P-value Formula 

The P-value formula is short for probability value. P-value defines the probability of getting a result that is either the same or more extreme than the other actual observations. The P-value represents the probability of occurrence of the given event. The P-value formula is used as an alternative to the rejection point to provide the least significance for which the null hypothesis would be rejected. The smaller the P-value, the stronger is the evidence in favor of the alternative hypothesis given observed frequency and expected frequency. 

What is P-value Formula?

P-value is an important statistical measure, that helps to determine whether the hypothesis is correct or not. P-value always only lies between 0 and 1. The level of significance(α) is a predefined threshold that should be set by the researcher. It is generally fixed as 0.05. The formula for the calculation for P-value is:

Step 1: Find out the test static Z is

\(Z = \frac{\hat{p}-p 0}{\sqrt{\frac{p 0(1-p 0)}{n}}}\)

Where,

• \(\hat{p}=\)Sample Proportion
• \(\mathrm{P0}=\) assumed population proportion in the null hypothesis
• N = sample size

Step 2: Look at the Z-table to find the corresponding level of P from the z value obtained.

P-value Formula

The formula to calculate the P-value is:

\(Z = \frac{\hat{p}-p 0}{\sqrt{\frac{p 0(1-p 0)}{n}}}\)

Where,

\(\hat{p}=\)Sample Proportion
\(\mathrm{P0}=\) assumed population proportion in the null hypothesis

P-value Table

The below-mentioned P-value table helps in determining the hypothesis according to the p-value. 

Examples Using P-value Formula 

Example 1: A statistician is testing the hypothesis H0: μ = 120 using the approach of alternative hypothesis Hα: μ > 120 and assuming that α = 0.05. The sample values that he took are as n =40, σ = 32.17 and x̄ = 105.37. What is the conclusion for this hypothesis?

Solution:

We know that,
\(\sigma_{\bar{x}}=\dfrac{\sigma}{\sqrt{n}}\)
Now substitute the given values
\(\sigma_{\bar{x}}=\dfrac{32.17}{\sqrt{40}}=5.0865\)

As per the test static formula, we get

t = (105.37 – 120) / 5.0865

Therefore, t = -2.8762

Using the Z-Score table, finding the value of P(t > -2.8762)

we get,

P (t < -2.8762) = P(t > 2.8762) = 0.003

Therefore,

If P(t > -2.8762) =1 - 0.003 =0.997

P- value =0.997 > 0.05

As the value of p > 0.05, the null hypothesis is accepted.

Therefore, the null hypothesis is accepted.

Example 2: P-value is 0.3105. If the level of significance is 5%, find if we can reject the null hypothesis.

Solution: Looking at the P-value table, the p-value of 0.3105 is greater than the level of significance of 0.05 (5%), we fail to reject the null hypothesis.

Example 3: P-value is 0.0219. If the level of significance is 5%, find if we can reject the null hypothesis.

Solution: Looking at the P-value table, the p-value of 0.0219 is less than the level of significance of 0.05, we reject the null hypothesis.