AP statistics help!!?

AIDS

In an effort to curb certain diseases, especially autoimmune (AIDS), San Francisco has a program whereby drug users can exchange used needles for fresh ones. As reported in the Journal of the American Medical Association (January 12, 1994, p. 115), 35% of 5644 intravenous drug users in San Francisco admitted to sharing needles. Is this sufficient evidence to say that the rate of sharing needles has dropped from the pre-needle exchange rate of 66%?


P < .001 so this is very strong evidence that the rate has dropped.

P = .0063 so this is strong evidence of a drop in rate.

P is between .01 and .05 so there is moderate evidence

P is between .05 and .10 so there is some evidence of a drop in rate.

P = .31 so there is no real evidence of a drop in rate.

P < .001 so this is very strong evidence that the rate has dropped.


Hypothesis Test for proportions:

Let X be the number of success in n independent and identically distributed Bernoulli trials, i.e., X ~ Binomial(n, p)

To test the null hypothesis of the form
H0: p = p0, or
H0: p 鈮?p0, or
H0: p 鈮?p0

Assuming that n*p0 > 10 and n * (1-p0) > 10 (some will say the necessary condition here is > 5, I prefer this more conservative assumption so that the approximations in the tail of the distribution are more accurate) then

find the test statistic z = (pHat - p0) / sqrt(p0 * (1-p0) / n)

where pHat = X / n

The p-value of the test is the area under the normal curve that is in agreement with the alternate hypothesis.
H1: p 鈮?p0; p-value is the area in the tails greater than |z|
H1: p < p0; p-value is the area to the left of z
H1: p > p0; p-value is the area to the right of z

If the p-value is less than or equal to the significance level 伪, i.e., p-value 鈮?伪, then we reject the null hypothesis and conclude the alternate hypothesis is true. If the p-value is greater than the significance level, i.e., p-value > 伪, then we fail to reject the null hypothesis and conclude that the null is plausible. Note that we can conclude the alternate is true, but we cannot conclude the null is true, only that it is plausible.

The hypothesis test in this question is:

H0: p 鈮?0.66 vs. H1: p < 0.66

The test statistic is:
z = ( 0.35 - 0.66 ) / ( 鈭?( 0.66 * (1 - 0.66 ) / 5644 )
z = -49.16361

The p-value = P( Z < z )
= P( Z < -49.16361 )
鈮? 0

Since the p-value is less than the significance level of 0.05 we reject the null hypothesis and conclude the alternate hypothesis p < 0.66 is true.