In more detail
The statistic is Σ (O − E)² / E. Sixty rolls of a die giving 5, 8, 9, 8, 10 and 20 against 10 expected each score 13.4 with 5 degrees of freedom, a p-value of about 0.02, so the die looks unfair.
A chi-square test compares observed counts with the counts expected under a hypothesis to see whether the gap is more than chance.
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The statistic is Σ (O − E)² / E. Sixty rolls of a die giving 5, 8, 9, 8, 10 and 20 against 10 expected each score 13.4 with 5 degrees of freedom, a p-value of about 0.02, so the die looks unfair.