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 age and price data for Corvettes from Exercise \(14.23\)

At the \(5%\) level of significance, do the data provide sufficient evidence to conclude that age and price of Corvettes are negatively linearly corelated?

Short Answer

Expert verified

The null hypothesis is rejected and it can be conclude that age and price of Corvettes and negatively linearly correlated.

Step by step solution

01

Step 1. Given information

The level of significance is \(0.05\) and the data is,

02

Step 2. Calculation

The hypothesis are,

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

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

The table is shown below.

The value of \(r\) is,

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

\(=\frac{13168-(41)\left ( \frac{3422}{10} \right )}{\sqrt{199-\frac{(41)^{2}}{10}}\sqrt{1196690-\frac{(3422)^{2}}{10}}}\)

\(=0.96787\)

The value of test statistic is,

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

\(=\frac{-0.96787}{\sqrt{\frac{1-(-0.96787)^{2}}{10-2}}}\)

\(=-10.887\)

The degree of freedom is,

\(dof=n-2\)

\(=10-2\)

\(=8\)

The curve is shown below.

Since, the value lies in the rejection region.

Thus, the null hypothesis is rejected and it can be conclude that age and price of Corvettes and negatively linearly correlated.

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Most popular questions from this chapter

In Exercises 14.12-14.21, we repeat the data and provide the sample regression equations for Exercises 4.48 -4.57.

a. Determine the standard error of the estimate.

b. Construct a residual plot.

c. Construct a normal probability plot of the residuals.

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role="math" localid="1652333468446" x3412y450-1 role="math" localid="1652333507650" y^=-3+x

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