correlation coefficient

can anyone of two pivotal quantity be used to calculate CI or certain conditions apply?

 

I assume you're asking about t-test and Fisher Z
So ans: apply t-test when H_0: rho=0 otherwise Fisher Z

 

ques 13 here r is greater than zero and they have used t test result to calculate CI....???

 

Which ques. 13?????

And we need H_0 : value of rho..... Find what ques need for population correlation coefficient(rho)
whatever is r (r is just a sample)

 

I think you misunderstood my question.....
Ques13(b)
What i asked is while solving the ques paper i calculated the CI for Rho using fisher z result and in solutions it has been calculated using t result...is both way correct?(both way gives approx same answer)

 

I did understand.
Actually you said, "r is greater than zero and they have used t test result" so I revert, "whatever is r, we bother about rho"

Now the main point, is any of the test is good for unknown population correlation coefficient with no clue?
I'd say Fisher Z is wider test to use.
I think both ways give you marks in exam.

PS: any experienced person give advice!!!!!!!!!!

 

Read page 11 of chapter 13 of the ActEd notes.

Two points to remember:

1. The t-test statistic only holds under \(\rho =0\). In other words, the given statistic will follow a t-distribution only when you assume \(\rho =0\). So, if you are doing a hypothesis test with Ho: \(\rho =0\), you can use it.

2. The Fisher test statistic works for any value of \(\rho\) (including \(\rho =0\)). In other words, the statistic will follow a Normal distribution whatever be the value of \(\rho\). Hence, this statistic can also be used to derive a Confidence Interval for \(\rho\).

Check Q11 of IAI May 2014 paper as well.

 

In IAI exam May 2009 , Q13 part (b)

CI is calculated by t-dustribution...... They assumes \(\rho =0\)

 

This is incorrect. The t-distribution cannot be used for deriving the confidence interval for \(\rho\).

The Normal test using the Fisher's z-transformation is needed. The confidence interval thus derived is (0.062, 0.767).

 

Thanks!