Point and Interval EstimatesSuppose we want to estimate a parameter, such as p or µ, basedon a finite sample of data. There are two main methods:1. Point estimate: Summarize the sample by a single numberthat is an estimate of the population parameter;2. Interval estimate: A range of values within which, we believe,the true parameter lies with high probability.Example. Suppose I wanted to estimate the mean height of allfemale students at UNC. I took a sample in this class and thesample mean was ¯x = 65.7 (inches). So the obvious thing todo is to take that as an estimate for the population mean. ButI didn’t have to use the sample mean. I could have taken thesample median (66) or the sample mode (also 66). It makessense to ask which is better.1What properties make a good point estimator?1. It’s desirable that the sampling distribution be centered aroundthe true population parameter. An estimator with this prop-erty is called unbiased.2. It’s desirable that our chosen estimator have a small standarderror in comparison with other estimators we might havechosen.The sample mean is exactly unbiased (whereas the sample me-dian may not be), and also, if the true population is normal,the sample mean has a smaller standard error than the samplemedian. Both of these would indicate that the sample mean ispreferable to the sample median as an estimator of the popula-tion mean. However there are other properties that could never-theless make the median preferable (e.g. it’s more resistant tooutliers).2In the case of a binomial proportion, the obvious point estimatoris the sample proportion. For example, consider our exampleabout President Obama’s popularity rating (class 18).In this example, 68% of respondents gave Obama a positive rat-ing after he had been in office for one month (the answer couldbe different if we repreated the poll now). The most natural in-terpretation of this is that 68% or 0.68 is a statistic which servesas an estimator of the true but unknown proportion of peoplewho would have approved of Obama if the whole population hadbeen surveyed. It seems obvious that we would use the sampleproportion as an estimator of the population proportion, but wedon’t have to.3Now let’s turn to interval estimates. The simplest way to intro-duce this is through an example.Example. In a college of 25,000 students, the administrationwould like to know for what proportion of students both parentshad completed college. A sample of 350 students was drawn atrandom and in that sample, 276 of the students said that boththeir parents had completed college.The sample proportion is276350= 0.789 (or 78.9%), so by thesame logic as in the last example, it makes sense to use thatnumber also as an estimator of the population proportion (inthis case the “population” is all 25,000 students at this college).But, how accurate is that?4The idea: Consider the intervalSample Proportion ± 1.96 × Standard Error (∗)Why would this work?5Sample Proportion ± 1.96 × Standard Error (∗)First let’s calculate Pr{−1.96 < X < 1.96} when X is a standardnormal random variable (mean 0, standard deviation 1).From the normal table, for z = 1.96 we have left-tail probability0.9750. For z = −1.96 we have left-tail probability 0.0250. Thedifference is 0.975–0.025=0.95.But given that the sample proportion has an approximately nor-mal distribution, this means that the probability that the sampleproportion lies within 1.96 standard errors of the true mean isalso 0.95. Or in other words, the probability that the interval (*)includes the true mean is 0.95. This is what we mean by sayingthat the interval we calculated is a 95% confidence interval.6A side comment. Earlier in the course, we said that there isa 95% chance that a normal random variable lies within twostandard deviations of the mean (this is part of the empiricalrule first discussed in Chapter 2, because although at that timewe didn’t use the words normal distribution, that’s actually whatthe empirical rule refers to). So why have we now replaced thenumber 2 with 1.96? Actually, 1.96 is more accurate. If werepeat the above normal probability calculations with z = ±2instead of z = ±1.96, the probability becomes .9772 − .0228 =.9544. That’s still quite close to 95%, and 1.96 is quite closeto 2, so in practice, it doesn’t make much difference whetherwe use 2 or 1.96 standard deviations. But at this stage of thecourse, we’re trying to be more precise about things than wewere earlier on, hence the change.7Confidence interval for a population proportionTo construct a confidence interval to measure the proportion p ofa population that has a particular characteristic (e.g. supportersof President Obama):Step 1: Take a sample of size n, calculate ˆp (pronounced p-hat)as the sample proportion of people who have that characteristic(e.g. saying they support Obama in a survey)Step 2: Calculate the standard error SE =rˆp(1−ˆp)n.Note: The formula should really berp(1−p)n, but we don’t knowp, so we use ˆp instead.Step 3: The 95% confidence interval (ˆp−1.96×SE, ˆp+1.96×SE).8Example from text. In one question of the GSS in 2000, 1154people were asked whether they would be willing to pay higherprices to support the environment. 518 said yes.Find a 95% confidence interval for p, the true proportion in thewhole population who would be willing to pay higher prices tosupport the environment.9Step 1: ˆp =5181154= 0.449 to 3 decimal places.Step 2: The standard error isq0.449×0.5511154= 0.0146.Step 3: 1.96 × 0.0146 = 0.029 to 3 decimal places.The 95% confidence interval is (0.420,0.478).In practice, we wouldn’t usually express this to three decimalplaces and simply say that we believe the true proportion ofpeople who support the proposition (i.e. who would be willingto pay higher prices to protect the environment) is between 42%and 48%.10A side comment. In another GSS survey people were askedwhether they would support legislation to force industry to adoptmore environment-friendly policies. This time close to 80% an-swered yes. Yet it seems likely that tighter regulation on industrywill result in higher prices for consumers (this will certainly betrue if you ask the industry representatives). Another case wherethe wording of a question arguably influences the answer to amuch greater extent than standard error calculations indicate.11Sample size conditionFor these calculations to be valid (standard error formula includ-ing ˆp in place