Bartlett's test is used to test the assumption that variances are equal (i.e., homogeneous) across groups. The test is easy to implement and produces valid results, assuming data points within groups are randomly sampled from a normal distribution.

Because Bartlett's test is sensitive to departures from normality, a normality test is prudent. Several ways to check for departures from normality are described at: How to Test for Normality: Three Simple Tests.

Note: Unlike Hartley's Fmax test, which also tests for homogeneity, Bartlett's test does not assume equal sample sizes across groups.

Bartlett's test is an actual hypothesis test, where we examine observed data to choose between two statistical hypotheses:

• Null hypothesis: Variance is equal across all groups.

  H₀: σ²_i = σ²_j for all groups

• Alternative hypothesis: Variance is not equal across all groups.

  H₀: σ²_i ≠ σ²_j for at least one pair of groups

Like many other techniques for testing hypotheses, Bartlett's test for homogeneity involves computing a test-statistic and finding the P-value for the test statistic, given degrees of freedom and significance level. If the P-value is bigger than the significance level, we accept the null hypothesis; if it is smaller, we reject the null hypothesis.

The steps required to conduct Bartlett's test for homogeneity are detailed below:

• Step 1. Specify the significance level ( α ).
• Step 2. Compute the sample variance ( s²_j ) for each group.

  s²_j = [ Σ ( X _{i, j} - X _j ) ² ] / ( n _j - 1 )

  where X _{i, j} is the score for observation i in Group j , X _j is the mean of Group j, n _j is the number of observations in Group j , and k is the number of groups.

• Step 3. Compute the pooled estimate of sample variance ( s²_p ).

  N = Σ n _i

  s²_p =  [ Σ ( n _j-1 ) s²_j ] / ( N - K )

  where n _j is the sample size in Group j , k is the number of groups, N is the total sample size, and s²_j is the sample variance in Group  j.

• Step 4. Compute the test statistic (T).

  A = ( N - k ) * ln( s²_p )

  B = Σ [ ( n _j - 1 ) * ln( s²_j ) ]

  C = 1 / [ 3 * ( k - 1 ) ]

  D = Σ [ 1 / ( n _j - 1 ) - 1 / ( N - k ) ]

  T = ( A - B ) / [ 1 + ( C * D ) ]

  where A is the first term in the numerator of the test statistic, B is the second term in the numerator, C is the first term in the denominator , D is the second term in the denominator, and ln is the natural logarithm.

• Step 5. Find the degrees of freedom ( df ), based on the number of groups ( k ).

  df = k - 1

• Step 7. Find the P-value for the test statistic. The P-value is the probability of seeing a test statistic more extreme (bigger) than the observed T statistic from Step 4.

  It turns out that the test statistic (T) is distributed much like a chi-square statistic with ( k-1 ) degrees of freedom. Knowing the value of T and the degrees of freedom associated with T, we can use Stat Trek's Chi-Square Calculator to find the P-value - the probability of seeing a test statistic more extreme than T.

• Step 7. Accept or reject the null hypothesis, based on P-value and significance level. If the P-value is bigger than the significance level, we accept the null hypothesis that variances are equal across groups. Otherwise, we reject the null hypothesis.

Bartlett's is designed to test the hypothesis of homogeneity among nonoverlapping sets of data. The number of groups is the number of datasets under consideration.

The significance level is the probability of rejecting the null hypothesis when it is true. Researchers often choose 0.05 or 0.01 for a significance level.

Sample size refers to the number of observations in a group. For each group, enter the number of observations in the space provided.

Note: Unlike some other tests for homogeneity (e.g., Hartley's Fmax test), Bartlett's test does not require equal sample sizes across groups.

In the fields provided, enter an estimate of sample variance for each group. To compute sample variance ( s²_j ) for each group, use the following formula:

s²_j = [ Σ ( X _{i, j} - X _j ) ² ] / ( n _j - 1 )

where X _{i, j} is the score for observation i in Group j , X _j is the mean of Group j, and n _j is the number of observations in Group j .

Bartlett's test computes a test statistic (T) to test for normality. The degrees of freedom ( df ) for a chi-square test of that statistic is:

df = N - k

where N is the total sample size across all groups, and k is the number of groups in the sample.

The test statistic (T) is the statistic used by Bartlett's test to make a decision about whether to accept or reject the null hypothesis of equal variances between groups. When T is very big, we reject the null hypothesis; when T is small, we accept the null hypothesis.

The calculator computes a T statistic, based on user inputs. The formulas that the calculator uses to compute a T statistic are given at Bartlett's Test for Homogeneity of Variance.

If you assume that the null hypothesis of equal variance is true, the P-value is the probability of seeing a test statistic (T) that is more extreme (bigger) than the actual test statistic computed from sample data.

Like many other techniques for testing hypotheses, Bartlett's test for homogeneity of variance involves computing a test-statistic and finding the P-value for the test statistic, given degrees of freedom and significance level.

If the P-value is bigger than the significance level, the calculator accepts the null hypothesis. Otherwise, it rejects the null hypothesis.