$X = $ the random variable that counts how many of the children in such samples are baseball players. so consider counting the, the, the number of side-effects for each drug among the 40 people. Generally speaking, we test one-sided claims with one-tailed tests. For simple claims about a proportions, like in Question 1, we model the sampling process as a Bernoulli process having a binomial distribution. So let's assume that the FDA gets really mad for this kind of drug if you have more than 10%. So we're using those assumptions, the IID assumptions to create the idea of a population, super population, that has a prevalence of side effects of p a. of course, we cannot, in general you can't know that unless your action is sending more people or going to great pains to actually sample independently from the population you're interested in. $$P(X \geq 9 \mid \rho = 60\%) = P(X = 9) + P(X = 10)$$ In the above calculation, $ P(X = 8) + P(X=9) + P(X= 10) $ are all done with the assumption that $\rho = 60\%$. Substituting in $n = 10$ and $\rho = 60\%$ we finally, actually calculate the p-value: $\text{p-value} = P(X \geq 8 \ \mid \rho = 60\%)$ The p-value is the area of the red bars. If the sample failed to provide statistical significance, for example, if in Question 6 we had $X = 104$, so that $\hat{p} = \dfrac{104}{200} = 52\%$, the p-value would be: $$\text{p-value } = P(X \geq 104 \mid \rho = 51\%) = .4162$$. The assumption that $\rho = 60\%$ is the null hypothesis $H_0$. The IID draw from a population. So this is actually the motivation for the Agresti Coull interval is that most people do 95% intervals and if we take our 1.96 and just round it up to two then and plug it into the score interval we get exactly the Agresti Coull interval. If the sample provided evidence contrary to the claim, like in Question 7 where we had $X = 100$ so that $\hat{p} = \dfrac{100}{200} = 50\%$ we wouldn’t bother to calculate the p-value (since the sample doesn’t support the claim). $= 10C8\, (.60)^8 (.40)^2 $ Quite far out in the tails, remember the three standard evasions covered the majority of the distribution. Video for Question 1 and 2 showing barplot. $+ 10C9\ (60\%)^9 \ (40\%)^{1} $ $\text{(null) } \ H_0: \rho = 55\%$, $\text{p-value} = P(X \geq 10 \ \mid \rho = 55\%)$ $\rho = $ “the true proportion of children in the population who play baseball”. It also is probably the most misunderstood concept in statistics. The Exact Binomial Test. Example. Serve as motivation for creating a confidence interval, as well. $$\text{p-value} = P(X \geq 8 \ \mid \rho = 60\%)$$. $\rho = $ “the true proportion of students in the population who are STEM majors”. Prevalance of side effects. The name of the hypothesis test that we use for this situation is “the exact binomial test“. Data: In a sample of 10 students 2 of them are STEM majors. if n is small then, then, then this term in front of the one, this term in front of the one half gets a little bit bigger and, well, the one half probably hopefully doesn't dominate but, but, but there's more a greater fraction placed on the, on the one half. $\text{(null) } \ H_0: \rho = 49\%$, $\text{p-value} = P(X \leq 3 \ \mid \rho = 49\%)$ Note about the above barplot. [NOISE] [NOISE] [NOISE] Okay, so let's put some context on this. Kinds of test actually think about what you're modeling is random and what your population model is that you're trying to do, because, you know, modeling is a modeling the, the calculations here are very simple. Get 40% off with code "grigorevpc". $= 10C8\, \rho^8 (1 – \rho)^2$ Note. In a formal hypothesis test we write the alternative and null hypotheses at the start of the problem. Statistics, Statistical Hypothesis Testing, Biostatistics. In some sense, the p-value measures how gullible you need to be in order to accept the claim as true based on the data in a sample. They, whenever a drug company says, oh, well, we want to test out this headache medication, they sign up. $+ 0.000766217865410401$ Finally, authors should name the type of hypothesis test that they used. The sample’s proportion $\hat{p}$ of STEM majors is $$\hat{p} = \dfrac{X(\text{our sample})}{n} = \dfrac{8}{10} = 80\%$$ which supports our claim because $80\% > 60\%$. $+ 0.00752286631493849$ $n = $ the size of our sample, so $n = 10$. The normal quantile times the standard error. $ = 0.08316$, which indicates NO statistical significance because the, (p-value = .08316) > ($\alpha = 0.05$). Also, rather than write p-value, authors typically just write p. The authors won’t usually say that the results are statistically significant: the p-value being less than 0.05 indicates that. Different scientific fields will use different levels of significance to determine if the p-value is small or large, and whether the study can be reported as being “statistically significant”. We'll talk about how you can do exact tests for two binomial proportions. And we'll talk about some of the ways. $\text{(claim) } \ H_A: \rho < 49\%$ $+ 10C3\ (49\%)^3\ (51\%)^{9}$ $X = $ the random variable that counts how many of the students in such samples are stem majors. But it's good, I think, whenever you're doing these. or. I want to point out one thing about the that, that the, this is not a numerator here by the way this is this quantity p hat times thing plus 1 half times this thing and then I ran out of space, ten on the next line I put plus or minus. And then we'll talk about comparing two binomial proportions. The above script also outputs the distribution of $X$ to the “R Console” window: The sample’s proportion $\hat{p}$ of baseball players is $$\hat{p} = \dfrac{X(\text{our sample})}{n} = \dfrac{10}{12} = 83.3\%$$ which supports our claim because $83.3\% > 55\%$. So here we have a table where there's 40 total people. $X(\text{our sample}) = 9$. P-value method of hypothesis testing for simple one-sided claims about a proportion. They probably adhere to their medication schedule very very precisely and, and other things like that, they're probably very good takers of medications. Then read through the first two or three Questions and their solutions. 1. $+ 12C0\ (49\%)^0 \ (51\%)^{12} $ There are 2 videos at the end of Question 1. $+ 10C0\ (60\%)^0 \ (40\%)^{10} $ So the scores should seem fairly familiar to you because it's just going to be constructed in the same way. Since the bars are very thin, if we put numbers on top of them, the result will be a mess. It doesn't use the Normal Approximation. Recall the binomial distribution formula: $$P(X = r) = nCr\ \rho^r \ (1 – \rho)^{n-r}$$. It gets more skewed, and because of that, you don't want that point right in the middle. The p-value method of hypothesis testing is probably the most common way to test a claim.