My question is as follows:

*An ad claims people prefer S coffee over P coffee. The person randomly samples 80 coffee drinkers and finds 43 prefer S coffee. The person concludes the claim is probably not true. How did the person justify this?

(a) 46% is not in the 95% CI (b) 46% is in the 95% CI (c) 50% is not in the 95% CI (d) 50% is in the 95% CI*

I set the Null Hypothesis as S =.5 (there is no difference in the people who like S coffee over P coffee. Therefore Ha = S > 0.5. (more people prefer S)

I compute the Z value using the proportions formula for z and get $z = 0.53$. I look this up in the normal z chart for a right-tailed test and find p-value to be about 30%. I would accept the Null based on such a high p-value. In other words I would think the ad's claim is not true. HOWEVER, it would be based on the z value, or corresponding p-value, and not any of the items listed in the question about 43% or 50% being in or out of the 95% CI.

What am I missing here?

  • 1
    $\begingroup$ Please add the self-study tag, and read the policy on this sort of question in the help center, plus the self-study tag-wiki. $\endgroup$ – Glen_b -Reinstate Monica Feb 27 '17 at 22:32
  • $\begingroup$ Curiously, the conclusion is not justified solely on the evidence given here, which leans (slightly) in favor of the claim because over half the sample preferred $S$. That contradicts the assertion the claim is "probably not true." One could state, with much better justification (albeit with a conceptually challenging triple negative!), that this sample of coffee drinkers was not large enough to disprove the counterclaim that people do not prefer $S$ over $P$. $\endgroup$ – whuber Feb 27 '17 at 23:17

You're correct that up until the point that that the answer options are offered, the question is framed as a hypothesis test, so that's the most obvious way to approach that set-up.

However, at the point you see the answer options it's clear that the question is asking you to look at the confidence interval for the proportion (to see if it contains the case specified under the null) rather than conduct the hypothesis test.

The confidence interval could be thought of as a set of values for the parameter that are (in a very particular sense) "consistent" with the data-set.

As soon as you see those options, it should be clear that if you don't think in terms of the confidence interval, you'll have trouble answering the question.

An aside that doesn't impact what you need to do here: The author(s) of the question seem to implicitly assume that the two procedures would yield the same conclusion, but in in the most commonly taught tests vs intervals for proportions it's not necessarily the case that they will yield the same conclusion. They nearly always give the same conclusion (and they do here), but it's not always so. It's quite possible to choose a different pair of procedures for this situation that will always correspond, though. In almost every other situation you'd learn in an elementary course, they always correspond. Many books and web-pages explicitly state that they always correspond (see here for one example of many; it's wrong on that point but you might find the explanations otherwise useful).

  • $\begingroup$ Glen, thanks, however how would you answer this question? $\endgroup$ – user163862 Feb 27 '17 at 22:21
  • $\begingroup$ I'm not going to explicitly say which option to choose -- see our policy in the help center under Homework (NB it doesn't matter if this is actually homework or something else - exam practice, your own study, etc, the policy still applies to explicitly answering that question -- we don't do that). However, you should be able to figure out the answer (without even needing to actually compute the interval, in fact). All the information (not already covered by the question) is already there in my answer; $\endgroup$ – Glen_b -Reinstate Monica Feb 27 '17 at 22:29
  • $\begingroup$ The book, which is often wrong lists (D) as the answer. IT would seem to meth at 50% would always be in the CI. So, unfortunately, I have learned nothing here. $\endgroup$ – user163862 Feb 28 '17 at 6:15
  • $\begingroup$ D is correct (and is the answer quite directly suggested by my words "see if it contains the case specified under the null" -- how much more explicit could I have been there, other than saying "choose D"?) Did you try calculating a CI for the sample proportion? I suspect you have an error in your thinking in relation to how that works. $\endgroup$ – Glen_b -Reinstate Monica Feb 28 '17 at 6:20
  • $\begingroup$ In particular, what's your CI if it had been 49 prefer S (out of 80)? $\endgroup$ – Glen_b -Reinstate Monica Feb 28 '17 at 6:26

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