ESTIMATION AND TESTING POPULATION PROPORTIONS 1 INTERVAL ESTIMATION OF THE POPULATION PROPORTION π Recall the binomial experiment in which we randomly select individuals from a population and record, for each individual, which of two categories they belong to. One of the two categories is defined as a “Success” and by default the other is a “Failure”. e.g. categories that are exhaustive and mutually exclusive: male vs female succeed vs. fail herbaceous vs. woody trained vs. untrained The population proportion of successes is denoted π and the sample proportion of successes is denoted trialsofnumber successesobservedofnumber ˆ=π. e.g. suppose we are studying ESP and we perform an experiment in which an individual is tested with three different playing cards. Each test consists of a card being held up and their guess as to which card it is. If they do not have ESP, then they have a 1/3 chance (33.0=π) of picking a card correctly. Now, suppose I run this experiment with the person being tested 25 times (n) and they get 10 correct. Then I observed a sample proportion of 40.02510ˆ==π correct responses.ESTIMATION AND TESTING POPULATION PROPORTIONS 2 If the sample size (n) is sufficiently large (both πn and 5)1( ≥−πn), then The sample proportion πˆ is approximately Normally distributed with mean πµπ=ˆ and variance n)1(2ˆππσπ−=. Since we don’t know the population proportion, estimate the unknown variance with ns)ˆ1(ˆ2ˆπππ−= and we check if the sample size is sufficiently large by checking that both πˆn and 5)ˆ1( ≥−πn. e.g. Parents of autistic children are often told that their child is autistic around 1-2 years of age, approximately the same age that children receive their MMR vaccinations (mumps, measles and rubella). As a result some parents claim that the vaccine causes autism. To test this, a study was done to estimate the rate of autism in children who receive the MMR vaccine. In a sample of 8,500 randomly selected children who did receive the MMR vaccine, the proportion with autism was .00282. Can we assume approximate normality? πˆ = n =ESTIMATION AND TESTING POPULATION PROPORTIONS 3 πˆn = )ˆ1(π−n = ns)ˆ1(ˆ2ˆπππ−= = A large sample 95% confidence interval for the population proportion π is n)ˆ1(ˆ96.1ˆπππ−± • Large-sample means that the sampling was done randomly and the sample size is sufficiently large to invoke the Central Limit Theorem. • 1.96 is the z-score, z*, that makes the following statement true: 0.95 = Pr(- z* < Z < + z*). We use this because we are using the CLT which states that sample proportions are normally distributed for large samplesESTIMATION AND TESTING POPULATION PROPORTIONS 4 The formula is easily adapted for other confidence levels. Simply replace 1.96 with the appropriate number from the table below. The z critical values for common confidence levels are: Confidence Level Z critical values 80% 1.28 90% 1.645 95% 1.96 98% 2.33 99% 2.58 99.9% 3.29 e.g. Autistic children. A 95% CI for the true proportion children who have received the MMR vaccine that are autistic is given by 00113.000282.0)0005752.0(96.100282.0)ˆ1(ˆ96.1ˆ±=±=−±nπππ We interpret this to mean that we are 95% confident that the true proportion children who have received the MMR vaccine that are autistic is within the interval (0.00169, 0.00395).ESTIMATION AND TESTING POPULATION PROPORTIONS 5 e.g. A researcher flew to the South Pacific and collected 150 fiddler crabs. For each crab she recorded whether the left or right pincer was dominant and observed that 20 crabs were left-pincered. Calculate a 90% C.I. to estimate the true proportion of left-pincered crabs on the island. 1) Is the sample size large enough to use our method? 2) =−===ns)ˆ1(ˆ and ,133.015020ˆˆππππ 3) 90% Confidence Î z = 4) The 90% C.I. then is 5) What if we had calculated a 95% C.I.? Would it be wider or shorter than the 90% C.I.?ESTIMATION AND TESTING POPULATION PROPORTIONS 6 6) What if she had seen 13.3% based on a sample of 300 crabs? Would the 90% C.I. be wider or shorter than the one based on 150 crabs? Defn: Confidence intervals can be written in the form point estimate ± MARGIN OF ERROR where the margin of error (ME) is the product of the critical value and the standard deviation of the point estimate. Suppose the scientist is planning to repeat the fiddler crab experiment and wants to calculate a 95% confidence interval with a margin of error of no more than 2.5%. How big a sample size should she take in the new experiment? Margin of Error (ME) = 025.0)ˆ1(ˆ96.1 =−nππ From the earlier experiment an estimate for πˆ is 0.133 so we’ll use that.ESTIMATION AND TESTING POPULATION PROPORTIONS 7 Now we need to solve 025.0)133.1(133.96.1 =−n for n. General equation to estimate the needed sampled size for a specified margin of error (ME) when estimating a population proportion is: 2200)1(⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛−=MEznαππ where • 0π is hypothesized as the likely true proportion in the population (if completely unsure use 5.00=π) • 2αzis the z critical value for the desired confidence level (1-α)100% • ME is the desired margin of error (in decimals)ESTIMATION AND TESTING POPULATION PROPORTIONS 8 TESTING THE POPULATION PROPORTION π Let’s walk through one example, put the pieces into a testing procedure for proportions and then use the procedure in another example e.g. autistic children. Some parents claim that the MMR vaccine causes autism. To test this, a study was done to compare the rate of autism in children who receive the MMR vaccine to the known population rate for children who do not receive the vaccine. Among those who did not receive the vaccine, the proportion of children with autism is 0.0021. In a sample of 8,500 randomly selected children who did receive the MMR vaccine, the proportion with autism was .0028. Is this sufficient evidence to indicate that the vaccine is related to autism? 1. Hypotheses: H0: π = HA: π 2. Significance level: α = 3. If the Null Hypothesis is true (which is assumed to be true until proven otherwise), then The distribution of the sample proportion that we get from doing such an experiment, πˆ,
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