8.2 2.13.2018

Activity Everybody grab a die You’re going to roll the die 40 times Counting what proportion of the time you get a 1 or a 6 So if you get 5 ones and 7 sixes, then your proportion is 12/40, or 0.3

Activity This proportion is your point estimate ( 𝑝 ) Let’s check our conditions for normality Since this is a proportion np≥10 40∗ 𝑝 ≈13.3 N(1-p)≥10 40∗(1− 𝑝 )≈26.7 We often will need to check independence, but since out population of dice rolls is infinitely large, this is not an issue in this case

Activity Because it is normal, we can calculate the standard deviation In this case, let’s pretend that we don’t know the exact value for p in the population In this case, we could calculate it, but in most cases we won’t know it 𝑝(1−𝑝) 𝑛 𝑝 (1− 𝑝 ) 𝑛 ≈.0745

Activity Now we’re ready to construct a confidence interval for the proportion of times we get a 1 or a 6 Remember the general form for a confidence interval: Let’s start with a 95% confidence interval The “critical value” means how many standard deviations away from the mean we need to go to capture that much area So for 95% confidence interval, it is 2 standard deviations

Activity So we plug in 𝑝 for our statistic (point estimate) We plug in 2 for our critical value And the standard error for the standard deviation Everybody’s answer will be a little bit different .333± 2 .0745 .333±.149 This is one way to write my interval Or .184--.482 Or (.184, .482) So I am 95% confident that the true proportion is between .184 and .482

Activity The true value in the population is 1/3 or .333333 Did anybody get an interval that did not include this value? Meaning that .3333 is OUTSIDE of your interval On average, we should get one person (ish)

Activity So we are 95% confident that the true proportion is between .184 and .482 This is a pretty wide range As discussed last class, we have 2 options to make this range narrower: Increase sample size (roll more than 40 times) Decrease the confidence level Let’s do option #2

Activity Now we want an 80% confidence interval The point estimate and the standard error do not change But the critical value does We want to know how many standard deviations I need to go away from the mean (of the sampling distribution) to capture 80% of the data Any thoughts on how we can do that?

Activity We can use Table A or InvNorm

Activity So our critical value now is 1.28 .333± 1.28 .0745 .333±.095 .333± 1.28 .0745 .333±.095 ( .238, .428) Narrower than before We are 80% confident that the true value for p is between .238 and .428 Did anyone get an interval that did not contain .333? Now try a 99% confidence interval

Activity .333± 2.576 (.0745) .333±.192 .141--.525 Yours will be different Did anyone get an interval that did not contain .333?

Formal Definition

Using our Calculator We can also ask our calculator to do these for us HOWEVER, you need to be ABLE to do them by hand Because you might be asked a questions like “what is the critical value for this interval?” or “what is your standard error?” STAT—TESTS-- “1-PropZInt…” (NOT “1-PropZTest”) Let’s replicate our 99% interval X=# of successes N=sample size C-Level= confidence level So we will use .99 as our confidence level

An Example The EPA recommends taking remedial action in any home that has radon levels above 4 picocuries per liter (pCi/L). A random sample of 200 houses in Pennsylvania revealed that 127 of them had radon levels above 4 pCi/L. Construct a 95% confidence interval for the proportion of houses that have radon levels above 4 pCi/L

An Example The EPA recommends taking remedial action in any home that has radon levels above 4 picocuries per liter (pCi/L). A random sample of 200 houses in Pennsylvania revealed that 127 of them had radon levels above 4 pCi/L. Construct a 95% confidence interval for the proportion of houses that have radon levels above 4 pCi/L, then interpret this interval in context (.56828, .70172) “We are 95% confident that the true proportion of houses in Pennsylvania with high radon levels is between 0.56828 and 0.70172” OR “We are 95% confident that the interval 0.56828 to 0.70172 captures the true proportion of houses in Pennsylvania with high radon levels”

Determining the Correct Sample Size We may sometimes want a particular margin of error and a particular confidence level And choose our sample size in order to get this margin of error Because we have not taken the sample yet, we don’t even know 𝑝 yet But we do know the critical value

Determining the Correct Sample Size In order to be conservative, we will assume that 𝑝 =0.50 This will always give us a big enough sample, regardless of what p-hat actually is Because 𝑝 ∗ 1− 𝑝 is maximized at 0.5

Example Let’s think back to our Pennsylvania example. Let’s say we wanted to do the same study in Colorado. Assume that the proportion of houses with radon in Pennsylvania tells us nothing about the proprtion of houses in Colorado. We want a 95% confidence interval with a margin of error of 5% (.05) or less ME=(Crit value)(St error) .05= 1.96 ( .5 .5 𝑛 ) Solve for n N=384.16 So we need a sample of at least 385 people

You try You want to estimate the true proportion of students at a certain school that have a tattoo with 95% confidence and a margin of error of at most 0.10. How many people should be surveyed?

You try You want to estimate the true proportion of students at a certain school that have a tattoo with 95% confidence and a margin of error of at most 0.10. How many people should be surveyed? At least 97 people