Question

Use the Statistical Applet: P ‑Value for a Test of One Proportion to answer the question.You have taken a sample of ????=1000 individuals and asked them if they typically eat breakfast. Because eating breakfast is considered healthy, you would like to know if the proportion in the population who eat breakfast is more than two‑thirds, about 67%If you only asked the question of ????=20 people and found ????=15 , what is the P ‑value?P-value=

Answer 1

`z=\frac{0.785 -0.7}{\sqrt{(0.7(1-0.7))/(750)}}=5.08`

`p_v =P(z>5.08)=1.88x10^(-7)`

So the p value obtained was a very low value and using the significance level assumed
`\alpha=0.05` we have
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of interest is highr than 0.67 or 67% .

Explanation:

1) Data given and notation

n=20 represent the random sample taken

X=15 represent the people who eat breakfast

`\hat p=(15)/(20)=0.75` estimated proportion of people who eat breakfast

`p_o=0.67` is the value that we want to test

`\alpha=0.05` represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

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

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion is higher than 67%:

Null hypothesis:
`p\leq 0.67`

Alternative hypothesis:
`p > 0.67`

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{(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`.

3) Calculate the statistic

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

`z=\frac{0.785 -0.7}{\sqrt{(0.7(1-0.7))/(750)}}=5.08`

4) 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 next step would be calculate the p value for this test.

Since is a right tailed test the p value would be:

`p_v =P(z>5.08)=1.88x10^(-7)`

So the p value obtained was a very low value and using the significance level assumed
`\alpha=0.05` we have
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of interest is highr than 0.67 or 67% .