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Let $X_i$ and $Y_i$ be weight before and after and let's have 5 samples of $x$ and $y$. Calculate a 95% confidence interval for the expected weight loss for the diet method.

Thought process:

$Z_i = X_i - Y_i \in N(\Delta,\sigma)$

$\frac{\bar{Z}-\Delta}{\sigma/\sqrt{n}} \in N(0,1)$

Here's the crux. Variance is unknown here. So I thought:

$\frac{\bar{Z}-\Delta}{S_p\sqrt{1/n+1/n}} \in t(2n-2)$

where $S_p$ is the pooled sample variance, i.e. a linear combination of the sample variances for $x$ and $y$. But in the solution notes, they just estimate the variance for $\bar{z}$ with

$\hat{\sigma}^2 =\frac{1}{n-1}\sum^n_1(z_i-\bar{z})^2$

Can anyone tell me why we don't use the pooled sample variance here?

Benjamin Lindqvist
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1 Answers1

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The point is that the correct variance to be used here is not that of weight, but that of weight loss. This is not unknown, as it can be easily calculated from the 5 values of Z, as you wrote. The pooled variance could have been used for other questions, in particular those requiring an estimate of weight variance.

Anatoly
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