Test Statistic and P-value Calculation???

#1
A 1995 survey indicated that 7% of the adolescents in this age group had tried marijuana. Assuming 7% was the true value in 1995, you want to perform a test to determine whether the proportion of kids in this age group that have tried marijuana has changed since 1995. Let α = 0.1.

What is the test statistic?
What is the p value?

I previously calculated the test statistic as 1.282 using a t-table. But I was told this is incorrect.

Help please!!
 
#3
This was the first part of the question, which I got right:

A recent survey on drug use and health measured the prevalence of drug use in the United States. This survey is designed to provide estimates on the use of illicit drugs, alcohol, and tobacco among members of United States households aged 12 and older. In addition, numerous demographic variables like gender,age, race, etc. are recorded. Assume that this is a legitimate statistical survey and that valid inferences for the US population may be drawn from its results.

In the 2007 survey, 90 out of 760 respondents aged 12 to 15 years old indicated that they had tried marijuana. Investigators would like to estimate the true proportion for the population of 12 to 15 year olds in the US that have tried marijuana. To do this, they wish to construct a 90% confidence interval.

Does this help?
 
#4
I'm hung up on this problem because there isn't very much information given about the sample population. I don't know how to go about determining P and Z without sample size.
 

Dason

Ambassador to the humans
#5
Yes that does help. Can you see what information that you just posted in relevant that is necessary to answer the question in the original post?

Also what do you mean "without sample size". They give you the sample size right there in the info you just posted.
 
#6
Ahhh, I think this is starting to make more sense. I misunderstood the question when I first read it. So, if the 90/760 smoked, then then df=759, so could I calculate the test statistic using a t-table?

I get 1.282 when I do this still. Which isn't the correct answer.
 
#8
I used a t-table with degrees of freedom=759 and alpha = 0.1.

I may be closer now... Can I use the point estimate for smokers/non-smokers so calculate the standard deviation?
I got square root of ((p(1-p))/n) = 0.0117.

If I can get the proper test statistic I can figure the rest of it out, but I feel like I'm going in circles right now.
 

Dason

Ambassador to the humans
#10
We typically don't use a t-test for proportions and instead just use a normal distribution using Z = \(\frac{\hat{p} - p_0}{SE(p_0)}\) where \(p_0\) is the value we're comparing to. We typically use this value \(p_0\) when calculating the standard error for the test test statistic instead of using the point estimate.
 
#13
I was using .118 for the denominator.

In that case, the answer should be 5.184, correct?
Which would make the p-value less than one.
So, I would FTR the null.
 

Dason

Ambassador to the humans
#15
I'm guessing by FTR you mean 'fail to reject'? Are you saying you think a p-value < .001 will cause you to fail to reject the null?
 
#16
Excuse me. I would reject the null hypothesis as the p-value is less than the significance level.

(Did my calculations for the test statistic look correct?)
 

Dason

Ambassador to the humans
#17
It's fairly close but I think you have some rounding errors. You don't want to round anything until the final result.