Answer to Question #274369 in Statistics and Probability for Thebe

Question

a.     Suppose the correlation coefficient is 0.8 and number of observation is 62. What is the calculated test statistics? Is this significant correlation using a 1% level of significant?

Answer 1

Null hypothesis: p = 0

Alternative hypothesis: $p >0$ , where p - correlation coefficient

test statistic for correlation test is calculated the following way

$t={\frac {r\sqrt{n-2}} {\sqrt{1-r^2}}}$ , where r - data correlation coefficient, n - sample size

In the given case we have

$t={\frac {0.8*\sqrt{62-2}} {\sqrt{1-0.8^2}}}={\frac {6.19} {0.6}}=10.32$

If t > Cr then we should reject null hypothesis, where Cr is such value, that

$P(T(n-2)>Cr)=a$ , where T(n-2) - Student's criteria with n-2 degrees of freedom, a - level of significance

In the given case we have

$P(T(60)>Cr)=0.01\implies Cr=2.66$

we obtained that t > Cr, then we should reject the null hypothesis and admit that, based on the data, there is a significant correlation at 0.01 level of significance