Calculate the p-Value from Z-Score in R
Calculate the p-Value from Z-Score in R, In statistics, we frequently want to know the p-value associated with a specific z-score obtained from a hypothesis test.
We can reject the null hypothesis of our hypothesis test if this p-value is less than a certain level of significance.
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In R, we may use the pnorm() function to find the p-value associated with a z-score, which has the following syntax.
pnorm(q, mean = 0, sd = 1, lower.tail = TRUE)
where:
q: The z-score
mean: The normal distribution’s mean. The default value is 0.
sd: The normal distribution’s standard deviation. The default value is 1.
lower.tail: If TRUE, the probability in the normal distribution to the left of q is returned. The probability to the right is returned if FALSE. TRUE is the default value.
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For a left-tailed test, right-tailed test, and two-tailed test, the following examples show how to get the p-value associated with a z-score.
Test with one tail on the left
In a left-tailed hypothesis test, let’s say we wish to calculate the p-value associated with a z-score of -1.8.
To find p-value
pnorm(q=-1.8, lower.tail=TRUE) 0.03593032
0.03593032 is the p-value. We would reject the null hypothesis based on a significance level of = 0.05 because the p-value is less than 0.05.
Test with the right tail
In a right-tailed hypothesis test, let’s say we wish to calculate the p-value associated with a z-score of 1.8.
To find p-value
pnorm(q=1.8, lower.tail=FALSE) 0.03593032
0.03593032 is the p-value. We would reject the null hypothesis if we used a significance level of = 0.05 because the p-value is less than 0.05.
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Test with two tails
In a two-tailed hypothesis test, let’s say we wish to calculate the p-value associated with a z-score of 1.24.
To find the p-value for the two-tailed test
2*pnorm(q=1.24, lower.tail=FALSE) [1] 0.2149754
We simply multiplied the one-tailed p-value by two to get the two-tailed p-value.
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0.2149 is the p-value. We would fail to reject the null hypothesis of our hypothesis test if we used a significance level of = 0.05 because the p-value is not less than 0.05.
