In a flowering plant species, snapdragons, variation in flower color is determined by a single gene.

Given that:
The first cross experiment ends up with 40 offspring's, including 10 red, 21 pink, and 9 white.
red:pink:white-flowered offspring's is 1:2:1
a) 1)
Null Hypothesis H0: The ratio of red:pink:white-flowered offspring's is 1:2:1.
Alternative Hypothesis H0: The ratio of red:pink:white-flowered offspring's is not 1:2:1.
H0 =Distribution is in ratio 1:2:1
Ha = Not distributed in ratio 1:2:1
Significance level ${\displaystyle \left(\alpha \right)=0.05}$
$\begin{array}{|ccc|}\hline \text{ Offspring's }& \text{ Expected }& \text{ Observed }\\ \text{Red }& 10& 10\\ \text{ Pink }& 20& 21\\ \text{ White }& 10& 9\\ \hline\end{array}$
${\displaystyle {\chi }^2=\left((10-10)\frac{2}{10}\right)+\left((21-20)\frac{2}{20}\right)+\left((9-10)\frac{2}{10}\right)}$
${\displaystyle {\chi }^2=0+0.05+0.1}$
${\displaystyle {\chi }^2=0.15}$
Degree of freedom = N - 1 (where N = number of quantities under observation)
= 2
Now, check the value of ${\displaystyle {\chi }^2}$ for degree of freedom 2
We get,
${\displaystyle {\chi }^2}$ calculate < ${\displaystyle {\chi }^2}$ tabular
Probability of getting a value of ${\displaystyle {\chi }^2\ge 0.15}$ is more than 90%. (from table)
So, we fail to reject null hypothesis
2) x <- c(10,21,9)
p <- c(1/4,2/4,1/2)
chi-square test (x, p = p, rescale. P = TRUE)
Running the R code, we get output as below.
> chi-square test (x, p = p, rescale. P = TRUE)
Chi-squared test for given probabilities data: x
X-squared = 5.125, df = 2, p-value = 0.07711172
P-value = 0.07711172
Since, p-value is greater than 0.05 significance level, we fail to reject null hypothesis H0.
We conclude that there is no strong evidence to reject the claim that ratio of red:pink:white-flowered offspring's is 1:2:1.
b) 1)
Since, p-value is less than 0.05 significance level, we reject null hypothesis H0.
We conclude that there is strong evidence from the sample data that ratio of red:pink:white-flowered offsprings is not 1:2:1.
2)
x <- c(1000,2100,900)
p <- c(1/4,2/4,1/2)
chi-square test(x, p = p, rescale. P = TRUE)
Running the R code, we get output as below.
> chi-square test(x, p = p, rescale. P = TRUE)
Chi-squared test for given probabilities data: x
X-squared = 512.5, df = 2, p-value < 2.2204e-16
P-value = 2.2204e-16
(x, p = p, rescale.p = TRUE)
Running the R code, we get output as below.
> chi-square test(x, p = p, rescale. P = TRUE)
Chi-squared test for given probabilities data: x
X-squared = 512.5, df = 2, p-value < 2.2204e-16
P-value = 2.2204e-16
(x, p = p, rescale. P = TRUE)
Running the R code, we get output as below.
> chi-square test(x, p = p, rescale. P = TRUE)
Chi-squared test for given probabilities data: x
X-squared = 512.5, df = 2, p-value < 2.2204e-16
P-value = 2.2204e-16