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To illustrate finding a $$p$$-value, consider a sample of $$25$$ individuals (which provides $$24$$ degrees of freedom) resulting in a computed $$t$$ score of $$1.50$$. The probability of obtaining a $$t$$ score at least this extreme depends on the proportion of expected scores in the corresponding distribution that are far from zero. Defining extreme in both directions, the $$p$$-value is the proportion of expected $$t$$ scores that are $$1.50$$ or greater, combined with those that are $$-1.50$$ or lower. In this scenario, that combined proportion is $$.14$$, meaning the final $$p$$-value is $$.14$$.

Example of Finding a p-Value for a t Score

If a sample results in a computed t score of 1.50, what expected t scores are combined to determine the p-value when defining 'extreme' in both directions?

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