9.4: Two Variance or Standard Deviation F-Test (2024)

Solution

Start by drawing the curve and finding the area in each tail. For this case, it would be an area of \(\alpha\)/2 in each tail. Then use technology to find the F-scores. Most technology only asks for the area to the left of the F-score you are trying to find. In Excel the function for F_\(\alpha\)/2 is F.INV(area in left-tail,df₁,df₂).

There is only one function, so use areas 0.025 and 0.975 in the left tail. For this example, we would have critical values F_0.025 = F.INV(0.025,6,14) = 0.1888 and F_0.975 = F.INV(0.975,6,14) = 3.5014. See Figure 9-11.

Figure 9-11

We have to calculate two distinct F-scores unlike symmetric distribution where we could just do ±z-score or ±t-score.

Note if you were doing a one-tailed test then do not divide alpha by two and use area = \(\alpha\) for a left-tailed test and area = 1 – \(\alpha\) for a right-tailed test.

Solution

Draw the curve and shade in the top 5% of the upper tail since \(\alpha\) = 0.05, see Figure 9-12. When using technology, you will need the area to the left of the critical value that you are trying to find. This would be 1 – \(\alpha\) = 0.95. Then identify the degrees of freedom. The first degrees of freedom are the numerator df, therefore df₁ = 2. The second degrees of freedom are the denominator df, therefore df₂ = 12. Using Excel, we would have =F.INV(0.95,2,12) = 3.8853.

Figure 9-12

Solution

In Excel, use the function F.DIST(x,deg_freedom1,deg_freedom2,cumulative). Always use TRUE for the cumulative. The F.DIST function will find the probability (area) below F. Since we want the area above F we would need to also use the complement rule. The formula would be =1-F.DIST(3.894,3,18,TRUE) = 0.0263.

TI-84: The TI-84 calculator has a built in F-distribution. Press [2^nd] [DISTR] (this is F5: DISTR in the STAT app in the TI-89), then arrow down until you get to the Fcdf and press [Enter]. Depending on your calculator, you may not get a prompt for the boundaries and df. If you just see Fcdf( then you will need to enter each the lower boundary, upper boundary, df₁, and df₂ with a comma between each argument. The lower boundary is the 3.394 and the upper boundary is infinity (TI-83 and 84 use a really large number instead of ∞), then enter the two degrees of freedom. Press [Paste] and then [Enter], this will put the Fcdf(3.894,1E99,3,18) on your screen and then press [Enter] again to calculate the value.

Figure 9-13.

Figure 9-13

Solution

1. The claim is “IQ scores of university students (Group 1) have a smaller variance than IQ scores of community college students (Group 2).”

This is a left-tailed test; therefore, the hypotheses are: \(\begin{aligned}
&H_{0}: \sigma_{1}^{2}=\sigma_{2}^{2} \\
&H_{1}: \sigma_{1}^{2}<\sigma_{2}^{2}
\end{aligned}\).

2. We are using the F-test because we are performing a test about two population variances. We can use the F-test only if we assume that both populations are normally distributed. We will assume that the selection of each of the student groups was independent.

The problem gives us s₁ = 10, n₁ = 28, s₂ = 12, and n₂ = 25.

The formula for the test statistic is \(F=\frac{s_{1}^{2}}{s_{2}^{2}}=\frac{10^{2}}{12^{2}}=0.6944\).

3. The critical value for a left-tailed test with a level of significance \(\alpha\) = 0.05 is found using the invF program or Excel. See Figure 9-16.

Using Excel: The critical value is F_\(\alpha\) =F.INV(0.05,27,24) = 0.5182.

Figure 9-16

4. Decision: Compare the test statistic F = 0.6944 with the critical value F_\(\alpha\) = 0.5182, see Figure 9-16. Since the test statistic is not in the rejection region, we do not reject H₀.

5. Summary: There is not enough evidence to support the claim that the IQ scores of university students have a smaller variance than IQ scores of community college students.

Solution

The populations are independent and normally distributed.

The hypotheses are \(\begin{aligned}
&\mathrm{H}_{0}: \sigma_{1}^{2}=\sigma_{2}^{2} \\
&\mathrm{H}_{1}: \sigma_{1}^{2} \neq \sigma_{2}^{2}
\end{aligned}\)

Use technology to compute the standard deviations and sample sizes. Enter the Sacramento data into list 1, then do 1-Var Stats L1 and you should get s₁ = 163.2362 and n₁ = 17. Enter the Portland data into list 2, then do 1-Var Stats L2 and you should get s₂ = 179.3957 and n₂ = 16. Alternatively, use Excel’s descriptive statistics.

The test statistic is

The p-value would be double the area to the left of F = 0.82796 (Use double the area to the right if the test statistic is > 1).

Using the TI calculator Fcdf(0,0.82796,16,15).

In Excel we get the p-value =2*F.DIST(E8,E7,F7,TRUE) = 0.7106.

Since the p-value is greater than alpha, we would fail to reject H₀.

There is no statistically significant difference between variance of electricity use between Sacramento and Portland.

Excel: When you have raw data, you can use Excel to find all this information using the Data Analysis tool. Enter the data into Excel, then choose Data > Data Analysis > F-Test: Two Sample for Variances.

Enter the necessary information as we did in previous sections (see below) and select OK. Note that Excel only does a one-tail F-test so use \(\alpha\)/2 = 0.10/2 = 0.05 in the Alpha box.

We get the following output. Note you can only use the critical value in Excel for a left-tail test.

Excel for some reason only does the smaller tail area for the F-test, so you will need to double the p-value for a two-tailed test, p-value = 0.355275877*2 = 0.7106.