Transcript Chi Squared (X²)

CHI SQUARED (X²)
By: Christina Patellos & Bismah Warraich
WHAT IS CHI SQUARE?
Chi Square is a statistical test that allows us to
determine differences between our observed and
expected measurements. The initial purpose is to
establish whether or not the two variables in the
experiment have a significant relationship or
differences in measurements are due to chance or
sample errors.
 This statistical test is usually used when we are
dealing with distinct data.

WHAT IS CHI SQUARE USED FOR?
 Chi
square is used to test the validity of a
null hypothesis.
 A null hypothesis is the hypothesis that
states there is no significant difference
between observed and expected values.
More so, any differences that occur would
be because of sampling or experimental
error.
HOW DO WE CALCULATE CHI SQUARE?
 There
is a formula!
 X² = Σ (o-e)²
e
 o stands for observed values
 e stands for expected values
 Σ stands for the ‘sum of’
DEGREE OF FREEDOM
•Degree of freedom (df) is used to accept or deny the null
hypothesis.
• Df is equivalent to the number of categories minus one.
•Here’s an example of a degree of freedom chart:
LET’S TRY TO APPLY IT TO AN EXAMPLE…
In fern plants purple leaves (R) are dominant to
yellow leaves (r). Plants from the offspring of a
cross between a purple plant and yellow plant
were used in a lab. A student counts 329 purple
and 299 yellow leaves on one fern.
 Calculate the chi-squared value for the null
hypothesis that the purple parent was
heterozygous for purple leaves. Give your answer
to the nearest tenth.

DON’T FORGET THE PUNNET SQUARE!

A cross between a heterozygous purple leaved
fern plant (Rr) and a yellow leaved fern plant (rr)
would yield offspring that display a 1:1 ratio
between purple and yellow leaves.
r
r
R
r
Rr
rr
Rr
rr
LET’S MAKE A TABLE!

Of the 628 leaves,it would be expected that 314
would be purple and 314 would be yellow
(Expected values).
Phenotype
Observed
Value
Expected
Value
Obs - Exp
(Obs-Exp)²
(Obs-Exp)² /
Exp
Purple
leaves
329
314
15
225
0.72
Yellow
leaves
299
314
-15
225
0.72
Σ
x² = 1.44
NOW TRY A PROBLEM OUT ON YOUR OWN!
RULES:
 Get into teams of four (turn to the table behind
you)
 Work out the problem AS A TEAM
 Whichever team correctly finishes the problem
first gets a special prize 

PRACTICE PROBLEM






Watermelons have genes for bitter taste (Su) and explosive
rind (e). Explosive rinds are recessive. Non-bitter
watermelons are also recessive and have the genotype
(susu). A farmer wants to determine if the genes assort
independently. She performs a testcross between a
bitter/nonexplosive hybrid and a plant homozygous
recessive for both traits. The following offspring are
produced:
bitter/non-explosive – 88
bitter/explosive – 68
non-bitter/non-explosive – 62
non-bitter/explosive – 81
Hint: A cross between Susu/Ee x susu/ee watermelon
plants is expected to produce offspring with a 1:1:1:1
phenotypic ratio.