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