### Learning Outcomes

Discuss two provides for the F distribution: one-means ANOVA and the test of two variances

Here are some facts about the F circulation.

You are watching: Which of the following is not a property of the f distribution?

The curve is not symmetrical yet skewed to the appropriate.Tbelow is a different curve for each collection of dfs.The F statistic is better than or equal to zero.As the levels of liberty for the numerator and for the denominator gain bigger, the curve approximates the normal.Other uses for the F distribution encompass comparing two variances and two-way Analysis of Variance. Two-Way Analysis is past the scope of this chapter. ### try it

MRSA, or Staphylococcus aureus, deserve to cause a significant bacterial infections in hospital patients. This table reflects assorted colony counts from various patients who might or may not have MRSA.

Conc = 0.6Conc = 0.8Conc = 1.0Conc = 1.2Conc = 1.4
916223027
6693147199168
9882120148132

Plot of the information for the various concentrations: Test whether the expect number of colonies are the same or are different. Construct the ANOVA table (by hand or by utilizing a TI-83, 83+, or 84+ calculator), uncover the p-value, and also state your conclusion. Use a 5% meaning level.

While tbelow are distinctions in the spreads in between the groups, the differences carry out not appear to be big enough to reason worry.

We test for the equality of expect variety of colonies:

H0 : μ1 = μ2 = μ3 = μ4 = μ5Ha: μiμj some ij

The one-means ANOVA table results are displayed in below.

Source of VariationSum of Squares (SS)Degrees of Freedom (df)Mean Square (MS)F
Factor (Between)10,2335 – 1 = 4displaystylefrac10,2334=2,558.25displaystylefrac2,558.254,194.9=0.6099
Error (Within)41,94915 – 5 = 10
Total52,18215 – 1 = 14displaystylefrac41,94910=4,194.9 Distribution for the test: F4,10Probability Statement: p-value = P(F > 0.6099) = 0.6649.

Compare α and also the p-value: α = 0.05, p-worth = 0.669, α Sorority 1Sorority 2Sorority 3Sorority 42.172.632.633.791.851.773.783.452.833.254.003.081.691.862.552.263.332.212.453.18

Using a definition level of 1%, is there a distinction in expect qualities among the sororities?

Solution:

Let μ1, μ2, μ3, μ4 be the population suggests of the sororities. Remember that the null hypothesis clintends that the sorority groups are from the same normal distribution. The alternate hypothesis claims that at least 2 of the sorority groups come from populaces with different normal distributions. Notice that the four sample sizes are each 5.

Note

This is an example of a well balanced design, bereason each aspect (i.e., sorority) has the same variety of observations.

H0: μ1 = μ2 = μ3 = μ4

Ha: Not all of the suggests μ1, μ2, μ3, μ4 are equal.

Distribution for the test: F3,16

wbelow k = 4 teams and n = 20 samples in total

df(num)= k – 1 = 4 – 1 = 3

df(denom) = nk = 20 – 4 = 16

Calculate the test statistic: F = 2.23

Graph: Probability statement: p-worth = P(F > 2.23) = 0.1241

Compare α and the p-value: α = 0.01

p-value = 0.1241

α Using a Calculator

Placed the data right into lists L1, L2, L3, and L4. Press STAT and also arrow over to TESTS. Arrow down to F:ANOVA. Press ENTERand Enter (L1,L2,L3,L4).

The calculator displays the F statistic, the p-worth and also the worths for the one-way ANOVA table:

F = 2.2303

p = 0.1241 (p-value)

Factor df = 3

SS = 2.88732

MS = 0.96244

Error df = 16

SS = 6.9044

MS = 0.431525

### attempt it

Four sporting activities groups took a random sample of players regarding their GPAs for the last year. The results are shown below:

GPAs for Four Sports Teams

3.62.14.02.0
2.92.62.03.6
2.53.92.63.9
3.33.13.22.7
3.83.43.22.5

Use a definition level of 5%, and recognize if tright here is a distinction in GPA among the teams.

With a p-worth of 0.9271, we decrease to disapprove the null hypothesis. There is not sufficient proof to conclude that tright here is a distinction among the GPAs for the sporting activities groups.

Using a Calculator

To calculate the p-value:

Press 2nd DISTRArrow down to Fcdf(and also pressENTER.Get in 0.134, E99, 2, 12)Press ENTER

The p-worth is 0.8759.

### try it

Another fourth grader likewise flourished bean plants, yet this time in a jelly-favor mass. The heights were (in inches) 24, 28, 25, 30, and 32. Do a one-method ANOVA test on the four groups. Are the heights of the bean plants different? Use the same strategy as shown in Example 2.

F = 0.9496p-value = 0.4402

From the sample information, the proof is not enough to conclude that the intend heights of the bean plants are various.

## References

File from a fourth grade classroom in 1994 in a personal K – 12 school in San Jose, CA.

Hand also, D.J., F. Daly, A.D. Lunn, K.J. McConmeans, and also E. Ostrowski. A Handbook of Small Datasets: Documents for Fruitfly Fecundity. London: Chapmale & Hall, 1994.

Hand, D.J., F. Daly, A.D. Lunn, K.J. McConmeans, and also E. Ostrowski. A Handbook of Small Datasets. London: Chapmale & Hall, 1994, pg. 50.

Hand, D.J., F. Daly, A.D. Lunn, K.J. McConmethod, and E. Ostrowski. A Handbook of Small Datasets. London: Chapman & Hall, 1994, pg. 118.

“MLB Standings – 2012.” Available digital at http://espn.go.com/mlb/standings/_/year/2012.

Mackowiak, P. A., Wasserman, S. S., and Levine, M. M. (1992), “A Critical Appraisal of 98.6 Degrees F, the Upper Limit of the Regular Body Temperature, and Other Legacies of Carl Reinhold August Wunderlich,” Journal of the Amerihave the right to Medical Association, 268, 1578-1580.

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## Concept Review

The graph of the F distribution is constantly positive and skewed ideal, though the shape have the right to be mounded or exponential relying on the combination of numerator and denominator levels of freedom. The F statistic is the ratio of a measure of the variation in the team means to a comparable measure of the variation within the teams. If the null hypothesis is correct, then the numerator should be little compared to the denominator. A tiny F statistic will result, and also the area under the F curve to the right will be big, representing a big p-worth. When the null hypothesis of equal group suggests is incorrect, then the numerator have to be huge compared to the denominator, offering a large F statistic and a small area (little p-value) to the ideal of the statistic under the F curve.

When the data have unequal team sizes (unwell balanced data), then methods need to be offered for hand calculations. In the situation of well balanced information (the teams are the same size) however, streamlined calculations based upon group implies and also variances might be used. In exercise, of course, software is usually employed in the evaluation. As in any evaluation, graphs of miscellaneous sorts must be used in conjunction through numerical techniques. Almeans look of your data!