69.10 Lithium. Lithium carbonate is a drug used to treat bipolar mental disorders. The average dose in well-maintained patients is 1.3 mquL with standard deviation mEq/L. A random sample of 25 patients on lithium demonstrates a mean level [E] of 1.4 mquL. Test to see if this m is signicantly higher than that of a well-maintained patient population. Use a two-sided alternative. Make sure you include all the steps of hypothesis testing. The p-value is a probability. Explain to a nonstatistician, what is it the probability of i_n this problem? [Don't use the term null." hypothesis]. 69.1!) Since n=25 is sufficiently ???'i', even if the it's are not normal. the distribution of the xbars should be approx. ????? from the Central Limit Thm. Observations are likely independent. Hypotheses: p=1.3 meqL in the population VS two-sided alternative Ha: p313 meqL in the population In words: Hg: The average dose in this population is p=1.3 mEqiL Ha: The average dose in this population doesn't equal 1.3 meqL in this population. Notice the hypotheses have the words "average\" and "population\" in them. Test Stat and P value: Our test statistic is z = stut DSN'E Two sided podium!\" [|Z| .5 1. 67) = 2 * [1.0475 =?????' display 2*(1-normprobt1.666'i}| ] . 09557'40? Interpretation: If t2 mean lithium dose were m ??????, we would expect to observe a sample mean of C"??? mquL (or something more extreme] about ?????? 96 of the time, by chance. Our data E"??? (re or are not] unusual if the true average is 1.3 mEqIL. Thus the true average in the populauon could be 1.3 mquL. Conclusion: We ???? (Can or cannot} reject the null hypothesis. We fail to reject the null hypothesis. (This is not evidence that the mean lithium dose is 1.3 meq'L. Remember: We CANNOT conclude the true average is 1.3 mquL. ]