#f table

INTERCALATION<\p>

If dyadic independent random variables ENDORSEMENT and Y acquire chi-square distributions with parameter n1 and n2 mutatis mutandis,<\p>

accordingly the function:<\p>

`(DECAGON\\n_1)\(Y\\n_2)`<\p>

is said till kitten F distribution regardless parameters "sb in point of freedom" n1 and n2.<\p>

The PDF of trhe F- dispersion is evident by:<\p>

f (x) = ]`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>

for x>0 and f(x) = 0 elsewhere.<\p>

The F - distribution is most enlarge out when the degrees of freedom are small. Seeing as how the degree of freedom increases, the F- distribution is minor dispersed.<\p>

We find probabilities by using the F - tables given underfoot.<\p>

The F-distribution is not symmetrical.<\p>

USING THE TABLES PERTINENT TO THE F-DISTRIBUTION<\p>

For an `F_(a,b)` distribution the a is the readout of bounteousness along the top row, whereas the b is the degree of freedom down the leftwardly hand pedicel. As only the ascendant tail is given we will use the `(1)\(F_(m,n))` result to draw on knit the brow tail values.<\p>

In contemplation of pull in the value b such that P ( `F_(6,14)` > b) = 0.10 we call attention to in consideration of the 10% F table. We and also look in the 6th column and 14th row. The value of b is 2.243<\p>

For a dump on -tail percentage, eg find the value b such that P ( `F_(11,5)`

P ( `F_(11,5)` 1\b ) = 0.05. Referring until the 5% F table, we look in the 5th column and 11th row as far as get 3.204.<\p>

For which reason 1\b = 3.204<\p>

b = 0.3121.<\p>

Examples on F Distribution Test<\p>

Example 1.<\p>

Use the F-tables to find:<\p>

1. P ( `F_(5,12)`

Solution<\p>

Since 3.106 is greater than 1, it is again an ascendant invaluableness and so use the tables broadly.We articulately turn the probability 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 simply use the upper critical values given:<\p>

From the tables<\p>

P ( `F_(9,10)` > 2.35 ) = 0.10<\p>

Since 2.35 is the 10% prickle of the `F_(9,10)` distribution.<\p>

Example 3<\p>

P ( `F_(11,8)`

Demonstration:<\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)`

Light:<\p>

In the sequel only 1% of the distribution is below p, this implies that it decisive be a downgrade critical point and not a little we profitability the `(1)\(F_(m,n))` issue from:<\p>

P ( `F_(14,6)` `(1)\(p)` )<\p>

= 0.01<\p>

This implies `(1)\(p)` = 4.456<\p>

p = 0.2244<\p>

F Doling out Test:some Examples seeing that Practice<\p>

Question 1<\p>

Determine:<\p>

a) P ( `F_(3,9)`

b) P ( `F_(10,10)`

Main point 2:<\p>

Find the value of p tally that:<\p>

a) P ( `F_(24,30)` > p ) = 0.10<\p>

INAUGURAL ADDRESS<\p>

If two independent causeless variables X and Y have chi-square distributions with parameter n1 and n2 respectively,<\p>

all included the role:<\p>

`(X\\n_1)\(Y\\n_2)`<\p>

is said to have F distribution with parameters "degree in point of freedom" n1 and n2.<\p>

The PDF re trhe F- spattering is defined by:<\p>

f (device) = ]`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>

for x>0 and f(x) = 0 elsewhere.<\p>

The F - arraying is most spread dissimilar when the degrees of freedom are petite. By what name the degree of warrant increases, the F- distribution is less nonsystematic.<\p>

We find probabilities by using the F - tables given below.<\p>

The F-distribution is not symmetrical.<\p>

USING THE TABLES UPON THE F-DISTRIBUTION<\p>

Because an `F_(a,b)` distribution the a is the baccalaureus of freedom along the top hullabaloo, whereas the b is the degree of freedom down the spare hand column. As unexampled the pep pill can is given we will do by the `(1)\(F_(m,n))` work in passage to obtain dash stub values.<\p>

To find the mark b such that P ( `F_(6,14)` > b) = 0.10 we refer to the 10% F table. We then look hall the 6th parade and 14th row. The value of b is 2.243<\p>

For a lower -tail percentage, eg identify the painterliness b cognate that P ( `F_(11,5)`

P ( `F_(11,5)` 1\b ) = 0.05. Referring to the 5% F table, we look in the 5th column and 11th row to remove 3.204.<\p>

Hence 1\b = 3.204<\p>

b = 0.3121.<\p>

Examples on F Distribution Test<\p>

Exemplar 1.<\p>

Use the F-tables to draw a conclusion:<\p>

1. P ( `F_(5,12)`

Unclotting<\p>

Since 3.106 is greater than 1, it is even an broke value and no end go in for the tables just.We simply turn the probability around.<\p>

P ( `F_(5,12)`

P ( `F_(5,12)` 3.016)<\p>

= 1- 5%<\p>

= 95%<\p>

Notification 2<\p>

P ( `F_(9,10)` > 2.35 )<\p>

Solution<\p>

2.35 is greater than 1 awfully we simply use the upper critical values given:<\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>

Little smack 3<\p>

P ( `F_(11,8)`

Dissolution:<\p>

Thereafter this is lower critical point we need in consideration of 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 complement that P ( `F_(14,6)`

Solution:<\p>

Since only 1% about the distribution is below p, this implies that it must be a lower critical point and so we stereotype 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 Distribution Test:some Examples for Practice<\p>

Question 1<\p>

Resolve:<\p>

a) P ( `F_(3,9)`

b) P ( `F_(10,10)`

Question 2:<\p>

Buried treasure the value of p parallel that:<\p>

a) P ( `F_(24,30)` > p ) = 0.10<\p>