Determining Null Hypothesis and use of Confidence Interval My statistics test review question asks the following:
"An ad claims that more people prefer S coffee to P coffee. A random sample of 90 coffee drinkers is conducted and out of these 90, 48 of them like S better.  However, the survey person concludes that the ad's claim is 'probably false'.  How did the surveyor justify this? "
The choices all have to do with whether or not 47% is or is not in the Confidence interval or whether 50% is or is not in the Confidence Interval.
My first challenge with this test his how even to determine the Null Hypothesis.   I thought that the null would be"status quo", which in this case would seem to be that people like S about the same as P, that Ho is $S=P$ and that the alternative would be Ha is $S>P$. In other words, I would think the null would be about the same number of people prefer S and P, and the alternative would be S>P.
However, if I'm trying to disprove a claim, don't I need to have the Null BE the claim?  Does that work even in proportions?
Finally, I have read in my class text that I can't use a CI unless it is a two-sided test, and this is clearly a right-tailed test.
I would find the z value by using $z=phat-pnull/((\sqrt(pnull)\cdot(1-pnull)/90)$ and then evaluate if statistically significant (outside the CI).
Accordingly, NONE of the answers seem to make sense since I would be evaluating the z value NOT some ratio.  
What am I missing about this question?  Am I reading in too much?
 A: The question is not clear, but I will show a one-sided hypothesis test
and a couple of two-sided confidence intervals. None of these procedures supports the claim that
more people prefer S than prefer P.
You have $n = 90$ subjects, presumably chosen at random from among
coffee customers. Of these $X = 48$ prefer S. So the estimated
proportion preferring S is $\hat p = 48/90 =  0.5333.$
One-sided test. To test the null hypothesis $H_0: p = .5$ against the alternative
$H_a: p > .5,$ one can use the test statistic
$$Z = \frac{\hat p - p_0}{\sqrt{p_0(1-p_0)/n}} 
= \frac{.5333 - .5}{\sqrt{.5(.5)/90}} =  0.6325.$$
One would reject $H_0$ if $Z > 1.645,$ which is not true.
So the data are consistent with $H_0$ and the claim that $p > .5$
is not supported by the data.
Wald confidence interval. One kind of 95% confidence interval for $p$ is of the form
$$\hat p \pm 1.96\sqrt{\frac{\hat p(1-\hat p)}{n}},$$
which computes to $0.5333 \pm 0.1031$ or $(0.4303, 0.6364).$
It contains $0.5,$ and so, again here, data are consistent
with S and P being equally popular.
Agresti confidence interval. A somewhat more accurate 95% CI uses $\tilde p = (X+2)/(n+4)$
and $\tilde n = n+ 4.$ Using the formula
$$\tilde p \pm 1.96\sqrt{\frac{\tilde p(1-\tilde p)}{\tilde n}},$$
it gives the 95% CI $(0.4310, 0.6328),$ which also includes $0.5$
and leads to the same interpretation as above.
One may quibble about whether to use a one- or two-sided test,
or about the style of CI to use. But $n = 90$ observations are
not nearly enough to settle which brand of coffee is more popular
in the general population based on a sample that has the two brands
so closely matched. 
[Note: In order to distinguish meaningfully between
$p = .533$ and $p = .50,$ a sample of more than 900 subjects would be
required. The rule of thumb for that would be $n \approx 1/.033^2 > 900.$]
