F-test

9.1.4 F-test

An important chapter in the investigation of relations between h random variables is based on the analysis of standard deviation or variance (see ANOVA, Section 9.1.6).

During its application we usually have to assume the normal distribution of h random variables and the equality of their standard deviations. In statistics the application of the F-test is considerably widespread.

Consider the normally distributed and independent random variablesXandY. Their expected values and standard deviations σ_X := D(X), σ_Y := D(Y) are unknown. We wish to decide about the hypothesis H₀ : σ_X =σ_Y at a significance level of 1−α. We choose the two-sided hypothesis H₁ : σ_X 6=σ_Y as alternative hypothesis, in other cases the one-sided hypothesis H₁⁰ :σ_Y < σ_X (or H₁⁰⁰ :σ_X < σ_Y).

It is known that in case the null hypothesis holds, the test statistic F := M SS_X^∗

M SS_Y^∗

has an F-distribution with parameters n_X −1 , n_Y −1, where n_X and n_Y are the two sample sizes (for details see [VIN][page 125]. (The notation F is used locally for the test statistic.)

The question of critical regions is more complicated than before, partly because the density function of the test statistic F is not symmetrical (cf. Fig. 9.7).

We keep from the previous practice – at a significance level of 1−αand seemingly in a straightforward way – that we specify such critical values a₁ and a₂ as boundaries of the region (interval) of acceptance that the area under the density function on the ”left of”a₁ be equal to the area on the ”right of”a₂ (cf. Fig.9.7):

P(F < a₁) =P(a₂ < F) = α 2 (here we ignore the notation of the parameters for simplicity).

Finally, when testing the hypothesis at a significance level of 1−αwe in fact examine if the observation

f := mss^∗_X mss^∗_Y

for the random variable F falls into the region of acceptance (a₁, a₂). However, it is known from the literature that the relation a₂ = _a¹

1 happens to be valid (which implies

Figure 9.7: Graph of the probability density function of the test statistic F of a given parameter with indication of the critical values a₁ and a₂ = _a¹

1 at a significance level of 1−α.

and its reciprocal is greater than a2”. From this follows that the probability of this latter event isα.

So, a good decision strategy for the originally two-sided hypothesis may be the fol-lowing one-sided test. Suppose that the ratio which is greater than 1 exceeds the value Ff1,f2,α, where f1, f2 are the parameters belonging to the numerator and denominator of the ratio, respectively. Then the hypothesis H₀ : σ_X = σ_Y is rejected against the alternative hypothesis H₁ : σ_X 6= σ_Y in an actually one-sided test test at a significance level of 1−α, otherwise we accept it. The critical value Ff1,f2,α can be looked up from the corresponding table (Table III). For example, if f₁ = 5, f₂ = 7, then at a significance level of α = 0.05 the critical value is F_5,7,0.05= 3.97.

In a one-sided test, if the significance level is 1−α and the null hypothesis is for instance H₀⁰ :σ_X < σ_Y, and the alternative hypothesis is H₁⁰ :σ_Y < σ_X, then the routine procedure is as follows. If mss^∗_Y < mss^∗_X, then we rejectH₀⁰ and acceptH₁⁰. However, if mss^∗_X ≤mss^∗_Y, then we accept H₀⁰ if the ratio ^mss_mss^∗Y∗

X

exceeds the critical valueF_f1,f2,α in Table III, where the parameters f₁ and f₂ belong to Y and X, respectively.

The F-test is by far not a robust test, i.e., it cannot be applied to distributions significantly different from a normal one.

Example 9.6. 10female and 8male cockroaches of the same age were deprived of food, and their times of survival x and y (in days) were measured. The mean survival times were x = 8.5 and y = 4.8. For the observation of the females the corrected empirical

variance was mss^∗_x = 3.6, while for the males mss^∗_y = 0.9. By assuming approximately normal distribution of the survival times within each sex (!), let us make a two-sided F-test at a significance level of 1−α = 0.95 in order to check the null hypothesis that the standard deviations are the same, i.e., H₀ : σ_X = σ_Y by the two-sided alternative hypothesis H₁ :σ_x 6=σ_y.

Solution: The critical value depends on whether the relation 1 < ^mss

∗ y

mss^∗_x or 1 < ^mss_mss^∗x∗ y

holds. Since now the latter one is true, the pair of parameters fx, fy is 9, 7. The corresponding critical value for a one-sided test and for 1−α = 0.95 is F_9,7,0.05 = 3.68 (see Table III).

The observed value of the test statistic is ^3.6_0.9 = 4.0. The latter one exceeds the critical value of 3.68. According to this, we reject the null hypothesis that the standard deviations are the same at the given significance level 0.95.

If the null hypothesis and the alternative hypothesis were H₀⁰ :σ_x < σ_y, H₁⁰ :σ_y < σ_x then due to mss^∗_y < mss^∗_x, we would at once reject H₀⁰ and accept H₁⁰. However, for the hypotheses H₀⁰⁰ : σ_y < σ_x, H₁⁰⁰ : σ_x < σ_y at the same significance level H₀⁰⁰ would be accepted, since mss^∗_y < mss^∗_x and ^mss_mss^∗x∗

y = 4.0>3.68.

Example 9.7. When examining the glutamic acid concentration in the urine of two primate species (gorilla (x) and chimpanzee (y)), the results are contained in Table 9.3.

sample size (number of individuals)

n

corrected empirical variance

mss^∗

chimpanzee n_x = 37 mss^∗_x = 0.0107

gorilla n_y = 6 mss^∗_y = 0.124

Table 9.3: Some intermediate results of the calculation when applying the F-test, see Example 9.7.

We assume the normality of the measurements within each species. Let us check the null hypothesis that the standard deviations are the same for the two species by using the two-sided F-test at a significance level of 1−α= 0.99 against the alternative hypothesis H1 :σx 6=σy.

Solution: We can find the required critical value at the pair of parameters (5, 36).

According to Table III, F_5,36,0.01 = 3.60. The observation with regard to the test statistic F is: f = _0.0107^0.124 = 11.589. This value is greater than the critical value 3.60. Therefore, at a significance level of 0.99, we reject the null hypothesis that the standard deviations are equal.