91影视

In a sample of offspring, there were 428 green peas and 152 yellow peas. It is claimed that 25% of offspring peas will be yellow.

The null hypothesis is written as follows.

The proportion of yellow offspring is equal to 25%.

$H_0:p=0.25$

The alternative hypothesis is written as follows.

The proportion of yellow offspring is not equal to 25%.

$H_1:p鈮�0.25$

The test is two-tailed.

The sample size is equal to

$$\begin{gathered} n=428+152 \\ =580 \end{gathered}$$

The sample proportion of yellow offspring is computed below.

$$\begin{gathered} \hat{p}=\frac{\text{Number}\text{of}\text{yellow}\text{offspring}}{\text{Sample}\text{Size}} \\ =\frac{152}{580} \\ =0.262 \end{gathered}$$

The population proportion of yellow offspring is equal to 0.25.

The value of the test statistic is computed below.

$$\begin{gathered} z=\frac{\hat{p}-p}{\sqrt{\frac{pq}{n}}} \\ =\frac{0.262-0.25}{\sqrt{\frac{0.25\left(1-0.25\right)}{580}}} \\ =0.671 \end{gathered}$$

Thus, z=0.671.

Referring to the standard normal table, the critical value of z at$伪=0.01$ for a two-tailed test is equal to 2.5758.

Referring to the standard normal table, the p-value for the test statistic value of 0.671 is equal to 0.5022.

As the p-value is greater than 0.01, the decision is to fail to reject the null hypothesis.

There is not enough evidence to reject the claim that the proportion of yellow peas is equal to 25%.

It can be concluded that Mendel鈥檚 claim of 25% offspring with yellow peas is accurate.