Question

Example 4: T-test for comparing experimental data to a "known" value ("Case Olympic athletes cannot have caffeine levels in their urine above concentrations of 12.00 ug/mL. Five measurements of a sample from one athlete gave a mean concentration of 12.12 ug/mL with a standard deviation of 0.09 ug/mL. At the 95% confidence level, does the athlete's caffeine level exceed the acceptable level? (Another way of asking this is, "Does the maximum acceptable value of 12.00 ug/mL fall within the 95% confidence interval")

Answer 1

T-test for one sample mean and Sd. 0:09 Sample mean ł = 12012 ad Sample Standard deviation (SD) Sample size=5 »n Ho ? 120 Null and Alternative Hapo thesus S Hoa u z 12.0 Hii M p 12.0 Nall Alternate This Corrsponds to a two tailed T test Rejection Region d=0:05 Critical value for a two-tailed test is te= 20776 The rejection region is R2 {t: 1t|720776} Statistics to X-llo 12:12 - 12.0 = 2981 - slun 000g/wn It is observed that 1+) = 2.981tc=91976 therefore null hypothesis is Rejected P value is p = 0·0407 Since p= 0.0407 20:05 Therefore null hypothesis is rejected Conclusion: The null hypothesis is rejected. Therefore the population mean ll is different than 12:0, at 0.05 Significance level t stats: 2.98) 9. Pvalue=0.0407 0-05- 40 V50