Statistical Sciences 2244A/B Lecture Notes - Lecture 11: Binomial Distribution, Color Blindness, Random Variable
Stats 2244
Hypothesis Tests I
Introduction to P values
The Problem
- If we had a sex biased population (more of one sex), you might expect a higher proportion of
red/green blindness depending on the sex ratio
- ANSWER: B
o Reasonable to predict a lower proportion bc we have a previous reason to expect it to be
less
o males are more likely to have red green color blindness than females – so if we have
fewer males than what is expected, then it can be reasonable to assume P<0.0425
- ANSWER:
o The sample size should make our estimate a better approximation of the parameter
o Larger samples result in sample means that are closer to the population mean (same
thing is true about proportions)
o The key thing about the p hat value is that its a statistic
▪ A statistic is never a definitve value – they just describe samples
â–Ş The value you get for your statistic depends on the sample you take!!
▪ This makes us say that we dont have enough yet
o Remember that this is one sample from the population and its one statistic in a whole
possibility of statistics
o We could have just as easily got a sample with 10 successes out of 200 →this would
have gave us a sample proportion (p hat) of 0.5
o We know that sample size influences the variation in the possible statistic (recall the
formula for the sigma of proportinos)
o When you take a single sample and get a single sample estimate – we have to think
about the sampling variability and how likely we are to get a value like that
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Document Summary
If we had a sex biased population (more of one sex), you might expect a higher proportion of red/green blindness depending on the sex ratio. This is a binomial distribution with 200 trials and a probability of success of 0. 0425. Using r, we can calculate the probability that we get proportion of 0. 02, 4 times (we can also do this ourselves) The likelihood of going into this population and getting a sample where 4 of them are red green colour bind is 0. 02. You don(cid:495)t expect to get anything more than (cid:884)(cid:882) (cid:523)bc looking at the graph, the probability of. We have to have a way to juddge where the values that are likely are location getting those is small: we can use relative position (where you are relative to something else) Including 0, that makes 201: the discrete value that this variable could take on is (cid:882),(cid:883),(cid:884),(cid:885),(cid:886),(cid:887) (cid:884)(cid:882)(cid:882)