Confidence Intervals: Confidence Interval for a Population Proportion 1.5 2008/06/06 16:38:08 GMT-5 2008/07/18 15:20:39.925 GMT-5 Barbara Illowsky illowskybarbara@deanza.edu Susan Dean deansusan@deanza.edu Connexions cnx@cnx.org elementary statistics During an election year, we see articles in the newspaper that state confidence intervals in terms of proportions or percentages. For example, a poll for a particular candidate running for president might show that the candidate has 40% of the vote within 3 percentage points. Often, election polls are calculated with 95% confidence. So, the pollsters would be 95% confident that the true proportion of voters who favored the candidate would be between 0.37 and 0.43 ( 0.40 - 0.03 , 0.40 + 0.03 ) .Investors in the stock market are interested in the true proportion of stocks that go up and down each week. Businesses that sell personal computers are interested in the proportion of households in the United States that own personal computers. Confidence intervals can be calculated for the true proportion of stocks that go up or down each week and for the true proportion of households in the United States that own personal computers.The procedure to find the confidence interval, the sample size, the error bound, and the confidence level for a proportion is similar to that for the population mean. The formulas are different.How do you know you are dealing with a proportion problem? First, the underlying distribution is binomial. (There is no mention of a mean or average.) If X is a binomial random variable, then X~B(n,p) where n = the number of trials and p = the probability of a success. To form a proportion, take X, the random variable for the number of successes and divide it by n, the number of trials (or the sample size). The random variable P' (read "P prime") is that proportion, P'=Xn (Sometimes the random variable is P̂, read "P hat".)When n is large, we can use the normal distribution to approximate the binomial. X ~ N ( n ⋅ p , n ⋅ p ⋅ q ) If we divide all values of the random variable by n, the mean by n, and the standard deviation by n, we get a normal distribution of proportions with P', called the estimated proportion, as the random variable. (Recall that a proportion = the number of successes divided by n.) X n = P ' ~ N ( n ⋅ p n , n ⋅ p ⋅ q n ) By algebra, n ⋅ p ⋅ q n = p ⋅ q n P' follows a normal distribution for proportions: P ' ~ N ( p , p ⋅ q n ) The confidence interval has the form (p'-EBP,p'+EBP). p ' = x n p ' = the estimated proportion of successes (p' is a point estimate for p, the true proportion) x = the number of successes. n = the size of the sampleThe error bound for a proportion is EBP = z α 2 ⋅ p ' ⋅ q ' n q ' = 1 - p ' This formula is actually very similar to the error bound formula for a mean. The difference is the standard deviation. For a mean where the population standard deviation is known, the standard deviation is σ n .For a proportion, the standard deviation is p ⋅ q n .However, in the error bound formula, the standard deviation is p ' ⋅ q ' n .In the error bound formula, p' and q' are estimates of p and q. The estimated proportions p' and q' are used because p and q are not known. p' and q' are calculated from the data. p' is the estimated proportion of successes. q' is the estimated proportion of failures.When a study gives a margin of error of "+ or - 3 percentage points", this is determined before the survey is done. Since p' and q' are unknown, the most conservative choice is p'=0.5 and q'=0.5, because these values give the largest standard deviation, error bound, and confidence interval.For the normal distribution of proportions, the z-score formula is as follows.If P ' ~ N ( p , p ⋅ q n ) then the z-score formula is z = p ' - p p ⋅ q n Suppose that a sample of 500 households in Phoenix was taken last May to determine whether the oldest child had given his/her mother a Mother's Day card. Of the 500 households, 421 responded yes. Compute a 95% confidence interval for the true proportion of all Phoenix households whose oldest child gave his/her mother a Mother's Day card.Note:The first solution is step-by-step. The second solution uses the TI-83+ and TI-84 calculators. Let X = the number of oldest children who gave their mothers Mother's Day card last May. X is binomial. X ~ B(500, 421500).To calculate the confidence interval, you must find p', q', and EBP. n = 500 x = the number of successes = 421 p ' = x n = 421 500 = 0.842 q ' = 1 - p ' = 1 - 0.842 = 0.158 Since CL = 0.95 , then α = 1 - CL = 1 - 0.95 = 0.05 α 2 = 0.025 .Then z α 2 = z .025 = 1.96 using a calculator, computer, or standard normal table.Remember that the area to the right = 0.025 and therefore, area to the left is 0.975.The z-score that corresponds to 0.975 is 1.96. EBP = z α 2 ⋅ p ' ⋅ q ' n = 1.96 ⋅ [ ( .842 ) ⋅ ( .158 ) 500 ] = 0.032 p ' - EBP = 0.842 - 0.032 = 0.81 p ' + EBP = 0.842 + 0.032 = 0.874 The confidence interval for the true binomial population proportion is (p'-EBP,p'+EBP) =(0.810,0.874).We are 95% confident that between 81% and 87.4% of the oldest children in households in Phoenix gave their mothers a Mother's Day card last May.We can also say that 95% of the confidence intervals constructed in this way contain the true proportion of oldest children in Phoenix who gave their mothers a Mother's Day card last May. TI-83+ and TI-84: Press STAT and arrow over to TESTS. Arrow down to A:PropZint. Press ENTER. Enter 421 for x, 500 for n, and .95 for C-Level. Arrow down to Calculate and press ENTER. The confidence interval is (0.81003, 0.87397). For a class project, a political science student at a large university wants to determine the percent of students that are registered voters. He surveys 500 students and finds that 300 are registered voters. Compute a 90% confidence interval for the true percent of students that are registered voters and interpret the confidence interval. x=300 and n=500. Using a TI-83+ or 84 calculator, the 90% confidence interval for the true percent of students that are registered voters is (0.564, 0.636).Interpretation:We are 90% confident that the true percent of students that are registered voters is between 56.4% and 63.6%. Ninety percent (90 %) of all confidence intervals constructed in this way contain the true percent of students that are registered voters. Binomial Distribution A discrete random variable (RV) which arises from the Bernoulli trials with the next additional requirements. There are fixed number, n, of independent trials. “Independent” means that the result to any trial (for example, trial 1) in no way affects the answer to all the following trials, and all trials are conducted under the same conditions. Under these circumstances the binomial RV X size 12{X} {} is defined as the number of success in n trials. The notation is: X~ B ( n , p ); the domain is the mean is μ np , and the variance is σ 2 = df. The probability to have exactly x successes in n trials is P ( X = x ) = n x p x q n − x . Confidential Interval An interval estimate for unknown population parameter. This depends on: The desired confidence level. What is known for the distribution information (for ex., known variance).Gathering from the sampling information. Confidence Level The percent expression for the probability that the confidence interval contains the true population parameter. That is, for ex., if CL=90%, then in 90 out of 100 samples the interval estimate will enclose the true population parameter. Error Bound for a Population Mean (EBM) The margin of error. Depends on the confidence level, sample size, and known or estimated population standard deviation. Normal Distribution A continuous random variable (RV) with pdf=1σ2πe−(x−μ)2/2σ2 size 12{ ital "pdf"= { {1} over {σ sqrt {2π} } } e rSup { size 8{ - \( x - μ \) rSup { size 6{2} } /2σ rSup { size 6{2} } } } } {}, where μ is the mean of the distribution and σ is its standard deviation. Notation: X ~ N μ σ 2 . If μ=0 and σ=1, the RV is called standard normal distribution, or z-score.