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P-value

In statistical problems, a researcher often checks the significance of the observed result which is known as the test static. For this, a hypothesis test is being utilized. In general, the P value is the measure of significance testing. The concept of P value has been used everywhere in statistical analysis. In a hypotheses test, P value determines the statistical significance. In practical life, at times, it is the P values on the basis of which it is decided what projects should get funding. In business, misinterpretation of P value may cost credibility and financial loss.

The P value is a very important concept, still it creates confusion among most of the statistics students. This article is all the about the P value, such as what is its definition, what is its interpretation, how to use them to find significance level etc.

 

Definition

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The P value is said to be the probability of getting a result that is either same as or more extreme than the actual observation was. The basic approach of P-value is to determine likely or unlikely of a test static by finding the probability of obtaining a more extreme result than the one that was already observed, assuming the null hypothesis true. In hypothesis testing, the p-value is quite commonly used, especially in significance testing of null hypothesis.

What does P Value Represent or Interpret

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In statistical analysis, the P values represent the following interpretations:

(i) P-value $\leq$ 0.05

This is considered as a 
small p-value which indicates that the null hypothesis is very unlikely. Therefore, it is rejected.

(ii) P-value > 0.05

This is referred to a large p-value which represents that the null hypothesis is very likely. Therefore, it fails to reject or null hypothesis is accepted.

(iii) P-value $\sim$ 0.05

i.e P-values is near the cutoff. So, it is said to be marginal. Therefore, the null hypothesis needs attention.

In order to analyze the statistical data, the p-value must be reported so that the researcher could draw conclusions.

Significance Level Alpha

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In statistics and probability theory, the significance level is denoted by $\alpha$.  The significance level is also called alpha level. They are being commonly utilized in hypothesis tests. Significance level alpha is said to be a probability of adopting a wrong decision, provided that the null hypothesis is true. Ideally, the hypothesis tests use an alpha level of 5% (or 0.05) or 1% (or 0.01).

Basically, there are two types of errors in hypothesis tests:
  
Type I Errors: This error is a false positive error which occurs in the case of wrong rejection of a null hypothesis when it is true.

Type II Errors: This is a false negative error that occurs on failure to reject a null hypothesis when it is false.

The alpha level is the probability of making type I error, i.e. rejecting a true null hypothesis.

P-value Hypothesis Testing

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P-value hypothesis testing is a method of testing a test static in order find a better and more accurate result than the observed result. In this method, the researcher chooses a model called the null hypothesis, before performing the experiment. Also, a threshold value for p is also chosen. Generally, it is taken as 1% or 5% which is referred as the significance level and is represented as $\alpha$. When in hypothesis testing, we get a p-value which is less than $\alpha$, then we reject the null hypothesis. On the other hand, if the p-value is not less than $\alpha$, we fail to reject the null hypothesis.

Examples

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Let us have a look at examples.

Example 1: A statistician wants to test a hypothesis $H_{0}$:  $\mu$ = 120 using alternative hypothesis $H_{a}$ : $\mu$ > 120 and assuming $\alpha$ = 0.05.

For this, he took the sample and obtained the following values:

$\bar{x}$ = 105.37, n = 40, $\sigma$ = 32.17.

What would be the conclusion?

Solution:
 We have

$\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$

$\sigma_{\bar{x}} = \frac{32.17}{\sqrt{40}}$
= 5.0865

Now,

t = $\frac{105.37−120}{5.0865}$

= -2.8762

We want the value of P(t > -2.8762). Using table, we get

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

Hence, P(t > -2.8762) = 1 - 0.003 = 0.997

Here, P-value 0.997 > $\alpha$ 0.05.

Thus, the conclusion is "fail to reject $H_{0}$".
Example 2: To test the hypothesis $H_{0}$: $\mu$ = 15, a statistician takes alternative hypothesis $H_{a}$: $\mu$ $\neq$ 15. He takes the results of a random sample as follows:

n = 20, $\bar{x}$ = 17.5, $\sigma$ = 5.9. He takes $\alpha$ = 0.05. Find P-value and tell what should be the conclusion 
null hypothesis?

Solution:
 $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$

$\sigma_{\bar{x}} = \frac{5.9}{\sqrt{20}}$ = 1.3193

t = $\frac{17.5−15}{1.3193}$

= 1.895

We want P(t > 1.895). Using table, we get

P(t > 1.895) = 0.0367

This is a two-tailed value, therefore, we have to multiply this by 2, i.e P-value = 0.0734.

0.0734 > 0.05

Hence, the conclusion is "fail to reject $H_{0}$".
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