Question

Random samples of sizes n₁ = 32 and n₂ = 40 are to be drawn from two independent populations.

μ₁ = 12.3      μ₂ = 9.8

σ₁ = 2.9         σ₂ = 2.4

1. P(Xbar1 - Xbar 2 < 2)
2. P(S₁²/ S₂² > 2)

Answer 1

a.

$P(\bar{X_1}-\bar{X_2}<2)=P(\frac{(\bar{X_1}-\bar{X_2})-(\mu_1-\mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}}<\frac{2-(12.3-9.8)}{\sqrt{\frac{2.9^2}{32}+\frac{2.4^2}{40}}})$

= P(Z < 2 - 2.5 0.6378 = P(Z < -0.78) = D(-0.78) = 1 – 0(0.78)
, $Z\to N(\mu=0,\sigma=1)$

$=1-0.7823=0.2177$ ; From standard normal distribution table

b.

$P(\frac{s_1^2}{s_2^2}>2)=P(\frac{\frac{(n_1-1)s_1^2}{\sigma_1^2}}{\frac{(n_2-1)s_2^2}{\sigma_2^2}}>2*\frac{\frac{n_1-1}{\sigma_1^2}}{\frac{(n_2-1)}{\sigma_2^2}})$

$=P(\frac{\chi^2_{n_1-1}}{\chi^2_{n_2-1}}>2*\frac{\frac{32-1}{2.9^2}}{\frac{(40-1)}{2.4^2}})$

$=P(\frac{\frac{\chi^2_{n_1-1}}{n_1-1}}{\frac{\chi^2_{n_2-1}}{n_2-1}}>2*\frac{\frac{32-1}{2.9^2(32-1)}}{\frac{(40-1)}{2.4^2(40-1)}})$

$=P(F_{n_1-1,n_2-1}>2*\frac{2.4^2}{2.9^2})$

$=P(F_{32-1,40-1}>1.3698)$

$=P(F_{31,39}>1.3698)$

$=0.1750$ ; The Excel function is , =FDIST(1.3698,31,39)