SOLUTION: n=10 x=755 y=819 xy=62085 x^2=68025 y^2=68109 If the level of significance is 0.01 and r=0.24 how to compute test statistic? (step 3 in correlation process) Test Statist

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Question 1169528: n=10
x=755
y=819
xy=62085
x^2=68025
y^2=68109
If the level of significance is 0.01 and r=0.24 how to compute test statistic? (step 3 in correlation process)
Test Statistic:n=? __?__ -test

Answer by CPhill(1987) (Show Source):
You can put this solution on YOUR website!
Absolutely! Let's break down how to compute the test statistic for a correlation coefficient.
**Understanding the Problem**
We are given:
* $n = 10$ (sample size)
* $r = 0.24$ (correlation coefficient)
* Significance level ($\alpha$) = 0.01
We need to calculate the test statistic for the correlation coefficient.
**Formula for the Test Statistic**
The test statistic for a correlation coefficient is calculated using the following formula:
$$t = \frac{r\sqrt{n-2}}{\sqrt{1-r^2}}$$
**Step-by-Step Calculation**
1. **Calculate $r^2$:**
$$r^2 = (0.24)^2 = 0.0576$$
2. **Calculate $1 - r^2$:**
$$1 - r^2 = 1 - 0.0576 = 0.9424$$
3. **Calculate $\sqrt{1 - r^2}$:**
$$\sqrt{1 - r^2} = \sqrt{0.9424} \approx 0.97077$$
4. **Calculate $\sqrt{n - 2}$:**
$$\sqrt{n - 2} = \sqrt{10 - 2} = \sqrt{8} \approx 2.8284$$
5. **Calculate $r\sqrt{n - 2}$:**
$$r\sqrt{n - 2} = 0.24 \times 2.8284 \approx 0.6788$$
6. **Calculate the test statistic $t$:**
$$t = \frac{0.6788}{0.97077} \approx 0.6992$$
**Result**
Therefore, the test statistic is approximately 0.6992.
**Final Answer**
Test Statistic: n=10 t-test = 0.6992