Testing Claims About Proportions. In Exercises 9–32, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. Use the P-value method unless your instructor specifies otherwise. Use the normal distribution as an approximation to the binomial distribution, as described in Part 1 of this section.

Mendelian Genetics When Mendel conducted his famous genetics experiments with peas, one sample of offspring consisted of 428 green peas and 152 yellow peas. Use a 0.01 significance level to test Mendel’s claim that under the same circumstances, 25% of offspring peas will be yellow. What can we conclude about Mendel’s claim?

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%.

\({H_1}:p \ne 0.25\).

The test is two-tailed.

The sample size is equal to

\(\begin{array}{c}n = 428 + 152\\ = 580\end{array}\).

The sample proportion of yellow offspring is computed below.

\[\begin{array}{c}\hat p = \frac{{{\rm{Number}}\;{\rm{of}}\;{\rm{yellow}}\;{\rm{offspring}}}}{{{\rm{Sample}}\;{\rm{Size}}}}\\ = \frac{{152}}{{580}}\\ = 0.262\end{array}\].

The population proportion of yellow offspring is equal to 0.25.

The value of the test statistic is computed below.

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

Thus, z=0.671.

Referring to the standard normal table, the critical value of z at\(\alpha = 0.01\)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.

Asthe 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’s claim of 25% offspring with yellow peas is accurate.