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Magazine Chapter 14: Q. 14.119 (page 585)
In Exercises \(14.118-14.127\). we repeat the data from Exercises \(14.12-14.21\) and specify an alternative hypothesis for a correlation \(t-\)test. For each exercise, decide at the \(10%\) significance level. whether the data provide sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis.
\(H_{a}:p<0\)
Short Answer
Since, the value do not lie in the rejection region.
Thus, the null hypothesis is not rejected.
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,
\(H_{0}:\rho=0\)
\(H_{a}:\rho<0\)
The table is shown below.
The value of \(r\) is,
\(r=\frac{\sum x_{i}y_{i}-\sum x_{i}\sum \frac{y_{i}}{n}}{\sqrt{\sum x_{i}^{2}-(\sum x_{i}^{2})}\sqrt{\sum y_{i}^{2}-(\sum y_{i})^{2}}}\)
\(=\frac{-22-(6)\left ( \frac{-9}{3} \right )}{\sqrt{14-\frac{(6)^{2}}{3}}\sqrt{41-\frac{(-9)^{2}}{3}}}\)
\(=-0.7559\)
The value of test statistic is,
\(t=\frac{r}{\sqrt{\frac{1-r^{2}}{n-2}}}\)
\(=\frac{-0.7559}{\sqrt{\frac{1-(-0.7559)^{2}}{3-2}}}\)
\(=-1.15\)
The degree of freedom is,
\(dof=n-2\)
\(=3-2\)
\(=1\)
The curve is shown below.
Since, the value do not lie in the rejection region.
Thus, the null hypothesis is not rejected.
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