Stats Homework: Remission Probability

Jul 20th, 2016
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Let X be a random variable representing the remission time in weeks for all patients using the new drug. Assume that the distribution of x is normal. A previously used drug treatment has a mean remission time of 12.5 weeks.

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Remission Time (Weeks)Remission Time (Weeks)1073223226161120196MeanStandard ErrorMedianModeStandard DeviationSample VarianceKurtosisSkewnessRangeMinimum17Maximum35Sum6103432251396Count17.095242.1820751669.99952499.99048-1.002730.5531862963535921x= raw score to be convertm= mean of the raw scoresSD= standard deviation of tTscore= 10x / SD + (50 - 10x= 12.5m= 17.0952SD= 9.9995Tscore= 10(12.5) / 9.9995 +Let X be a randomGiven:Population Mean (: 12.5 (The mean reDoes the data indicTwo tail test:ALTENATIVE HY(The mean remisstime using the new12.5)NUL HYPOTHSISwe should use the tis unknown.Test statistics:Degree of freedomThe critical region:Compare the testThe critical regionThus, we don rejecOr:The P-Value is .04( do not reject the nAt the 1% level ox= raw score to be convertedm= mean of the raw scoresSD= standard deviation of the raw scoresTscore= 10x / SD + (50 - 10m / SD)m= 17.0952SD= 9.9995Tscore= 10(12.5) / 9.9995 + (50 - 10(17.0952 / 9.9995)Let X be a random variable representing the remission time in weeks for all patients using the nePopulation Mean (: 12.5 (The mean remission)Does the data indicate that the mean remission time using the new drug is different from 12.5 wTwo tail test:ALTENATIVE HYPOTHESIS:(The mean remission (time using the new drug is different from 12.5 (NUL

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