Question

ind the value of the test statistic z using z equals StartFraction ModifyingAbove p with caret minus p Over StartRoot StartFraction pq Over n EndFraction EndRoot EndFractionz= p−p pq n. The claim is that the proportion of peas with yellow pods is equal to 0.25​ (or 25%). The sample statistics from one experiment include 530530 peas with 139139 of them having yellow pods.

Answer 1

Answer:

$z=\frac{0.262 -0.25}{\sqrt{\frac{0.25(1-0.25)}{530}}}=0.638$  

Step-by-step explanation:

Data given and notation

n=530 represent the random sample taken

X=139 represent the yellow pods in the random sample

$\hat p=\frac{139}{530}=0.262$ estimated proportion of yellow pods

$p_o=0.25$ is the value that we want to test

$\alpha$ represent the significance level

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 is 0.25 or 25%:  

Null hypothesis:$p=0.25$  

Alternative hypothesis:$p \neq 0.25$  

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

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (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:  

$z=\frac{0.262 -0.25}{\sqrt{\frac{0.25(1-0.25)}{530}}}=0.638$