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When Population variance is unknown and n<30 we use t-test.
x=sample mean, u=population mean, s=sample standard deviation,
In my book and on many online websites i found t-test formula to be t=(x-u)/(s/sqrt(n))
But also on some websites it is given that t=(x-u)/(s/sqrt(n-1))
So, which formula should i use for my exams. Any help about this!
It depends on your cacluation of $s$, your estimate of the standard deviation
If you calculate $$s=\sqrt{\frac1{n-1}\sum\limits_1^{n} \left(x_i-\bar{x}\right)^2}$$ then you should be using $$t=\dfrac{\bar{x}-\mu}{s / \sqrt{n} }$$
while if you calculate $$s=\sqrt{\frac1{n}\sum\limits_1^{n} \left(x_i-\bar{x}\right)^2}$$ then you should be using $$t=\dfrac{\bar{x}-\mu}{s / \sqrt{n-1} }$$
as they both end up as $$t=\dfrac{\frac1{n}\sum\limits_1^{n} x_i-\mu}{\sqrt{\frac{\sum\limits_1^{n} \left(x_i-\bar{x}\right)^2}{n(n-1)} }}$$
