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- Select the data from which you want to
**calculate p value**(i-e chi-square, z, t, f critical values). - Input the value according to the
**selected data**. - Enter
**Significance Level(α)**,**Degree of freedom**and**Hypothesis**In The Input Box. - Put the
**Degrees Of Freedom**In The Input Box. - Hit The
**Calculate**Button. - Use The
**Reset**Button To calculate New Values.

This **P-value calculator** is a calculus tool that helps to compute the probability level using the test value, degree of freedom, and significance level.

You can find the significance level of p-values through this calculator using different hypothesis tests e.g from t value, z score, and chi-square.

**Definition of p-value is:**

“The probability of getting a sample similar or extreme than our estimated data under the null hypothesis.”

In simple words, how probable or how likely it is that one gets the same sample data as we just got from the experiment, considering the null hypothesis is true.

P-value is easily calculable using the calculator above. But for your convenience, the steps to find the p-value manually with the z-score test are given ahead.

Make the hypotheses.

Hold an experiment and collect data.

Put the values in the z score formula.

Find the score of z on the normal distribution chart.

This is the probability value and it is the area under the curve after the z value to the extreme.

**Example:**

A consumer rights company wants to test the null hypothesis i.e a nuts pack has exactly **78 **nuts against the alternative hypothesis i.e nuts are not **78**.

For a sample of **100 **packets, the mean amount of nuts is **76 **with a standard deviation of **13.5**. While the population mean is **80**. Find the probability value for a two-tailed test.

**Solution:**

__ Step 1:__ Write both hypotheses.

h0 = a pack contains 78 nuts

ha = a pack does not contain 78 nuts

__ Step 2:__ Write the data for test statistics.

n = 100

͞x = 76

s = 13.5

μ_{0} = 80

__ Step 3:__ Find the z score value.

(Since the sample size is greater than 30, population and sample standard deviations are the same.)

Using the formula;

z = (76 - 80) / (13.5/√100)

z = -4 / 1.35

**z = -2.96**

__ Step 4:__ Look for this value on the z-table.

The value for -2.96 is 0.0015. But since the test type was two-tailed, you will have to multiply this value by 2 to get the area under the curve for both tails.

= 2 x (0.0015)

**= 0.003**

This p-value is less than the standard significance level i.e 0.05. Therefore not enough evidence to reject the null hypothesis.