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I am slightly confused regarded the below problem:

I assume you have to do the 2 sample t proportion test, but don't I need two n values, one for each year? I only see the 2007 year, so I don't know what I would use for my $n_1$ value.

Could someone please walk me through the solution to this problem?

According to a study conducted in the United States in 2000, $13.6\%$ of all Americans 18-25 years old used marijuana or hashish. In 2007, a random sample of $1283$ Americans of the same age group was interviewed, out of which $205$ said that they used these drugs. At the $5\%$ significance level, test whether the data provides sufficient evidence to conclude that the proportion of Americans 18-25 years old that used marijuana and hashish in 2007 had changed since the study of 2000.

The possible answers are given below:

a) $H_0:p=0.136;H_a:p\neq0.136;z=2.49;=p-value=0.9872$. There is not enough evidence to conclude that the proportion of Americans 18-25 years old that used marijuana and hashish in 2007 had changed since the study of 2000.

b) $H_0:p=0.136;H_a:p\neq0.136;z=2.49;=p-value=0.0128$. There is sufficient evidence to conclude that the proportion of Americans 18-25 years old that used marijuana and hashish in 2007 had changed since 2000.

c) $H_0:p=0.136;H_a:p\neq0.136;z=2.49;p-value=0.0064$. There is sufficient evidence to conclude that the proportion of Americans 18-25 years old that used marijuana and hashish in 2007 had changed since 2000.

d) $H_0:p=0.136;H_a:p>0.136;z=2.49;p-value=0.0064$. There is sufficient evidence to conclude that the proportion of Americans 18-25 years old that used marijuana and hashish in 2007 had changed since 2000.

e) $H_0:p=0.136;H_ap\neq0.136;z=2.33;p-value=0.0198$. There is sufficient evidence to conclude that the proportion of Americans 18-25 years old that used marijuana and hashish in 2007 had changed since 2000.

Thanks in advance!

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  • $\begingroup$ hmm I think you are smarter than this question. My guess is they want you to do a one-sample T-test and test the hypothesis that the proportion is $13.6\%.$ You are right to be asking what the sampling error of the last study was when they use loaded language like 'to conclude that the proportion has changed since last study' but they're asking you (without having the courtesy to tell you) to assume the last study exactly reflected the population. $\endgroup$ – spaceisdarkgreen Apr 16 '17 at 6:40
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Hints:

  • One or two-sided test?
  • If population proportion $p=0.136$, what is the expected number or proportion from a sample of $1283$? What is the corresponding variance and standard deviation?
  • What is the $z$-value corresponding to $205$ out of $1283$? (subtracting the mean and dividing the difference by the standard deviation)
  • Using a normal approximation to the binomial, from the normal distribution what is the corresponding $p$-value?
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