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The number of peas from genetics experiments is recorded.

The number of sample values $n=580$.

The confidence interval is$99\%$

The requirements for the test are:

1. The samples are selected randomly and normally distributed.
2. There are two categories of the outcome, either green or yellow.
3. The counts of successes and failures are 428 and 152respectively which are greater than 5

All the conditions are satisfied. Hence we can construct a confidence interval for the population proportion.

a.

Thesample proportion of the green peas is:

$$\begin{gathered} \hat{p}=\frac{x}{n} \\ =\frac{428}{580} \\ =0.74 \end{gathered}$$

Therefore, the sample proportion is 0.74.

Then,

$$\begin{gathered} \hat{q}=1-\hat{p} \\ =1-0.74 \\ =0.26 \end{gathered}$$

At confidence interval, $伪=0.01$.

Using the standard normal table,

$$\begin{gathered} z_{crit}=z_{\frac{伪}{2}} \\ =z_{0.005} \\ =2.575 \end{gathered}$$

The margin of error is given by,

$$\begin{gathered} E=z_{crit}脳\sqrt{\frac{\hat{p}\hat{q}}{n}} \\ =2.575脳\sqrt{\frac{0.74脳0.26}{580}} \\ =0.0472 \end{gathered}$$

The margin of error is 0.0472.

The formula for the 99% confidence interval is given by,

$$\begin{gathered} CI=\hat{p}-E<p<\hat{p}+E \\ =\left(0.74-0.0472<p<0.74+0.0472\right) \\ =(0.6907<p<0.7851) \end{gathered}$$

In percentage,99% confidence interval is between ..

b.

It is claimed that the 99% confidence interval of the percentage of green peas is between 69.07% and 78.51%.

The interval includes 75% claimed value.

Therefore, we can expect that 75% of the offspring peas would be green.

Given the percentage of green peas is not 75%, the result contradicts Mendel鈥檚 theory.