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Section 5

[ "article:topic", "authorname:dkiernan" ]
  • Page ID
    2894
  • F-Test for Comparing Two Population Variances

    One major application of a test for the equality of two population variances is for checking the validity of the equal variance assumption \((\sigma_1^2=\sigma_2^2)\) for a two-sample t-test. First we hypothesize two populations of measurements that are normally distributed. We label these populations as 1 and 2, respectively. We are interested in comparing the variance of population 1 \((\sigma_1^2)\) to the variance of population 2 \((\sigma_2^2)\).

    When independent random samples have been drawn from the respective populations, the ratio

    $$\frac {S^2_1/S_2^2}{\sigma_1^2/\sigma ^2_2}$$

    possesses a probability distribution in repeated sampling that is referred to as an F distribution and its properties are:

    • Unlike Z and t, but like \(\chi^2\), F can assume only positive values.
    • The F distribution, unlike the Z and t distributions, but like the (\chi^2\) distribution, is non-symmetrical.
    • There are many F distributions, and each one has a different shape. We specify a particular one by designating the degrees of freedom associated with \(S_1^2\) and \(S_2^2\). We denote these quantities by \(df_1\) and \(df_2\), respectively.

    Image37109.GIF

    Figure 5. The F-distribution.

    Note: A statistical test of the null hypothesis \(\sigma_1^2 = \sigma_2^2\) utilizes the test statistic \(S_1^2/S_2^2\). It may require either upper tail or lower tail rejection region, depending on which sample variance is larger. To alleviate this situation, we are at liberty to designate the population with the larger sample variance as population 1 (i.e., used as the numerator of the ratio \(S_1^2/S_2^2\)). By this convention, the rejection region is only located in the upper tail of the F distribution.

    Null hypothesis:  \(H_0:\sigma_1^2 = \sigma_2^2\)

    Alternative hypothesis:

    • \(H_a: \sigma_1^2 > \sigma_2^2\) (one-tailed), reject \(H_0\) if the observed F > Fα
    • \(H_a: \sigma_1^2 \ne \sigma_2^2\) (two-tailed), reject \(H_0\) if the observed F > Fα/2.

    Test statistic: \(F = \frac {S_1^2}{S^2_2}\) assuming \(S_1^2>S_2^2\),

    where the F critical value in the rejection region is based on 2 degrees of freedom \(df_1 = n_1 – 1\) (associated with numerator \(S_1^2\)) and \(df_2 = n_2 – 1\) (associated with denominator \(S_2^2\)).

    Example \(\PageIndex{1}\):

    A forester wants to compare two different mist blowers for consistent application. She wants to use the mist blower with the smaller variance, which means more consistent application. She wants to test that the variance of Type A (0.087 gal.2) is significantly greater than the variance of Type B (0.073 gal.2) using α = 0.05.

    Type A Type B
    \(S_1^2\) = 0.087 \(S^2_2\)=0.073
    \(n_1\)= 16 \(n_2\) = 21

    Solution

    \(H_0: \sigma_1^2 = \sigma_2^2\)

    \(H_1:\sigma_1^2 > \sigma_2^2\) 

    The critical value \((df_1 = 15\)  and  \(df_2 = 20)\) is 2.20.

    The test statistic is:

    $$F = \frac {S_1^2}{S_2^2} = \frac {0.087}{0.073}=1.192$$

    The test statistic is not larger than the critical value (it does not fall in the rejection zone) so we fail to reject the null hypothesis. While the variance of Type B is mathematically smaller than the variance of Type A, it is not statistically smaller. There is not enough statistical evidence to support the claim that the variance of Type A is significantly greater than the variance of Type B. Both mist blowers will deliver the chemical with equal consistency.

    Software Solutions

    Minitab

    087_1.tif

    087_2.tif

    Test and CI for Two Variances

    Method

    Null hypothesis

    Variance(1) / Variance(2) = 1

    Alternative hypothesis

    Variance(1) / Variance(2) > 1

    Significance level

    Alpha = 0.05

    Statistics

    Sample

    N

    StDev

    Variance

    1

    16

    0.295

    0.087

    2

    21

    0.270

    0.073

    Ratio of standard deviations = 1.092

    Ratio of variances = 1.192

    Tests

     

    Test

    Method

    DF1

    DF2

    Statistic

    p-value

    F Test (normal)

    15

    20

    1.19

    0.351

    Excel

    086_1.tif

    086_2.tif

    F-Test Two-Sample for Variances

     

    Type A

    Type B

    Mean

    11.07188

    11.10595

    Variance

    0.08699

    0.073379

    Observations

    16

    21

    df

    15

    20

    F

    1.185483

     

    \(P(F\le f)\) one-tail

    0.355098

     

    F Critical one-tail

    2.203274