A genetic experiment involving peas yielded one sample of of

A genetic experiment involving peas yielded one sample of offspring consisted of 424 green peas and 152 yellow peas. Use a 0.01 significance level to test the claim that under the same circumstance, 24% of offspring peas will be yellow. What is the test statistic and what is the P-value?

Solution

Total peas =424+152 = 576

The test hypothesis:

Ho: p=0.24 (i.e. null hypothesis)

Ha: p not equal to 0.24 (i.e. alternative hypothesis)

The test statiistic is

Z=(phat-p)/sqrt(p*(1-p)/n)

=(152/576-0.24)/sqrt(0.24*(1-0.24)/576)

= 1.34

It is a two-tailed test.

So the p-value= 2*P(Z> 1.34) = 0.1802 (from standard normal table)

Since the p-value is larger than 0.01, we do not reject the null hypothesis.

So we can not conclude that 24% of offspring peas will not be yellow