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Testing Differences in Population VariancesPowerPoint Presentation

Testing Differences in Population Variances

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Recap – Testing Means

- 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.

The F-distribution-I

- 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

The F-distribution-II

- 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.

The Two-Sample F-test for Variances-I

- 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.

The Two-Sample F-test for Variances-II(Finding the rejection region for the test)

- 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.

The Two-Sample F-test for Variances-III(Finding the rejection region for the test)

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

Example-A-I

- 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.

Example-A-II

- 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:

Example-B-IExample-B-II Test the claim that the lengths in the Reserve are more variable than those in the fished area (assume

- 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)

Confidence intervals for Test the claim that the lengths in the Reserve are more variable than those in the fished area (assume

- 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).

Confidence intervals for Test the claim that the lengths in the Reserve are more variable than those in the fished area (assume (Example-1)

- 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

Confidence intervals for Test the claim that the lengths in the Reserve are more variable than those in the fished area (assume (Example-2)

- The 95% confidence interval is given by:

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