Mathematics, 06.05.2020 05:33 zah16

A recent poll of 2000 randomly selected 18-25-year-olds revealed that 232 currently use marijuana or hashish. According to a publication, 12.5 % of 18-25-year-olds were current users of marijuana or hashish in 1997.
a) Do the data provide sufficient evidence to conclude that the percentage of 18-25-year-olds who currently use marijuana or hashish has changed from the 1997 percentage of 12.5%? Use α=0.05 significance level.

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Answer from: DASASDAEDWEDA

$z=\frac{0.116 -0.125}{\sqrt{\frac{0.125(1-0.125)}{2000}}}=-1.217$

We are conducting a bilateral test then the p value would is:

p_v =2*P(z

Since the p value obtained is higher than the significance level used of 0.05 we have enough evidence to FAIL to reject the null hypothesis and there is no evidence to conclude that percentage of 18-25-year-olds who currently use marijuana or hashish has changed from the 1997 percentage of 12.5%

Step-by-step explanation:

Information given

n=2000 represent the random sample of people

X=232 represent the people between 18-25-year-old who currently use marijuana or hashish

$\hat p=\frac{232}{2000}=0.116$ estimated proportion of people between 18-25-year-old who currently use marijuana or hashish

p_o=0.125 is the value that we want to verify

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic

p_v represent the p value

Hypothesis to test

We want to verify if the percentage of 18-25-year-olds who currently use marijuana or hashish has changed from the 1997 percentage of 12.5%.:

Null hypothesis: p=0.125

Alternative hypothesis: $p \neq 0.125$

The statistic for a proportion z test is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

Replacing into the previous formula we got:

$z=\frac{0.116 -0.125}{\sqrt{\frac{0.125(1-0.125)}{2000}}}=-1.217$

P value

We are conducting a bilateral test then the p value would is:

p_v =2*P(z

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A recent poll of 2000 randomly selected 18-25-year-olds revealed that 232 currently use marijuana or...