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Testing Differences in Population Variances

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Testing Differences in Population Variances

QSCI 381 – Lecture 42

(Larson and Farber, Sect 10.3)

- We used different tests when comparing means depending on whether we could assume that the population variances for the two populations were the same.
- Today we identify a test which can be used to test for differences between two population variances.

- Let and represent the sample variances of two populations. If both populations are normal and the population variances and are equal, then the sampling distribution of:
is called an .

F-distribution

- Properties of the F-distribution:
- Positively skewed.
- The curve is determined by the degrees of freedom for the numerator and that for the denominator.
- The area under the curve is 1.
- The mean value is approximately 1.
- F-values are always larger than 0.

- A two-sample F-test is used to compare two population variances and when a sample is randomly selected from each population. The populations must be independent and normally distributed. The test statistic is:
- where and represent the sample variances with . The degrees of freedom for the numerator is d.f.N=n1-1 and the degrees of freedom for the denominator is d.f.D=n2-1.

- Specify the level of significance .
- Determine the degrees of freedom for the numerator, d.f.N.
- Determine the degrees of freedom for the denominator, d.f.D.
- Determine whether this a one-tailed or a two-tailed test.
- One-tailed – look up the F-table for d.f.N and d.f.D.
- Two-tailed – look up the /2 F-table for d.f.N and d.f.D.

One-tailed

Two-tailed

F0=2.901

F0=3.576

In EXCEL:

FINV(prob,dfN,dfD)

d.f.N=5; d.f.D=15; =0.05

- We sample two populations. The sample variances for the two populations are 9.622 (n1=46) and 10.352 (n2=51). Test the claim that the two variances are equal (=0.1).
- H0: ; Ha:
- Determine the critical value and the rejection region.
- This is a two-tailed test.
- We reverse the order of samples 1 and 2 so that . Therefore d.f.N=50; d.f.N=45.
- The critical value is F(0.05,50,45) = 1.626.

- The test statistic is:
- We fail to reject the null hypothesis

30 fish are sampled from a Marine Reserve and a fished area. Test the claim that the lengths in the Reserve are more variable than those in the fished area (assume =0.05). The data are:

- H0: ; Ha:
- =0.05; d.f.N=29; d.f.D=29. This is a one-sided test so the rejection region is F>1.861 = FINV(0.05,29,29)
- The test statistic is:
- We reject the null hypothesis (the data provide support for the claim)

- When and are the variances of randomly selected, independent samples from normally distributed populations, a confidence interval for is:
where FL is the left-tailed critical F-value and FR is the right-tailed critical F-value (based on probabilities of /2).

- Find a 95% confidence interval for for example A.
- The lower and upper critical points for the F-distribution are computed:
- FINV(0.975,50,45)=0.564
- FINV(0.025,50,45)=1.788

- The 95% confidence interval is given by: