Confidence Intervals on a Difference in Mean

Three Cases

1. Where ${\sigma }_{x_1}$ and ${\sigma }_{x_2}$ are known

$$[L,U{)}_{1-\alpha }=({\bar{x}}_1-{\bar{x}}_2)\pm (Z_{\frac{\alpha }{2}})({\sigma }_{{\bar{x}}_1-{\bar{x}}_2})$$

where $({\sigma }_{{\bar{x}}_1-{\bar{x}}_2}^2)$ = Var(${\bar{x}}_1-{\bar{x}}_2$) = $\text{Var}({\bar{x}}_1)+\text{Var}({\bar{x}}_2)$.

2. ${\sigma }_{x_1}$ and ${\sigma }_{x_2}$ are unknown (*we will use $t_{\frac{\alpha }{2}})$
   • If we have n₁ and n₂, what is V? (see notes for expression for V)
   • Simplifying assumption that ${\sigma }_{x_1}$ = ${\sigma }_{x_2}$ = ${\sigma }_x$ (unknown), we get an improved estimate of ${\sigma }_x$ by pooling the data

$$S_p^2=\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{(n_1+n_2-2)}$$

- This is the weighted average of S<sub>1</sub><sup>2</sup> and S<sub>2</sub><sup>2</sup>

Where 

$$S_1^2=\frac{1}{n_1-1}\sum _{i=1}^{n_1}(X_{1_i}-{\bar{x}}_1{)}^2$$

and 

$$S_2^2=\frac{1}{n_2-1}\sum _{i=1}^{n_2}(X_{2_i}-{\bar{x}}_2{)}^2$$

If $u_1$ = $n_2$, then Sp<sup>2</sup> = $S_p^2 = \frac{(n_1 - 1)S_1^2 + (n_2 - 1)S_2^2}{(2(n-1)}$ = $\frac{1}{2}(S_1^2 + S_2^2)$

Now $\nu$ = n<sub>1</sub> + n<sub>2</sub> -2 

3. Paired Data

Solution: Take differences: Let D_i = $X_{1_i}-X_{2_i}$, & d_i = $x_{1_i}-x_{2_i}$

Now compute

$$\bar{d}=\frac{1}{n}\sum _{i=1}^n{d}_i$$

Where $\bar{d}$ is an estimate of ${\mu }_D={\mu }_{1_i}-{\mu }_{2_i}$. We will also need $S_d^2=\frac{1}{n-1}{\sum }_{i=1}^n(d_i-\bar{d}{)}^2$

Now,

$$[L,U{)}_{1-\alpha }=(\bar{d}\pm (t_{\frac{\alpha }{2},n-1})\frac{S_d}{\sqrt{n}}$$

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In order for two pieces of data to be paired, they must have some sort of correlation.