Sorry, I'm just on Khan academy and can't seem to grasp the essence of statistics. Hope to find some help out here.
Both are samples, but why when looking for confidence interval of a:
- sample proportion, we take $\hat{p} \pm Z^*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$
- sample mean, we take take $\bar{x} \pm t^*{\frac{S}{\sqrt{n}}}$
What I understand: That both are trying to determine the confidence level that Population mean falls between an interval.
I can see from google that:
Z-scores are based on your knowledge about the population's standard deviation and mean. T-scores are used when the conversion is made without knowledge of the population standard deviation and mean.
But I don't think I properly understand it: what throws me off the statement is that the confidence interval formula $\bar{x} \pm t^*{\frac{S}{\sqrt{n}}}$ still uses S(sampling standard deviation).
Question:
- Why can't we just take $\bar{x} \pm Z^*\sqrt{\frac{\bar{x}(1-\bar{x})}{n}}$ as we did $\hat{p} \pm Z^*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ ?
- Both are samples, why does one estimate with $\hat{p}(1-\hat{p})$ while the other uses S (sample standard deviation)?
- What is the difference between sample proportion and sample mean? Is the former Bernuli distribution and another not?
- Is the sample proportion confidence interval of a single sample VS sample mean which is the average of multiple samples?
- Both are samples, why do we n-1 during t test, but not z test? I get that samples need to -1 to compensate for accuracy, but why didn't we do it for z tests?
Don't feel pressure to answer all the questions, just what you can, perhaps the collective can help me out of this rut. Appreciate it your time reading this!