91Ó°ÊÓ

Confidence Interval obtained for the each of the two differences ${\mathrm{μ}}_1-{\mathrm{μ}}_2$and ${\mathrm{μ}}_1-{\mathrm{μ}}_3=95\%$.

The probability that a population parameter will fall between a set of values for a particular proportion of the time is referred to as a confidence interval.

No, it is not possible to be $95\%$sure in both outcomes at the same time, meaning that both discrepancies are contained inside their respective confidence ranges.

Let role="math" localid="1652193214871" $E_1$be the interval event with probability based on the difference ${\mathrm{μ}}_1-{\mathrm{μ}}_2$.

$P\left(E_1\right)=0.95$and the role="math" localid="1652193228446" $E_2$be the event of the interval based on the difference ${\mathrm{μ}}_1-{\mathrm{μ}}_3$with probability

$P\left(E_2\right)=0.05$

The simultaneous occurrence of role="math" localid="1652193390006" $E_1$and role="math" localid="1652193298728" $E_2$events is symbolized by $P\left(E_1âˆ\copyright E_2\right)$, however the events $E_1$and role="math" localid="1652193407288" $E_2$are not independent in reality.

Using the multiplication formula, $P\left(E_1âˆ\copyright E_2\right)=P\left(E_1\right)P\left(E_2âˆ\pounds E_1\right)$can be calculated.

The information concerning $P\left(E_2âˆ\pounds E_1\right)$is not mentioned in the circumstance. As a result, the events $E_1$and $E_2$cannot be calculated at the same time. Furthermore, $P\left(E_1\right)$has a value of $0.95$, and $P\left(E_2âˆ\pounds E_1\right)$must be less than$0.95$.