Sampling Distribution of Means (Continued)

$\displaystyle P\left[ \vert(\bar{X}-5)\vert \geq 0.027 \right]
$

$\displaystyle =P[(\bar{X}-5) \geq 0.027]+P[(\bar{X}-5) \leq -0.027]
$

$\displaystyle =2P(Z\geq2.7)=2*0.0035=0.007
$

Strongly refutes the conjecture!

Figure 2: Area for Example 8.7.
\includegraphics[scale=0.4]{figures/08-12}

$\displaystyle \mu_{\bar{X_A}-\bar{X_B}}=\mu_A-\mu_B=0
$

$\displaystyle \sigma_{\bar{X_A}-\bar{X_B}}^2=\frac{\sigma_A^2}{n_A}+\frac{\sigma_B^2}{n_B}
$

$\displaystyle z=\frac{(\bar{X_1}-\bar{X_2})-(\mu_1-\mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}}=\frac{1-0}{\sqrt{1/9}}=3.0
$

$\displaystyle P(Z>3.0)=1-P(Z<3.0)
$

$\displaystyle =1-0.9987=0.0013
$

Figure 3: Area for Example 8.8.
\includegraphics[scale=0.5]{figures/08-13}
Low probability. Assumption?

Solution: Since both $ n_1$ and $ n_2$ is greater than 30, the sampling distribution of $ \bar{X_1}-\bar{X_2}$ will be approximately normal.

$\displaystyle \mu_{\bar{X_1}-\bar{X_2}}=\mu_1-\mu_2=6.5-6.0=0.5
$

$\displaystyle \sigma_{\bar{X_1}-\bar{X_2}}^2=\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}
$

$\displaystyle =\frac{0.9^2}{36}+\frac{0.8^2}{49}=0.0356
$

$\displaystyle z=\frac{1-0.5}{\sqrt{0.0356}}=2.65
$

$\displaystyle P(\bar{X_1}-\bar{X_2} \geq 1.0)=P(Z> 2.65)
$

$\displaystyle =1-P(Z<2.65)=1-0.9960
$

$\displaystyle =0.004
$

Low probability value.

Figure 4: Area for Example 8.9.
\includegraphics[scale=1.2]{figures/08-14}

Cem Ozdogan 2010-05-17