Question

In Exercises \(14.128-14.133\). we repeat the information from Exercises \(14.22-14.27\) Presuming that the assumptions for regression inferences are met, perform the required correlation \(t-\)tests, using either the critical- value approach or the \(P-\)value approach.

Following are the data on age of fetuses and length of crown-rump from Exercise \(14.26\)

At the \(10%\) significance level, do the data provide sufficient evidence to conclude that age and crown-rump length are linearly correlated?

Short answer

The null hypothesis is not rejected and it can be conclude that age and crown-rump length are linearly correlated.

Step-by-step solution

Step 1. Given information

The level of significance is \(0.1\) and the data is,

Step 2. Calculation

The hypothesis are,

\(H_{0}:\rho=0\)

\(H_{a}:\rho \neq 0\)

The table is shown below.

The value of \(r\) is,

\(r=\frac{\sum x_{i}y_{i}-\sum x_{i}\sum \frac{y_{i}}{n}}{\sqrt{\sum x_{i}^{2}-(\sum x_{i}^{2})}\sqrt{\sum y_{i}^{2}-(\sum y_{i})^{2}}}\)

\(=\frac{32476-(178)\left ( \frac{1593}{10} \right )}{\sqrt{3522-\frac{(178)^{2}}{10}}\sqrt{302027-\frac{(1593)^{2}}{10}}}\)

\(=0.997473\)

The value of test statistic is,

\(t=\frac{r}{\sqrt{\frac{1-r^{2}}{n-2}}}\)

\(=\frac{0.997473}{\sqrt{\frac{1-(0.997473)^{2}}{10-2}}}\)

\(=39.71\)

The degree of freedom is,

\(dof=n-2\)

\(=10-2\)

\(=8\)

The curve is shown below.

Since, the value do not lie in the rejection region.

Thus, the null hypothesis is not rejected and it can be conclude that age and crown-rump length are linearly correlated.