Question

Assume that a cross is made between a heterozygous tall pea plant and a homozygous short pea plant. Fifty offspring are produced in the following frequency:
30 = tall
20 = short
Null hypothesis: The deviations from a 1:1 ratio (25 tall and 25 short) are due to chance.
1. What is the Chi-square value associated with the appropriate test of significance?
2. How many degrees of freedom are associated with this test of significance?

Answer 1

Answer:

1) $\chi^2 = \frac{(30-25)^2}{25} +\frac{(20-25)^2}{25} =1 +1 =2$

2) $df = c-1 = 2-1$

Where c represent the number of categories c=2

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".

Assume the following dataset:

Tall =30 , Short =20

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

H0: The deviations from a 1:1 ratio (25 tall and 25 short) are due to chance.

H1: The deviations from a 1:1 ratio (25 tall and 25 short) are NOT due to chance.

Part 1

So then we know that the expected values would be 25 for each case

The statistic to check the hypothesis is given by:

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

And if we replace we got:

$\chi^2 = \frac{(30-25)^2}{25} +\frac{(20-25)^2}{25} =1 +1 =2$

Part 2

For this case the degreed of freedom are given by:

$df = c-1 = 2-1$

Where c represent the number of categories c=2

And we can calculate the p value given by:

$p_v = P(\chi^2_{1} >2)=0.157$

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

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