# P-Value

Every value you get from formula or physical observation needs to be checked for significance. In statistics, observed results are checked using the P-value.

The P-value is a probability value concept. It has two hypotheses. One is the null hypothesis, in which we believe there is no difference or change in the probability. This means that even though we observe a difference in the result, we choose to ignore it. Simplify by stating the difference is due to a random variable, and it will have no real effect on our event.

The second hypothesis, or known as the alternate hypothesis, assumes that the null hypothesis is false.  We then use a set of mathematical-statistical methods to calculate the probability of our event, assuming that the null hypothesis is correct. So, P-value is defined as the probability or the greater deviation that we are calculating from the null-hypothesis.

If the value obtained from the p-value formula is less than the predefined number, we reject the null hypothesis and adopt the alternative hypothesis, which says that the result is statistically significant. However, if the result is statistically insignificant, we accept the null hypothesis. Read on to know more

## Hypothesis Interpretation From the Calculated P-Value.

 P-Value Description Hypothesis Interpretation P-value ≤ 0.05 It indicates the null hypothesis is very unlikely. Rejected P-value > 0.05 It indicates the null hypothesis is very likely. Accepted or it “fails to reject”. P-value = 0.05 The P-value is near the cut-off. It is considered as marginal Redo the calculation.

### How to Find P Value Formula:

P-value is a statistical measure, which is a probability calculation of the event to check which hypothesis is correct. Hence, its value lies between 0 and 1. There is no direct p-value formula. We calculate test static and from using the z-value to find the p-value table.

Step 1: Find the value of test static Z

Z = , Where

= Sample proportion

= Assume proportion in the null hypothesis

N = Sample Size

Step 2: Use Z-table to find the corresponding P-value.

Once we find the z value, we open z-table to find p-value. But before we do that, we need the level of significance(α). This is a predefined threshold usually provided in the question. If not, you take it as 0.05. Using z and α, you find the corresponding p-value and choose the correct hypothesis.

### Real-Life P-Value Examples:

Let’s take an example to understand p value meaning. Imagine you are experimenting in the lab. Your friend accidentally mixed the wrong chemical, but you still get a result close to the ideal result. Now, you calculate the P-value for this experiment. If the value of the P-value is less (insignificant), you accept the null hypothesis that the wrong chemical has little or no effect on the experiment. However, if the value of the p-value is significant, you accept the alternative hypothesis. Sadly, you have to redo the experiment to get the proper result. This is a real-life p-value example.

### Solved Examples

1. A chemist wants to test his hypothesis H0: μ = 110 using the alternative hypothesis Hα: μ > 110 and assuming that α = 0.05. To test his hypothesis, he has some sample values such that n = 45, σ = 35.24 and x̄ = 108.39. Determine the conclusion for this hypothesis?

Solution: Now, we know that σ =

From question, σ = 35.24, n = 45.

Hence, σ = 35.24/ = 5.25

Now, we will calculate the static test value,

t = (108.39 – 110)/ 5.25 = -0.306

Now, we will use the z-value table, to find the value of P(t>-0.306)

Using the table, we get

P (t<-0.306) = P(t>0.306) = 0.379802.

Therefore,

If P(t>-0.306) =1- 0.379802 =0.620

P- value =0.620 > 0.05

Therefore, as we can see p>0.05, the null hypothesis is accepted or fails to reject.

Hence, the conclusion is “fails to reject H0.”

2. A person conducted a statistical test about the mean value of a variable Y.

H0: μ = μ0 vs H1: μ ≠ μ0

He obtains a test statistic of (-2.5.) Assume, 5% significance level, Find the p-value and also write your interpretation.

Solution: There is not much information about the variable Y, but we are checking statistical significance. Hence, it is a two-tailed test.

Given value of z = -2.5

So, P(z=-2.5) = 0.0062*2 = 0.0124 = 1.24% (We multiplied it by two because it is two-tailed test)

As we can see, the p-value is 1.24% is less than the level of significance (5%). Therefore, we can easily reject the null hypothesis from our value.