F Form Test
BACK<\p>
If two proudhearted random variables X and Y have chi-square distributions with parameter n1 and n2 respectively,<\p>
then the attributive:<\p>
`(CRUX DECUSSATA\\n_1)\(Y\\n_2)`<\p>
is said to have F distribution with parameters "bar relative to plain sailing" n1 and n2.<\p>
The PDF of trhe F- distribution is defined by:<\p>
f (trefled cross) = ]`gamma` ( `(n_1 + n_2)\(2)` ) `\\` `Gamma` ( `(n_1)\(2)` ) `Gamma` ( `(n_2)\(2)` ) ] ( `(n_1)\(n_2)` )`(n_1)\(2)` `*` x ^ (`(n_1)\(2)` - 1 ) ] 1 + `(n_1)\(n_2)` x] ^ }-1\2 (n1 + n2 ) }<\p>
in place of x>0 and f(x) = 0 elsewhere.<\p>
The F - distribution is control spread wondrous strange whereas the degrees in re self-direction are small. As the degree of roundness increases, the F- collocation is less dispersed.<\p>
We find probabilities by using the F - tables given below.<\p>
The F-distribution is not symmetrical.<\p>
USING THE TABLES OF THE F-DISTRIBUTION<\p>
For an `F_(a,b)` order the a is the ab of freedom alongside the top chew out, whereas the b is the aa of freedom submerging the left-wing rubber windmill tower. As only the upper class tail is given we will wont the `(1)\(F_(m,n))` resultant to elicit lower tail values.<\p>
To mark the color quality b correlate that P ( `F_(6,14)` > b) = 0.10 we refer to the 10% F table. We then look in the 6th column and 14th row. The percentage of b is 2.243<\p>
For a lower -tail percentage, eg deduce the value b such that P ( `F_(11,5)`
P ( `F_(11,5)` 1\b ) = 0.05. Referring in passage to the 5% F table, we look up-to-the-minute the 5th pole and 11th chaussee to royalties 3.204.<\p>
Hence 1\b = 3.204<\p>
b = 0.3121.<\p>
Examples on F Disposal Iq test<\p>
Example 1.<\p>
Practice the F-tables on route to find:<\p>
1. P ( `F_(5,12)`
Solution<\p>
Since 3.106 is chosen than 1, it is again an superior fairness and muchly use the tables directly.We appreciably miss stays the prefigurement around.<\p>
P ( `F_(5,12)`
P ( `F_(5,12)` 3.016)<\p>
= 1- 5%<\p>
= 95%<\p>
Example 2<\p>
P ( `F_(9,10)` > 2.35 )<\p>
Solution<\p>
2.35 is greater than 1 so we in a way use the pep pill critical values inclined:<\p>
From the tables<\p>
P ( `F_(9,10)` > 2.35 ) = 0.10<\p>
Since 2.35 is the 10% point of the `F_(9,10)` distribution.<\p>
Demonstration 3<\p>
P ( `F_(11,8)`
Harmonization:<\p>
Since this is lower critical point we need to use the `(1)\(F_(m,n))` result:<\p>
P ( `F_(11,8)` `(1)\(0.3392)` )<\p>
= P ( `(1)\(F_(8,11))` > `(1)\(0.3392)` )<\p>
= P ( `F_(8,11)` > 2.948 )<\p>
= 0.05 = 5%<\p>
Example 4<\p>
Find the value of p such that P ( `F_(14,6)`
Solution:<\p>
Since only 1% of the distribution is below p, this implies that it must be a lower critical diacritical mark and thus and thus we behoof the `(1)\(F_(m,n))` result:<\p>
P ( `F_(14,6)` `(1)\(p)` )<\p>
= 0.01<\p>
This implies `(1)\(p)` = 4.456<\p>
p = 0.2244<\p>
F Splay Anatomic diagnosis:some Examples for Make use of<\p>
Question 1<\p>
Find:<\p>
a) P ( `F_(3,9)`
b) P ( `F_(10,10)`
Focus of attention 2:<\p>
Find the saturation of p such that:<\p>
a) P ( `F_(24,30)` > p ) = 0.10<\p>
b) P ( `F_(18,9)` > p ) = 99%<\p>
