- #1
kcirick
- 54
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Question:
Consider n+1 mutually independent random variables x+i from a normal distribution N(\mu ,\sigma ^{2}). Define:
\bar{x} = \frac{1}{n} \sum_{i=1}^{n}{x_{i}} and s^{2}=\frac{1}{n}\sum_{i=1}^{n}{\left(x_{i} - \bar{x}\right)^{2}}
Find the constant c so that the statistic
t= c\frac{\bar{x} - x_{n+1}}{s}
follows a t-student law. Find the degrees of freedom. Justify.
What I have so far:
Not much, but for a statistic to follow Student T distribution with n-1:
t=\frac{\bar{x}-\mu}{s / \sqrt{n}}
Because we have n+1 random variables, and s has n degrees of freedome, the resulting student t distribution will have n degree of freedoms (if s is sample standard deviation, it should have n-1 degrees of freedom). I also just expanded the above equation:
t=c \frac {\frac{1}{n}\sum_{i=1}^{n}{x_{i}}-x_{n+1}}{\left(\frac{1}{n}\sum_{i=1}^{n}\left(x_{i} - \bar{x}\right)^{2}\right)^{\frac{1}{2}}}
But what next? Can someone help? thanks!
Consider n+1 mutually independent random variables x+i from a normal distribution N(\mu ,\sigma ^{2}). Define:
\bar{x} = \frac{1}{n} \sum_{i=1}^{n}{x_{i}} and s^{2}=\frac{1}{n}\sum_{i=1}^{n}{\left(x_{i} - \bar{x}\right)^{2}}
Find the constant c so that the statistic
t= c\frac{\bar{x} - x_{n+1}}{s}
follows a t-student law. Find the degrees of freedom. Justify.
What I have so far:
Not much, but for a statistic to follow Student T distribution with n-1:
t=\frac{\bar{x}-\mu}{s / \sqrt{n}}
Because we have n+1 random variables, and s has n degrees of freedome, the resulting student t distribution will have n degree of freedoms (if s is sample standard deviation, it should have n-1 degrees of freedom). I also just expanded the above equation:
t=c \frac {\frac{1}{n}\sum_{i=1}^{n}{x_{i}}-x_{n+1}}{\left(\frac{1}{n}\sum_{i=1}^{n}\left(x_{i} - \bar{x}\right)^{2}\right)^{\frac{1}{2}}}
But what next? Can someone help? thanks!
