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4 years ago

Name and describe the five schedules of control substances. Name a few drugs in each schedule.

Mathematics

1 answer:

zmey [24]4 years ago

6 0

Answer and Step-by-step explanation:

The U.S. Drug Enforcement Administration (DEA) has divided the sustances into five categories schedules, which they are:

Schedule 1 (I) drugs: substances with no accepted medical use so far and a high potential for abuse. This is the most dangerous schedule because they are considered to have a very high potential of severe psychological and physical dependence. Examples: Heroin, LSD, Methylenedioxymethamphetamine (ecstasy)

Schedule 2 (II) drugs: substances with very controlled medical use with a abuse potential very high but less than Schedule 1 drugs. They are considered very dangerous, because they can lead to a severe psychological and physical dependence. Examples: Cocaine

Methamphetamine, Ritalin.

Schedule 3 (III) drugs: substances that are defined as drugs with a moderate to low potential for physical and psychological dependence. Their abuse potential is less than Schedule 1 and 2, but higher than Schedule 4. Examples: Vicodin, Anabolic steroids, Testosterone.

Schedule 4 (IV) drugs: substances with a abuse potential low and their risk of dependence is also low. Examples: Xanax, Valium , Ativan.

Schedule 5 (V) drugs: substances abuse potential lower potential than Schedule 4 (IV) and they are made with limited amounts of some narcotics. They are used for analgesic purposes, antidiarrheal and less serious conditions. Examples: Lomotil, Robitussin

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Answer:

$t=\frac{669-634}{\frac{732}{\sqrt{1700}}}=1.971$

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If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 634 at 10% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=669$ represent the sample mean

$s=732$ represent the sample standard deviation

$n=1700$ sample size

$\mu_o =68$ represent the value that we want to test

$\alpha=0.$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is different from 634, the system of hypothesis would be:

Null hypothesis: $\mu = 634$

Alternative hypothesis: $\mu \neq 634$

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{669-634}{\frac{732}{\sqrt{1700}}}=1.971$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=1700-1=1699$

Since is a two sided test the p value would be:

$p_v =2*P(t_{(1699)}>1.971)=0.049$

Conclusion

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