Medical Marijuana. An issue with legalization of medical marijuana is "diversion," the process in which medical marijuana prescribed for one person is given, traded, or sold to someone who is not registered for medical marijuana use. Researchers S. Sautel et al. study the issue of diversion in the article "Medical Marijuana Use Among Adolescents in Substance Abuse Treatment" . The mean number of days that $120$adolescents in substance abuse treatment used medical marijuana in the last six months was $102.72$. Assume the population standard deviation is $32$days.

a. Find a $95\%$confidence interval for the mean number of days, $\mu$, of diverted medical marijuana use in the last 6 months of all adolescents in substance abuse treatment.

b. Repeat part (a) at a $90\%$confidence level.

c. Draw a graph similar to Fig. $8.6$on page $326$to display both confidence intervals.

d. Which confidence interval yields a more accurate estimate of$\mu$ ? Explain your answer.

Step by step solution

01

Given Information

Two confidence levels $90\%,95\%$are given.

$\sigma =32$

$n=120$

$\stackrel{-}{x}=102.72$

02

Determination of 95% confidence interval of μ

$95\%$confidence interval is given by:

$\left.\overline{x}\pm z_{\frac{\alpha }{2}}\left(\frac{\sigma }{\sqrt{n}}\right)\right)=102.72\pm 1.96\left(\frac{32}{\sqrt{120}}\right)$

role="math" localid="1652802679707" $=102.72\pm 5.7259$

$=\left(96.994,108.446\right)$

03

Determination of 90% confidence interval of μ

From table, $z_{\frac{\alpha }{2}}=1.645$

$90\%$confidence interval is given by

$\stackrel{-}{x}\pm z_{\frac{\alpha }{2}}\left(\frac{\sigma }{\sqrt{n}}\right)=102.72\pm 1.645\left(\frac{32}{\sqrt{120}}\right)$

$=102.72+4.8054$

$=\left(97.915,107.525\right)$

04

Representation of Confidence Level on Graph

Confidence level$90\%$ is shown as:

$95\%$Confidence level is shown as:

05

Determination of confidence interval that give more accurate estimate 

$90\%$ confidence interval has narrower interval. Hence, it gives more accurate estimate of$\mu$than$95\%$

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