Chapter 6: Confidence Intervals

Transcription

1 Chapter 6: Confidence Intervals Part II Eric D. Nordmoe Math 261 Department of Mathematics and Computer Science Kalamazoo College Spring 2009

2 Outline Using SPSS to Obtain Confidence Intervals for µ Sample Size Determination Sampling Variability of Proportions Confidence Interval for a Population Proportion

3 Review: Confidence Interval for µ If Y 1, Y 2,..., Y n are a random sample from a normally distributed population, then Ȳ ± t α/2 (n 1)s/ n is a 100(1 α)% confidence interval for µ and t α/2 (n 1) is the upper α/2 critical value for the t distribution on n 1 degrees of freedom.

4 More Data from a Classic Experiment A study of sleep aids among 10 patients in a psychiatric hospital obtained the following sleep hours data: Patient Drug ȳ 4.00 s 2.10

5 More Data from a Classic Experiment Compute a 95% confidence interval for the population mean. Interpret the interval in the context of the data.

6 Interpreting a Confidence Interval The interpretation of a confidence interval in context includes: The confidence level The parameter being estimated Reference to the context including units The population to which inference is being made

7 Test of Understanding Which of the following is true? We are 95% confident that the sample mean sleep hours for individuals in this population is between 2.50 and 5.50 hours. In another sample of size ten, there is a 95% probability that the sample mean X will be within 1.50 of In another sample of size ten, there is a 95% probability that the sample mean X will be within 1.50 of the unknown population mean µ.

8 Test of Understanding Which of the following is true? The probability is 95% that the sample mean is between 2.50 and 5.50 hours. The probability is 95% that the population mean µ is between 2.50 and 5.50 hours. In repeated random sampling, 95% of all intervals computed by this method would contain the true population mean µ. In the population, 95% of all individuals have sleep hours between 2.50 and 5.50 hours. We can be 95% confident that 95% of all individuals in the population have sleep hours between 2.50 and 5.50 hours.

9 Using SPSS to Obtain Confidence Intervals for µ Using SPSS Explore First, choose Explore from the Analyze menu and select the variable of interest. Open the Statistics dialog box and enter the desired confidence level:

10 Using SPSS to Obtain Confidence Intervals for µ Using SPSS Explore The interval x ± t α/2 (n 1)s/ n appears in the Descriptives section of the output.

11 Sample Size Determination Using the Desired SE When planning a study, the sample size may be selected to ensure the required precision: The SE is the primary measure of precision of estimation: SE = s n In practice, find the desired n as a function of the desired SE and guessed s: n = ( ) Guessed s 2 Desired SE

12 Sample Size Determination Using the Margin of Error Find n to achieve a desired margin of error (MOE) where: MOE = t α/2 (n 1) s n 2 s n is the half-width of a confidence interval. Set n = ( ) 2 Guessed s 2 Desired MOE to find the required n as a function of the guessed s and desired MOE.

13 Sample Size Determination Example How many patients would be required to obtain an SE =.5 hours if the guessed s = 1.8 hours? How many patients would be required to obtain an MOE =.5 hour if the guessed s = 1.8 hours?

14 Sampling from a Dichotomous Population Problem: Estimate the proportion of mutants in a population of organisms.

15 The Sampling Distribution of ˆp Estimation of p requires knowledge of the sampling distribution of ˆp where ˆp = Number of successes Number of trials = Y n If p is known, the sampling distribution of ˆp can be computed using the Binomial probability distribution since Pr{j successes} = Pr(Y = j) = n C j p j (1 p) n j. if the BINS conditions are met.

16 The Normal Approximation to the Sampling Distribution of ˆp For n large, the distribution of the number of successes Y is approximately normal: ( Y N np, ) np(1 p) approximately. Similarly, for n large, the distribution of the proportion of successes ˆp is approximately normal: ( ) p(1 p) ˆp N p, approximately. n

17 Example An estimate ˆp is obtained by taking a simple random sample of 30 from a population with proportion p =.39 of mutants. Find Pr(ˆp.45). Below what level would you expect the sample proportion of mutants to fall just 1% of the time?

18 Confidence Intervals for Proportions The Classic Version The standard error of ˆp is SE(ˆp) = ˆp(1 ˆp) Historically, for large n, the accepted 100(1 α)% confidence interval for p has been: n ˆp(1 ˆp) ˆp ± z α/2. n Recent studies have shown this interval has poor coverage properties. The actual confidence level is usually less than the nominal level.

19 Confidence Intervals for Proportions New and Improved Version A better method that works even for relatively small samples is to compute the Wilson estimator p = Y + 2 n + 4 the sample proportion from a fictitious sample with four more observations, two successes and two failures. An approximate 100(1 α)% confidence interval for p is: p ± z α/2 p(1 p) n + 4. Use this interval when the confidence interval is at least 90% and the sample size n is at least 10. Note that this recommendation differs somewhat from the suggestion in the Samuels-Witmer text.

20 Example In a 1992 study of Pet Birds as an Independent Risk Factor for Lung Cancer, researchers sampled 239 lung cancer patients in Berlin. Of these 239, some 98 reported keeping pet birds. Obtain a 95% confidence interval for the population proportion p who kept pet birds.

21 Sample Size Determination When planning a study, the sample size may be selected to ensure the required precision for p: Solving the previous confidence interval for n we obtain ( ) 2 n = p (1 p zα/2 ) 4 MOE where MOE is the desired half-width of the confidence interval and p is either A guessed value of p or p = 0.5 if no guess is available.

22 Example Find the required sample size to obtain a 95% confidence interval for the proportion p of lung cancer patients who have pet birds. The interval should have half-width no greater than MOE = 0.03.