11-1 TOPIC (11) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION As has been mentioned repeatedly, we use sample information to estimate the unknown information about the population we’re interested in. EXAMPLES 1. Rats raised in an enriched environment appear to have larger cortexes than those raised in unenriched environments. What is the average cortex weight, µ, for the population of cortex weights for every rat which could be raised in an enriched environment? We’ll grow a sample of rats in the enriched environment and use our sample data to estimate µ. 2. Fiddler crabs are known as such because they have asymmetrically sized pincers. Like human handedness, it is estimated that approximately 10% of fiddler crabs are “left-pincer”, i.e. the left pincer is larger than the right. What is the true proportion, π, of left-pincered crabs on a remote island in the South Pacific? We’ll fly to the S. Pacific, take a random sample of crabs and use the sample proportion p to estimate π.11-2 Several ways we could report the results: We could report a POINT ESTIMATE which is a single number from the sample. It has little context or meaning unless additional information is provided. For example, when reporting the sample mean it usually is better to also report something about its sampling variability, such as including the sample size n and the sample standard deviation s. So, Other Possibilities for Reporting: 1. Report x and s. Not useful since s is not the standard deviation of x 2. Report x and nσσx=. Usually can’t because we don’t know the value of σ 3. Report x and nsSEM =. • SEM is STANDARD ERROR OF THE MEAN • SEM is an unbiased estimate of xσ (under random sampling) • one common report is to list SEMx± but it’s not the best choice11-3 We would like to report a range or interval of plausible values for the population parameter we’re estimating. Such a report is called an INTERVAL ESTIMATE of the population parameter. When done right, the interval estimator includes our point estimate and an estimate of the accuracy of the point estimator. The particular method we’ll use is: Defn: A CONFIDENCE INTERVAL for a population characteristic is an interval of plausible values for that parameter. It is constructed so that the true value of the parameter is captured inside the interval with a chosen specified level of confidence. EXAMPLE Political polls are almost always reported as follows: “The proportion of voters who would vote for Bush today is 48% ± 4% with 95% confidence”. What that means is that based on their sample, the true proportion π of voters who would choose Bush is somewhere inside the interval 44% and 52%. The 95% confidence refers to the probability that the interval captures π.11-4 Defn: The CONFIDENCE LEVEL associated with a confidence interval is the probability that the interval estimate covers the true value of the parameter. One way of thinking of it is as the success rate of the method we are using to do the estimation! The method includes the choice of point estimator, the assumptions about the distribution of that estimator, and the sampling design. The confidence level is chosen by the researcher doing the reporting. Common levels in the life sciences are 90%, 95% and 99%. In social sciences you sometimes see 85% as well. When intervals are constructed in a certain way that uses knowledge of the sampling distributions of the point estimators, we can assign these probabilities and not just guess what they might be.11-5 µ • • • • • Important Point: the confidence level is not a probability about a particular interval including the true value. The interval either does or does not cover the true value (but you’ll never know that). The confidence level is a probability that the method you used will give an interval that includes the true value!!!11-6 1) INTERVAL ESTIMATION OF THE POPU-LATION MEAN µ Recall that we have a CLT for the sample mean as well so that if a sample is large enough the sample mean x is Normally distributed with a mean µµx= And a standard deviation nσσx= . ⇒ We should be able to construct a confidence interval estimate of the population mean µ in a way very similar to what we did for the population proportion: i.e. use point estimate ± z × std. dev. of estimate i.e. use nσx 96.1±K for a 95% C.I. for µ. Problem? Almost never know σ either. Since s is an unbiased estimator of σ, can we use nsx 96.1±K instead?11-7 Sort of. If n is really large (> 30) then yes. Other wise we need to use a slightly different C.I. method. Use of z-scores in the above Confidence Intervals is based on the fact that nσµxσµxzxx−=−= has a standard normal distribution when x is normally distributed. 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% 3.2911-8 CONFIDENCE INTERVALS FOR THE POPULATION MEAN µ have the general formula ⎟⎠⎞⎜⎝⎛±nsx value)critical z( where the z critical value is as in the earlier table. This interval is appropriate under the following: 1) sampling is random, and 2) the sample size is large so we can use the CLT11-9 EXAMPLE Suppose we wish to determine if the average cortex weight for rats is larger than usual when the rats are raised in an enriched environment. Assign 10 randomly selected newborn rats (hah!) to an enriched environment and raise them to adult stage. The cortexes had an average weight of 565=xmg and a standard deviation of 170=s mg. Calculate a 90% C. I. for the true mean cortex weight of rat raised in enriched environments. 1) since n is small we should check to see if the frequency distribution of the population sampled is a bell curve. • Can’t since we didn’t raise the entire population! • Can we use the 10 sample values? I.e. do a boxplot and make sure there are no extreme outliers or obvious skew (hard with 10 numbers). Since we weren’t given the raw data we can’t do this either. • It is not unreasonable to assume that weight is normally distributed or at least approximately so. 2) n=10 so n -1 = 9 90% confidence level + 9 df gives t-score = 1.83
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