kcirick
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Question:
Consider n+1 mutually independent random variables [itex]x+i[/itex] from a normal distribution [itex]N(\mu ,\sigma ^{2})[/itex]. Define:
[tex]\bar{x} = \frac{1}{n} \sum_{i=1}^{n}{x_{i}}[/tex] and [tex]s^{2}=\frac{1}{n}\sum_{i=1}^{n}{\left(x_{i} - \bar{x}\right)^{2}}[/tex]
Find the constant c so that the statistic
[tex]t= c\frac{\bar{x} - x_{n+1}}{s}[/tex]
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:
[tex]t=\frac{\bar{x}-\mu}{s / \sqrt{n}}[/tex]
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:
[tex]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}}}[/tex]
But what next? Can someone help? thanks!
Consider n+1 mutually independent random variables [itex]x+i[/itex] from a normal distribution [itex]N(\mu ,\sigma ^{2})[/itex]. Define:
[tex]\bar{x} = \frac{1}{n} \sum_{i=1}^{n}{x_{i}}[/tex] and [tex]s^{2}=\frac{1}{n}\sum_{i=1}^{n}{\left(x_{i} - \bar{x}\right)^{2}}[/tex]
Find the constant c so that the statistic
[tex]t= c\frac{\bar{x} - x_{n+1}}{s}[/tex]
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:
[tex]t=\frac{\bar{x}-\mu}{s / \sqrt{n}}[/tex]
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:
[tex]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}}}[/tex]
But what next? Can someone help? thanks!