Question

A genetic experiment involving peas yielded one sample of offspring consisting of 420 green peas and 174 yellow peas. Use a 0.01 significance level to test the claim that under the same​ circumstances, 23​% of offspring peas will be yellow. Identify the null​ hypothesis, alternative​ hypothesis, test​ statistic, P-value, conclusion about the null​ hypothesis, and final conclusion that addresses the original claim. Use the​ P-value method and the normal distribution as an approximation to the binomial distribution.

a. What is the test statistic?
b. What is the critical value?

Answer 1

a)
`z=\frac{0.293 -0.23}{\sqrt{(0.23(1-0.23))/(594)}}=3.649`

b) For this case we need to find a critical value that accumulates
`\alpha/2` of the area on each tail, we know that
`\alpha=0.01`, so then
`\alpha/2 =0.005`, using the normal standard table or excel we see that:

`z_(crit)= \pm 2.58`

Since the calculated value is higher than the critical value we have enough evidence to reject the null hypothesis at 1% of significance.

Explanation:

Data given and notation

n=420+174=594 represent the random sample taken

X=174 represent the number of yellow peas

`\hat p=(174)/(594)=0.293` estimated proportion of yellow peas

`p_o=0.23` is the value that we want to test

`\alpha=0.01` represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

`p_v` represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion of yellow peas is 0.23:

Null hypothesis:
`p=0.23`

Alternative hypothesis:
`p \\eq 0.23`

When we conduct a proportion test we need to use the z statisitc, and the is given by:

`z=\frac{\hat p -p_o}{\sqrt{(p_o (1-p_o))/(n)}}` (1)

The One-Sample Proportion Test is used to assess whether a population proportion
`\hat p` is significantly different from a hypothesized value
`p_o`.

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

`z=\frac{0.293 -0.23}{\sqrt{(0.23(1-0.23))/(594)}}=3.649`

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided
`\alpha=0.05`. The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

`p_v =2*P(z>3.649)=0.00026`

So the p value obtained was a very low value and using the significance level given
`\alpha=0.05` we have
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis.

b) Critical value

For this case we need to find a critical value that accumulates
`\alpha/2` of the area on each tail, we know that
`\alpha=0.01`, so then
`\alpha/2 =0.005`, using the normal standard table or excel we see that:

`z_(crit)= \pm 2.58`

Since the calculated value is higher than the critical value we have enough evidence to reject the null hypothesis at 1% of significance.