Hi guys,
I'm a bit stuck in an exercise and I would appreciate some help.
Let \(\displaystyle X_1,X_2,...,X_n,X_{n+1}\) be a random sample of size n+1, n>1, from a distribution that is \(\displaystyle N(\mu,\sigma^2)\). Let \(\displaystyle \bar{X}=\sum_{i=1}^n X_i/n\) and \(\displaystyle S^2=\sum_{i=1}^n(X_i-\bar{X})^2 /(n-1)\). Show that in order for the statistic \(\displaystyle c(\bar{X}-X_{n+1})/S\) to have a t-distribution, the constant c must equal \(\displaystyle \sqrt{\frac{n-1}{n+1}}\).
Okay, we can view the statistic as a difference in means Confidence Interval. From the definition of the t-distribution we know that \(\displaystyle T=\frac{W}{\sqrt{V/r}}\) where \(\displaystyle W\sim N(0,1)\ and\ V\sim \chi^2(r)\).
We know that \(\displaystyle \bar{X}\sim N(\mu,\frac{\sigma^2}{n})\) and \(\displaystyle X_{n+1}\sim N(\mu,\sigma^2)\). Since those two are independent(because of the random sample) we can say that the total variance is equal to \(\displaystyle \sigma^2(\frac{n+1}{n})\). Since the means are equal our numerator then becomes \(\displaystyle \displaystyle\sqrt{n} \frac{\bar{X}-X_{n+1}}{\sigma\sqrt{n+1}}\).
My problem is now to find the chi squared distributed denominator that will give me the \(\displaystyle \sqrt{n-1}\) part of c. Since the variances are unequal I do not think we can use a pooled estimator. Any suggestions are greatly appreciated. Thanks.
EDIT: I have looked it up extensively in the meantime and it seems that everyone uses the coefficient \(\displaystyle \displaystyle\frac{\sqrt{n}}{\sqrt{n+1}}\) instead of \(\displaystyle \displaystyle \frac{\sqrt{n-1}}{\sqrt{n+1}}\). The first case is the one I am able to show and it conforms with my intuition but I do not know which one is correct. Could it be that my book is mistaken at this point? God, this is confusing indeed.
I'm a bit stuck in an exercise and I would appreciate some help.
Let \(\displaystyle X_1,X_2,...,X_n,X_{n+1}\) be a random sample of size n+1, n>1, from a distribution that is \(\displaystyle N(\mu,\sigma^2)\). Let \(\displaystyle \bar{X}=\sum_{i=1}^n X_i/n\) and \(\displaystyle S^2=\sum_{i=1}^n(X_i-\bar{X})^2 /(n-1)\). Show that in order for the statistic \(\displaystyle c(\bar{X}-X_{n+1})/S\) to have a t-distribution, the constant c must equal \(\displaystyle \sqrt{\frac{n-1}{n+1}}\).
Okay, we can view the statistic as a difference in means Confidence Interval. From the definition of the t-distribution we know that \(\displaystyle T=\frac{W}{\sqrt{V/r}}\) where \(\displaystyle W\sim N(0,1)\ and\ V\sim \chi^2(r)\).
We know that \(\displaystyle \bar{X}\sim N(\mu,\frac{\sigma^2}{n})\) and \(\displaystyle X_{n+1}\sim N(\mu,\sigma^2)\). Since those two are independent(because of the random sample) we can say that the total variance is equal to \(\displaystyle \sigma^2(\frac{n+1}{n})\). Since the means are equal our numerator then becomes \(\displaystyle \displaystyle\sqrt{n} \frac{\bar{X}-X_{n+1}}{\sigma\sqrt{n+1}}\).
My problem is now to find the chi squared distributed denominator that will give me the \(\displaystyle \sqrt{n-1}\) part of c. Since the variances are unequal I do not think we can use a pooled estimator. Any suggestions are greatly appreciated. Thanks.
EDIT: I have looked it up extensively in the meantime and it seems that everyone uses the coefficient \(\displaystyle \displaystyle\frac{\sqrt{n}}{\sqrt{n+1}}\) instead of \(\displaystyle \displaystyle \frac{\sqrt{n-1}}{\sqrt{n+1}}\). The first case is the one I am able to show and it conforms with my intuition but I do not know which one is correct. Could it be that my book is mistaken at this point? God, this is confusing indeed.
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