Question

One kind of plant has only blue flowers and white flowers. According to a genetic model, the offsprings of a certain cross have a 0.75 chance to be blue-flowering, and a 0.25 chance to be white-flowering, independently of one another. Two hundred seeds of such a cross are raised, and 142 turn out to be blue-flowering. We are interested in determining whether the data are consistent with the model or, alternatively, the chance to be blue-flowering is smaller than 0.75. For this question, find the appropriate test statistic.

Answer 1

Answer:

There is not enough evidence to support the claim that the chance of this cross to be blue-flowering is significantly smaller than 0.75 (P-value = 0.11).

Test statistic z=-1.225.

Step-by-step explanation:

This is a hypothesis test for a proportion.

The claim is that the chance to be blue-flowering is significantly smaller than 0.75.

Then, the null and alternative hypothesis are:

$H_0: \pi=0.75\\\\H_a:\pi$

The significance level is 0.05.

The sample has a size n=200.

The sample proportion is p=0.71.

$p=X/n=142/200=0.71$

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.75*0.25}{200}}\\\\\\ \sigma_p=\sqrt{0.000938}=0.031$

Then, we can calculate the z-statistic as:

$z=\dfrac{p-\pi+0.5/n}{\sigma_p}=\dfrac{0.71-0.75+0.5/200}{0.031}=\dfrac{-0.038}{0.031}=-1.225$

This test is a left-tailed test, so the P-value for this test is calculated as:

$\text{P-value}=P(z$

As the P-value (0.11) is greater than the significance level (0.05), the effect is  not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the chance to be blue-flowering is significantly smaller than 0.75.