This question is related to another question I asked, but this is more problem-specific.
I have a sequence of A/B currency exchanges for some days. With that data I can calculate the daily returns, and that's what I did. I need to calculate the confidence interval for the expected daily returns of the A/B currency exchange by using the 1.96 rule.
To calculate the confidence interval I would need the standard deviation of the population, but I don't have it. So, I think I need to use the sample standard deviation.
I calculated the sample variance as follows $$SV = \frac{1}{n - 1} \cdot \sum_i ({DR}_{i} - SM)^ 2$$
The sample mean came out to be $0.0083$ and the sample variance $0.6147$. Then I calculated the interval using the formulas
$$\left[L= SM - 1.96 * \sqrt{\frac{SV}{n}}, R= SM + 1.96 * \sqrt{\frac{SV}{n}} \right]$$
where $SM$ and $SV$ are respectively the sample mean and variance I found above. I found that $L = -0.0303$ and $R = 0.0469$.
First of all, is my procedure to find the interval correct? Apart from that, I am still not sure why this procedure would be correct anyway.
What's the exact relation between the number $1.96$ and the $95\%$? And what is this $95\%$ referred exactly to?
If you need the values of the daily returns, I can somehow provide the file.
Thanks for any help!
Yes, your method of calculating the confidence interval is correct.
What you calculated as: $$ SE = \sqrt{\frac{SV}{n}}$$ is known as the standard-error of your estimate of the mean. It's basically the standard deviation of your estimator.
Let's assume the true mean were $\mu$. What's referred to the t-stat is: $$ t = \frac{SM - \mu}{SE} $$ follows a t-distribution with $n-1$ degrees of freedom. By the CLT, the distribution of estimator SM converges to the normal distribution but there's a bit of extra complexity for the t-stat here in that we're dividing by the square root of our estimate of the variance, which is a random variable too. The t-distribution is basically the standard normal distribution with somewhat fatter tails.
Where does the 1.96 etc... come from? A $t$ distributed random variable with greater than approximately 1000 degrees of freedom will lie between -1.96 and 1.96 about 95 percent of the time. This can be seen using the inverse of the t-cdf. $F_t^{-1}(.975, 1000) \approx 1.9623 $ and $F_t^{-1}(.025, 1000) \approx -1.9623 $.