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Ber [7]
2 years ago
5

8.36 Football and Brain Size: Randomization Test Exercise 8.35 describes a study examining hippocampus volume in the brain betwe

en controls who have never played football, football players with no history of concussion, and football players with a history of concussion. The data are available in FootballBrain. Use StatKey or other technology to conduct an ANOVA randomization test to determine whether there is a significant difference in mean hippocampus volume between the three categories. What is the F-statistic for the observed data
Mathematics
1 answer:
erma4kov [3.2K]2 years ago
7 0

Answer:

Since the calculated F=  2.8657 does not lie in the critical region  F (2,2)≥ 19 we conclude that there no difference between the means.

Since the calculated F=  205.0819  lies in the critical region F(1,2) ≥ 18.51  we conclude that there is a difference between the standard deviations of the three groups.

The F- statistic for the observed data is 0.101  as calculated by F- test in excel.

Step-by-step explanation:

To conduct an ANOVA randomization test we perform anova without replication in excel.

The data for the test is

   

                            Mean    Standard Deviation Sample Size

Never

Played Football 7602.5          1074                   25

FB with no History   6459.2         779.9                  25

<u>FB with Concussion 5734.6       593.4                   25 </u>

<u>Total                 6598.8      1133.5                  75 </u>

<u />

Anova: Two-Factor Without Replication    

   

SUMMARY Count Sum Average Variance

Row 1 2 8676.5 4338.25 21310656.13

Row 2 2 7239.1 3619.55 16127224.25

Row 3 2 6328 3164 13215968.72

   

Column 1 3 19796.3 6598.766667 886871.7433

Column 2 3 2447.3 815.7666667 58708.90333

<u>ANOVA Table </u>  

Source           Sum of     Degrees     Mean

of Variation      Square    of freedom  Squares

<u>                             SS     df          MS              F P-value </u>

B/w Rows 1401945.703 2 700972.8517 2.8657 0.259

B/w Col      50164633.5 1 50164633.5 205.0819  0.00484

<u>Error     489215.59         2 244607.795                           </u>

Total 52055794.79 5    

Set up the null and alternate hypotheses :

H0 ; there's no difference between the means and

there's no difference between the standard deviations

against the claim

Ha: there's a difference between the means and

there's  a difference between the standard deviations

The significance level is chosen at 0.05

The test statistics to use are

F1= estimated variance from Between Means/  estimated variance from Error SS

F1= s1²/s3²

F1= estimated variance from Between Standard Deviations /  estimated variance from Error SS

F2- s2²/s3²

with υ1= (2,2)  and υ2(1,2) degrees of freedom.

The critical regions for F (2,2)≥ 19 for 0.05 significance level

and F(1,2) ≥ 18.51  for 0.05 significance level

Since the calculated F=  2.8657 does not lie in the critical region  F (2,2)≥ 19 we conclude that there no difference between the means.

Since the calculated F=  205.0819  lies in the critical region F(1,2) ≥ 18.51  we conclude that there is a difference between the standard deviations of the three groups.

The F- statistic for the observed data is 0.101  as calculated by F- test in excel.

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Step-by-step explanation:

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substitute the values

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tatiyna
<h2>Put x = \dfrac{39}{8} I write the sum of 6x and 2x is at least 39 .</h2>

Step-by-step explanation:

We have,,

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