[amp_mcq option1=”$$\frac{{\left( {1 – {r^2}} \right)}}{{\sqrt n }}$$” option2=”$$\sqrt {\frac{{\left( {1 – {r^2}} \right)}}{{\left( {n – 2} \right)}}} $$” option3=”$$0.6745\left( {\frac{{1 – {r^2}}}{{\sqrt n }}} \right)$$” option4=”$$\sqrt {\frac{{\left( {n – 2} \right)}}{{\left( {1 – {r^2}} \right)}}} $$” correct=”option2″]
The correct answer is: B. $\sqrt {\frac{{\left( {1 – {r^2}} \right)}}{{\left( {n – 2} \right)}}}$
The standard error of the correlation coefficient is a measure of how much the correlation coefficient is likely to vary from sample to sample. It is calculated as follows:
$$\text{SE}(r) = \sqrt {\frac{{\left( {1 – {r^2}} \right)}}{{\left( {n – 2} \right)}}}$$
where $n$ is the number of pairs of observations.
In this case, there are 25 paired observations, so $n = 25$. Substituting this into the formula, we get:
$$\text{SE}(r) = \sqrt {\frac{{\left( {1 – {r^2}} \right)}}{{\left( {25 – 2} \right)}}} = \sqrt {\frac{{\left( {1 – {r^2}} \right)}}{{23}}}$$
The other options are incorrect because they do not include the factor of $\frac{1}{n-2}$ in the denominator.