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A diet pill is given to 9 subjects over six weeks. The average difference in weight (follow up - baseline) is -2 pounds. What would the standard deviation have to be for the 95% T confidence interval to lie entirely below 0? Give your answer to two decimal places.
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95% CI = mean +- 1.96 SD/sqrt(9) < 0 where 95% CI is 95% confidence interval for the true mean and SD/sqrt(9) is the standard error of the mean.
-2+1.96 SD/3 < 0 ? then the SD is the answer?
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