Confidence Intervals for Population Means: Understanding Interpretation and Precision, Lab Reports of Statistics

Download Confidence Intervals for Population Means: Understanding Interpretation and Precision and more Lab Reports Statistics in PDF only on Docsity! Lab 10: Confidence intervals STT 421: Summer, 2004 Vince Melfi Confidence intervals provide a way to report an estimate of a population parameter (such as a mean µ or a proportion p) along with some information about the estimate’s precision. Although different settings lead to different formulas for computing confidence intervals, the basic interpretation is always the same. And understanding the interpretation is most important. In this lab we will investigate confidence intervals for the population mean µ in the most basic setting, where the population standard deviation σ is known. Our main focus will be understanding what we can and cannot claim based on a confidence interval. Computing a confidence interval The file u:\msu\course\stt\421\summer04\cipop.dat contains three variables, pop1, pop2, and pop3. We’ll treat each of these as the population of interest. Our goal is estimation of the population mean. Populations pop1 and pop2 are normally distributed, so the confidence interval procedures should work well for any sample size n. Population pop3 is skewed, so it will take a large sample size n to get accurate confidence levels. First we’ll read in the data and compute 95% and a 99% confidence intervals for the means of pop1 and pop2 based on a sample of size n = 10. data cipop; infile ’u:\msu\course\stt\421\summer04\cipop.dat’; input pop1 pop2 pop3; proc surveyselect data=cipop n=10 rep=1 out=ci1; id pop1; proc surveyselect data = cipop n=10 rep=1 out=ci2; id pop2; proc means data = ci1; proc means data = ci2; run; Remember that the formula for a confidence interval for µ when σ is known is x̄ ± z∗ σ√ n where z∗ is the appropriate percentile of a standard normal distribution. For a 95% confi- dence interval z∗ = 1.96, since 95% of the area under a standard normal density is between 1 −1.96 and 1.96, while for a 99% confidence interval z∗ = 2.576, because 99% of the area under a standard normal density is between −2.576 and 2.576. For pop1 we’ll use σ = 12, while for pop2 we’ll use σ = 3. The output of the two proc means statements gives you x̄, so you have all the ingre- dients necessary to compute the confidence intervals. Please do this, and put the resulting confidence limits in Table 1. Confidence interval lower limit upper limit 95%, pop1 95%, pop2 99%, pop1 99%, pop2 Table 1: Confidence intervals for the means of pop1 and pop2. Questions 1. For a given confidence level, the interval for pop2 is narrower than the interval for pop1. What component in the formula makes the interval narrower? How would you explain to someone who knows no statistics why it makes sense that the interval for pop2 should be narrower? 2. For a given population, the 95the 99% confidence interval. What component in the formula makes this true? How would you explain to someone who knows no statistics why it makes sense that the 95% interval should be narrower? 3. Without knowing the population mean µ, do you know whether your 99% interval contains this value? How would you explain, in nontechnical terms, the meaning of the confidence level 99%? 2 uplim95 = mean + 1.96 * (12/sqrt(10)); lowlim99 = mean - 2.576 * (12/sqrt(10)); uplim99 = mean + 2.576 * (12/sqrt(10)); obsnum = _N_; drop mean; data ci1works; set ci1limits; if(lowlim95 < 53 AND uplim95 > 53) THEN yes95 = 1; ELSE yes95 = 0; if(lowlim99 < 53 AND uplim99 > 53) THEN yes99 = 1; ELSE yes99 = 0; drop lowlim95 uplim95 lowlim99 uplim99 obsnum replicate; proc freq data = ci1works; run; Questions 1. What proportion of the 95% confidence intervals contained the population mean? 2. What proportion of the 99% confidence intervals contained the population mean? A skewed population If the population is not normally distributed, then the confidence level may be inaccurate for small sample sizes, because the distribution of the sample mean will not be well-approximated by a normal distribution. In other words, what we report as a 95% confidence interval may really have a coverage probability of only 87%. We’ll investigate this in the context of pop3, which is skewed and hence not normal, and we’ll make the sample size very small (n = 3). Note that the mean of this population is µ = 1 and the variance is σ = 1. We’ll also add some other confidence levels. proc surveyselect data=cipop n=3 rep=1000 out=ci3; id pop3; proc univariate data = ci3 noprint; output out = ci3means mean=Mean; var pop3; by replicate; data ci3limits; set ci3means; lowlim95 = mean - 1.96 * (1/sqrt(3)); uplim95 = mean + 1.96 * (1/sqrt(3)); 5 lowlim98 = mean - 2.326 * (1/sqrt(3)); uplim98 = mean + 2.326 * (1/sqrt(3)); lowlim99 = mean - 2.576 * (1/sqrt(3)); uplim99 = mean + 2.576 * (1/sqrt(3)); lowlim995 = mean - 2.807 * (1/sqrt(3)); uplim995 = mean + 2.807 * (1/sqrt(3)); obsnum = _N_; drop mean; data ci3works; set ci3limits; if(lowlim95 < 1 AND uplim95 > 1) THEN yes95 = 1; ELSE yes95 = 0; if(lowlim98 < 1 AND uplim98 > 1) THEN yes98 = 1; ELSE yes98 = 0; if(lowlim99 < 1 AND uplim99 > 1) THEN yes99 = 1; ELSE yes99 = 0; if(lowlim995 < 1 AND uplim995 > 1) THEN yes995 = 1; ELSE yes995 = 0; drop lowlim95 uplim95 lowlim98 uplim98 lowlim99 uplim99 lowlim995 uplim995 obsnum replicate; proc freq data = ci3works; run; Questions 1. What proportion of the 95% confidence intervals contained the mean µ = 1? 2. What proportion of the 98% confidence intervals contained the mean µ = 1? 3. What proportion of the 99% confidence intervals contained the mean µ = 1? 4. What proportion of the 99.5% confidence intervals contained the mean µ = 1? 5. Comment generally on whether the confidence levels we observed were close to the stated (95%, 98%, 99%, 99.5%) confidence levels. If they differed, were the observed confidence levels smaller (this would be a worse mistake, since we’d be claiming more than the data supported) or larger? 6