =0.10$, 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.