How do i test my claim, please help

Question

My claim was that more than 70 percent of my high school population drinks alcohol. My population was 60 people of which 46 answered yes.

please help me with my null and alternative hypotheses, test statistic, p-value, and conclusion.

Pleas i need someones help project is due in a day.

thanks---mark----

Answer 1

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 ≥ p0, or

H0: p ≤ 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 ≠ 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 α, i.e., p-value ≤ α, 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 > α, 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 ≤ 0.7 vs. H1: p > 0.7

The test statistic is:

z = ( 0.7666667 - 0.7 ) / ( √ ( 0.7 * (1 - 0.7 ) / 60 )

z = 1.126872

The p-value = P( Z > z )

= P( Z > 1.126872 )

=  0.1298982

Since the p-value is greater than the significance level of 0.05 we fail to reject the null hypothesis and conclude p ≤ 0.7 is plausible.

Consider the hypothesis test as a trial against the null hypothesis.  the data is evidence against the mean.  you assume the mean is true and try to prove that it is not true.  After finding the test statistic and p-value, if the p-value is less than or equal to the significance level of the test we reject the null and conclude the alternate hypothesis is true.  If the p-value is greater than the significance level then we fail to reject the null hypothesis and conclude it is plausible.  Note that we cannot conclude the null hypothesis is true, just that it is plausible.

If the question statement asks you to determine if there is a difference between the statistic and a value, then you have a two tail test, the null hypothesis, for example, would be μ = d vs the alternate hypothesis μ ≠ d

if the question ask to test for an inequality you make sure that your results will be worth while.  for example.  say you have a steel bar that will be used in a construction project.  if the bar can support a load of 100,000 psi then you'll use the bar, if it cannot then you will not use the bar.

if the null was μ ≥ 100,000 vs the alternate μ < 100,000 then will will have a meaningless test.  in this case if you reject the null hypothesis you will conclude that the alternate hypothesis is true and the mean load the bar can support is less than 100,000 psi and you will not be able to use the bar.  However, if you fail to reject the null then you will conclude it is plausible the mean is greater than or equal to 100,000.  You cannot ever conclude that the null is true.  as a result you should not use the bar because you do not have proof that the mean strength is high enough.

if the null was μ ≤ 100,000 vs. the alternate μ > 100,000 and you reject the null then you conclude the alternate is true and the bar is strong enough; if you fail to reject it is plausible the bar is not strong enough, so you don't use it.  in this case you have a meaningful result.  

Any time you are defining the hypothesis test you need to consider whether or not the results will be meaningful.