Question

According to the Mendelian theory of genetics, a certain garden pea plant should produce either white, pink, or red flowers with respective probabilities 1/4, 1/2, and 1/4. To test this theory, a sample of 564 peas was conducted with the result that 141 produced white, 291 produced pink, and 132 produced red flowers. Using the chi-square approximation, what conclusion would be drawn at the 5 percent level of significance?

Answer 1

`\chi^2 = ((141-141)^2)/(141)+((291-282)^2)/(282)+((132-141)^2)/(141)=0.862`

`p_v = P(\chi^2_(2) >0.861)=0.6502`

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(0.861,2,TRUE)"

Since the p value is higher than the significance level
`0.6502>0.05` we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we don't have significant differences for the proportions assumed.

Explanation:

Previous concepts

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Solution to the problem

Assume the following dataset:

White = 141, Pink = 291, Red= 132

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is no difference with the proportions assumed
`p_(white)=1/4, p_(pink)=1/2, p_(red)=1/4`

H1: There is difference with the proportions assumed
`p_(white)=1/4, p_(pink)=1/2, p_(red)=1/4`

The level of significance assumed for this case is
`\alpha=0.05`

The statistic to check the hypothesis is given by:

`\chi^2 = \sum_(i=1)^n ((O_i -E_i)^2)/(E_i)`

The table given represent the observed values, we just need to calculate the expected values with the following formula
`E_i = p_i *Total`

And the calculations are given by:

`E_(White) =\fra{1}{4} * 564=141`

`E_(Pink) =(1)/(2) *564=282`

`E_(Red) =(1)/(4)*564=141`

And now we can calculate the statistic:

`\chi^2 = ((141-141)^2)/(141)+((291-282)^2)/(282)+((132-141)^2)/(141)=0.862`

Now we can calculate the degrees of freedom for the statistic given by:

`df=categories-1=3-1=2`

And we can calculate the p value given by:

`p_v = P(\chi^2_(2) >0.861)=0.6502`

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(0.861,2,TRUE)"

Since the p value is higher than the significance level
`0.6502>0.05` we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we don't have significant differences for the proportions assumed.