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

Answer:

$\chi^2 = \frac{(141-141)^2}{141}+\frac{(291-282)^2}{282}+\frac{(132-141)^2}{141}=0.862$

$p_v = P(\chi^2_{2} >0.861)=0.6502$

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.

Step-by-step 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:

$\chi^2 = \sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

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:

$E_{White} =\fra{1}{4} * 564=141$

$E_{Pink} =\frac{1}{2} *564=282$

$E_{Red} =\frac{1}{4}*564=141$

And now we can calculate the statistic:

$\chi^2 = \frac{(141-141)^2}{141}+\frac{(291-282)^2}{282}+\frac{(132-141)^2}{141}=0.862$

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.