An investigation of the effectiveness of an antibacterial soap in reducing operating room contamination resulted in the table above. The new soap was tested in a sample of eight operating rooms in the London metropolitan area during the last year and the results are given in the table above.lf we test whether the contamination measurements are lower after using the soap we find the test statistic and critical values are respectively:

An investigation of the effectiveness of an antibacterial soap in reducing operating room contamination resulted in the table above. The new soap was tested in a sample of eight operating rooms in the London metropolitan area during the last year and the results are given in the table above.lf we test whether the contamination measurements are lower after using the soap we find the test statistic and critical values are respectively:

Name: Operating RoomADGHBefore666.59.010.311.28.16.311.6After6.8247.48.58.1
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Description: Operating RoomADGHBefore666.59.010.311.28.16.311.6After6.8247.48.58.16.1342.0An investigation of the effectiveness of an antibacterial soap in reducing operating room contamination resulted in the table above. The new soap was tested in asample of e

Linear Algebra: A Modern Introduction

4th Edition

ISBN:9781285463247

Author:David Poole

Publisher:David Poole

Chapter4: Eigenvalues And Eigenvectors

Section4.6: Applications And The Perron-frobenius Theorem

Problem 22EQ

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I II III n = 5 n = 5 n = 5 M = 1 M = 5 M = 6 N = 15 T = 5 T = 25 T = 30 G = 60 s² = 9.00 s² = 10.00 s² = 11.00 ∑X² = 430 SS = 36 SS = 40 SS = 44 Use an ANOVA with α = .05 to determine whether there are any significant differences among the three treatment means. Source SS df MS F Fcriticalcritical Between treatments Within treatments Total F Distribution Numerator Degrees of Freedom = 6 Denominator Degrees of Freedom = 16 0.01.02.03.04.05.06.07.08.09.010.011.012.0F Conclusion: Reject the null hypothesis; there are no significant differences among the three treatments. Fail to reject the null hypothesis; there are no significant differences among the three treatments. Reject the null hypothesis; there are significant differences among the three treatments. Fail to reject the null hypothesis; there are significant…
I II III n = 5 n = 5 n = 5 M = 1 M = 5 M = 6 N = 15 T = 5 T = 25 T = 30 G = 60 s² = 9.00 s² = 10.00 s² = 11.00 ∑X² = 430 SS = 36 SS = 40 SS = 44 Use an ANOVA with α = .05 to determine whether there are any significant differences among the three treatment means. Source SS df MS F Fcriticalcritical Between treatments Within treatments Total F Distribution Numerator Degrees of Freedom = 6 Denominator Degrees of Freedom = 16 0.01.02.03.04.05.06.07.08.09.010.011.012.0F Conclusion: Reject the null hypothesis; there are no significant differences among the three treatments. Fail to reject the null hypothesis; there are no significant differences among the three treatments. Reject the null hypothesis; there are significant differences among the three treatments. Fail to reject the null hypothesis; there are significant…