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Question given to me:

A drug to reduce blood pressure is administered to n = 300 patients. After 15 minutes, 276 had blood pressure in the normal range.The company wants to claim that the drug is effective for more than 90% of patients. Using a significance level of $\alpha = .05$, determine if they can do this, or if they should make the more conservative claim that the drug is at least 90% effective.

My instructor's "correct" hypotheses:

$campaign 1: H_{0}: \widehat{p}= .9, H_{a}: \widehat{p}> .9, campaign 2: H_{0}: \widehat{p}= .9, H_{a}: \widehat{p}< .9$

My main question is in the title; wouldn't "at least" just be inclusive of 90% and since it is continuous, be essentially the same as "more than"?

Jason
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    In general your null hypothesis should have enough information that you can exactly calculate probabilities (so it should exactly specify parameters) while your alternative hypothesis should be disjoint from your null hypothesis. So the inequalities in a proper test should be strict. – Ian Oct 27 '16 at 18:16
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    That said, there seems to be some disagreement between the symbolic specification of the tests and the text. I have no idea where the "<" came from in the second case. Also, as I said before you can't really have your alternative hypothesis be "the drug is at least 90% effective", it can only be "the drug is more than 90% effective", because it makes no sense for the alternative hypothesis to overlap with the null hypothesis. – Ian Oct 27 '16 at 18:16
  • I had the same idea, I'll have to check if there is a mistake in the question. – Jason Oct 27 '16 at 18:34
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    Not sure I understand the question; If $a$ is at least $b$, then $a\ge{b}$, but if $a$ is more than $b$, then $a>b$. The only difference is that if $a=b$, $a$ is at least $b$ but not more than $b$. – 4yl1n Apr 19 '20 at 17:10

1 Answers1

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Yes. I give a calculus-based explanation.

The (Riemann) integral doesn't care if you remove one point. In fact, it doesn't care if you remove countably many points. The integral will still be the same.

Now in these tests you use the normal distribution, a continuous distribution. The area is computed using an integral, approximated by your tables. It doesn't matter if you have the strict inequality or not, the result will still be the same.

Sean Roberson
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  • Thanks @Sean, however, my professor provided $H_{a}: \widehat{p}> .9$ for "more than" and $H_{a}: \widehat{p}< .9$ for "atleast". This must be a mistake? – Jason Oct 27 '16 at 18:12
  • Yow, I saw that. I feel like your instructor is wrong but I'm not quite confident. This makes me uneasy. – Sean Roberson Oct 27 '16 at 18:22