Statistics: Chi-square Test of Independence, Null Hypothesis

*Can not have a directional hypothesis with a chi-square test

The is that the relative proportions of one variable areindependent of the second variable; in other words, the proportions at onevariable are the same for different values of the second variable. In thevaccination example, the null hypothesis is that the proportion of children given thigh injections who have severe reactions is equal to the proportion of children given arm injections who have severe reactions.

The chi-square formula is a difficult formula to deal with. That’s mostly because you’re expected to add a large amount of numbers. The easiest way to solve the formula is by making a table.
Sample question: 256 visual artists were surveyed to find out their zodiac sign. The results were: Aries (29), Taurus (24), Gemini (22), Cancer (19), Leo (21), Virgo (18), Libra (19), Scorpio (20), Sagittarius (23), Capricorn (18), Aquarius (20), Pisces (23). Test the hypothesis that zodiac signs are evenly distributed across visual artists.

The shape of the chi-square distribution depends on the number of degrees of freedom. For an extrinsic null hypothesis (the much more common situation, where you know the proportions predicted by the null hypothesis before collecting the data), the number of degrees of freedom is simply the number of values of the variable, minus one. Thus if you are testing a null hypothesis of a 1:1 sex ratio, there are two possible values (male and female), and therefore one degree of freedom. This is because once you know how many of the total are females (a number which is "free" to vary from 0 to the sample size), the number of males is determined. If there are three values of the variable (such as red, pink, and white), there are two degrees of freedom, and so on.

Chi Square Statistics - Hobart and William Smith Colleges

In 1933, Pearson's son Egon and Jerzy Neyman developed a null hypothesis for the Chi Square test that was either accepted or rejected, thus forming the Chi Square Analysis.
The Null Hypothesis
Shown in this picture are the M&M's I received in a cup.

Null and Alternative Hypothesis | Real Statistics Using …

Mannan and Meslow (1984) studied bird foraging behavior in a forest in Oregon. In a managed forest, 54% of the canopy volume was Douglas fir, 40% was ponderosa pine, 5% was grand fir, and 1% was western larch. They made 156 observations of foraging by red-breasted nuthatches; 70 observations (45% of the total) in Douglas fir, 79 (51%) in ponderosa pine, 3 (2%) in grand fir, and 4 (3%) in western larch. The biological null hypothesis is that the birds forage randomly, without regard to what species of tree they're in; the statistical null hypothesis is that the proportions of foraging events are equal to the proportions of canopy volume. The difference in proportions is significant (chi-square=13.59, 3 d.f., P=0.0035).

Chi-Square Test of Independence - Statistics Solutions

2 value, or the collection of data divided by the expected results, we obtain is less than or equal to the Chi Square value, we must accept the null hypothesis.

Chi-Square Test of Independence

The result is chi-square=5.61, 1 d.f., P=0.018, indicating that you can reject the null hypothesis; there are significantly more left-billed crossbills than right-billed.

Chi-Square Independence Testing | Real Statistics Using …

2 value.
Based on this data of my individual test, I reject the null hypothesis because the calculated Chi Square sample size from my individual data was 13.97, which is a larger value than 11.07, the original Chi Square value for all experiments under 5 degrees of freedom.