P-Value (Definition, Formula, Table & Example) | Significant P-Value (2024)

In Statistics, the researcher checks the significance of the observed result, which is known as test static. For this test, a hypothesis test is also utilized. The P-valueor probability value concept is used everywhere in statistical analysis. It determines the statistical significance and the measure of significance testing. In this article, let us discuss its definition, formula, table, interpretation and how to use P-value to find the significance level etc. in detail.

Table of Contents:

• P-value Definition
• Table
• Formula
• Example
• FAQs

P-value Definition

The P-value is known as the probability value. It is defined as the probability of getting a result that is either the same or more extreme than the actual observations. The P-value is known as the level of marginal significance within the hypothesis testing that represents the probability of occurrence of the given event. The P-value is used as an alternative to the rejection point to provide the least significance at which the null hypothesis would be rejected. If the P-value is small, then there is stronger evidence in favour of the alternative hypothesis.

P-value Table

The P-value table shows the hypothesis interpretations:

Generally, the level of statistical significance is often expressed in p-value and the range between 0 and 1. The smaller the p-value, the stronger the evidence and hence, the result should be statistically significant. Hence, the rejection of the null hypothesis is highly possible, as the p-value becomes smaller.

Let us look at an example to better comprehend the concept of P-value.

Let’s say a researcher flips a coin ten times with the null hypothesis that it is fair. The total number of heads is the test statistic, which is two-tailed. Assume the researcher notices alternating heads and tails on each flip (HTHTHTHTHT). As this is the predicted number of heads, the test statistic is 5 and the p-value is 1 (totally unexceptional).

Assume that the test statistic for this research was the “number of alternations” (i.e., the number of times H followed T or T followed H), which is two-tailed once again. This would result in a test statistic of 9, which is extremely high and has a p-value of 1/2⁸ = 1/256, or roughly 0.0039. This would be regarded as extremely significant, much beyond the 0.05 level. These findings suggest that the data set is exceedingly improbable to have happened by random in terms of one test statistic, yet they do not imply that the coin is biased towards heads or tails.

The data have a high p-value according to the first test statistic, indicating that the number of heads observed is not impossible. The data have a low p-value according to the second test statistic, indicating that the pattern of flips observed is extremely unlikely. There is no “alternative hypothesis,” (therefore only the null hypothesis can be rejected), and such evidence could have a variety of explanations – the data could be falsified, or the coin could have been flipped by a magician who purposefully swapped outcomes.

This example shows that the p-value is entirely dependent on the test statistic used and that p-values can only be used to reject a null hypothesis, not to explore an alternate hypothesis.

P-value Formula

We Know that P-value is a statistical measure, that helps to determine whether the hypothesis is correct or not. P-value is a number that 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

\(\begin{array}{l}z = \frac{\hat{p}-p0}{\sqrt{\frac{po(1-p0)}{n}}}\end{array} \)

Where,

\(\begin{array}{l}\hat{p} = \text{Sample Proportion}\end{array} \)

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 Example

An example to find the P-value is given here.

Question: A statistician wants to test the hypothesis H₀: μ = 120 using the alternative hypothesis Hα: μ > 120 and assuming that α = 0.05. For that, he took the sample values as

n =40, σ = 32.17 and x̄ = 105.37. Determine the conclusion for this hypothesis?

Solution:

We know that,

\(\begin{array}{l}\sigma _{\bar{x}}=\frac{\sigma }{\sqrt{n}}\end{array} \)

Now substitute the given values

\(\begin{array}{l}\sigma _{\bar{x}}=\frac{32.17 }{\sqrt{40}}= 5.0865\end{array} \)

Now, using the test static formula, we get

t = (105.37 – 120) / 5.0865

Therefore, t = -2.8762

Using the Z-Score table, we can find the value of P(t>-2.8762)

From the table, 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

Therefore, from the conclusion, if p>0.05, the null hypothesis is accepted or fails to reject.

Hence, the conclusion is “fails to reject H_0.”

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