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Two research laboratories have independently produced drugs that provide relief to arthritis sufferers. The …first drug was tested on a group of 90 arthritis sufferers and produced an average of 8.5 hours of relief, and a sample standard deviation of 1.8 hours. The second drug was tested on 80 arthritis sufferers, producing an average of 7.9 hours of relief, and a sample standard deviation of 2.1 hours. At the 0.05 level of …significance, does the second drug provide a signifi…cantly shorter period of relief?

This asks for a confidence interval for the difference between 2 means with the standard deviations calculated from the samples. We therefore need to use a one-tailed t-distribution. The standard deviation, SD, for the difference is (with n_1 = # of samples for the first group = 90 and n_2 = 80 for the second)

SD_1,2 = ( SD_1^2/n_1 + SD_2^2/n_2)^1/2 = 0.30

The t-score is thus t = (8.5 - 7.9)/0.30 = 2.0.

With the number of degrees of freedom = (n_1 - 1) + (n_2 - 1) = 168, a table of critical t-values gives a p-value (=probability in the tail of the distribution) of approximately 0.025. Since this is smaller than 0.05, the second drug does provides a shorter period at the 5% level of confidence.

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26 years as professional physical scientist and project manager for elite research company providing academic quality basic and applied research for government and defense industry clients (currently retired).
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- Deployed vibration sensors in Arctic ice floes; analysis of data
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“A Numerical Model for Low-Frequency Equatorial Dynamics” (with Mark A. Cane), J. of Phys. Oceanogr., 14, No. 12, pp. 18531863, December 1984.**Education/Credentials**

MIT, MS Physical Oceanography, 1981
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