Estimation and Confidence Intervals

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1 Estimation and Confidence Intervals Sections Cathy Poliak, Ph.D. Office in Fleming 11c Department of Mathematics University of Houston Lecture Cathy Poliak, Ph.D. Office in Fleming 11c Sections (Department of Mathematics University of Lecture Houston 17 ) / 43

2 Outline 1 Proportions 2 Estimating Parameters 3 Introduction to Confidence Intervals 4 T-distribution 5 Confidence Interval for Population Mean 6 Examples of Confidence Intervals Cathy Poliak, Ph.D. cathy@math.uh.edu Office in Fleming 11c Sections (Department of Mathematics University of Lecture Houston 17 ) / 43

3 Sampling Distribution of The Sample Mean, X Suppose the a random variable, X, has a population mean, µ, and a population standard deviation of σ. If we take a sample of size n and determine the sample mean X, the distribution of the sample means will be: Normal if the original population has a normal distribution or the sample size is large (n > 30) (CLT). With expected value E( X) = µ And standard deviation SD( X) = n σ. Cathy Poliak, Ph.D. cathy@math.uh.edu Office in Fleming 11c Sections (Department of Mathematics University of Lecture Houston 17 ) / 43

4 Review Questions Suppose that the distribution of heights of all male students on your campus is Normal with mean 70 inches and standard deviation 2.8 inches. Which would be the best way to find the following? 1. If you choose one student at random, what is the probability that he is at most 71 inches tall? a. pnorm(71,70,2.8) b qnorm(70,71,2.8) c. 1 - pnorm(71,70,2.8) d. 1 - pnorm(70,71,2.8) 2. Find the value of x such that P(X x) = 0.1. a. pnorm(0.1,70,2.8) b. 1-pnorm(0.1,70,2.8) c. qnorm(1-0.1,70,2.8) d. 1 - qnorm(0.1,70,2.8) 3. Suppose you take a sample of 4 students, what is the probability that the sample mean of these 4 students is at most 71 inches tall? a. pnorm(71,70,2.8) b. pnorm(71,70,1.4) c. pnorm(71,35,0.7) d. pnorm(70,71,2.8) Cathy Poliak, Ph.D. cathy@math.uh.edu Office in Fleming 11c Sections (Department of Mathematics University of Lecture Houston 17 ) / 43

5 Sample Proportions The population proportion is p a parameter. In some cases we do not know the population proportion, thus we use the sample proportion, ˆp to estimate p. The sample proportion is calculated by: ˆp = X n X = the number of observations of interest in the sample or the number of "successes" in the sample. n = the sample size or number of observations. The mean of ˆp is µˆp = E(ˆp) = p. The standard deviation of ˆp is σˆp = SD(ˆp) = p(1 p) n. If np 10 and n(1 p) 10, then ˆp has an approximate Normal distribution. Cathy Poliak, Ph.D. cathy@math.uh.edu Office in Fleming 11c Sections (Department of Mathematics University of Lecture Houston 17 ) / 43

6 Voting Questions Suppose that 52% voted for a certain candidate. We take a random sample of 1500 likely voters. Determine the following probabilities. 1. What is the probability that the sample proportion is less than 0.50? 2. What is the probability that the sample proportion is within 3% of the population proportion? Cathy Poliak, Ph.D. cathy@math.uh.edu Office in Fleming 11c Sections (Department of Mathematics University of Lecture Houston 17 ) / 43

7 Example Political Polls gallup-daily-trump-job-approval.aspx senate/tx/texas_senate_cruz_vs_orourke-6310.html general_election_trump_vs_biden-6247.html Cathy Poliak, Ph.D. Office in Fleming 11c Sections (Department of Mathematics University of Lecture Houston 17 ) / 43

9 Estimation Estimation is the process of determining the value of a population parameter from evidence made available by a sample statistic. A point estimate is a single value that has been calculated from sample data to estimate the unknown population parameter. We like to use unbiased estimates to predict the parameter. Cathy Poliak, Ph.D. cathy@math.uh.edu Office in Fleming 11c Sections (Department of Mathematics University of Lecture Houston 17 ) / 43

10 Unbiased Estimators If we desired to estimate a parameter, we want to know that we are using a good estimator. The way we know that is if the statistics is an unbiased estimate of the parameter. A point estimator ˆθ is said to be an unbiased estimator of θ if E(ˆθ) = θ for every possible value of θ. If ˆθ is not unbiased, the difference E(ˆθ) θ is called the bias of ˆθ. Cathy Poliak, Ph.D. cathy@math.uh.edu Office in Fleming 11c Sections (Department of Mathematics University of Lecture Houston 17 ) / 43

11 Notation of Parameters and Statistics Name Sample Statistic Population Parameter mean x µ mu standard deviation s σ sigma correlation r ρ rho regression coefficient b β beta proportion ˆp p We are going to use the sample statistics to estimate the population parameters / 43

12 Standard Error The standard error of an estimator ˆθ is its standard deviation SE(ˆθ) = Var(ˆθ). Examples of standard errors SE( X) = σ n. SE(ˆp) = p(1 p) n. The problem is that often we do not know σ or p for example, we can use the estimators for these parameters. Then we have the estimated standard error. Again we need to know how these estimators ˆθ are being distributed / 43

13 What we use for estimating? A confidence interval is a range of possible values that is likely to contain the unknown population parameter we are seeking. First, we must have a level of confidence. Then based on this level we will compute a margin of error. Last, we can say that we are % confident that the true population parameter falls within our confidence interval / 43

14 Another Example Suppose the heights of the population of basketball players at a certain college are in question. A sample of size 16 is randomly selected from this population of basketball players and their heights are measured. The average heights is found to be 6.2 feet and the margin of error is found to be ±0.4 feet. If this margin of error was determined with a 95% confidence level, find and interpret the confidence interval / 43

15 Confidence Interval Interpretation From a sample of 1,000 TV homes in the U.S., 12.3% watch Thursday night football. From this sample a 99% confidence interval for the percent of TV homes in the U.S. that watch Thursday night football is 12.3% ± 3%. Which of the following is the best interpretation for this confidence interval? a) We are 99% confident that the percent of the 1,000 TV homes in the U.S. sample that watch Thursday night football is between 9.3% and 15.3%. b) We are 99% confident that the percent of all TV homes in the U.S. that watch Thursday night football is 12.3%. c) We are 99% confident that the percent of all TV homes in the U.S. that watch Thursday night football is between 9.3% and 15.3%. d) There is a 99% chance that the percent of all TV homes in the U.S. that watch Thursday night football is between 9.3% and 15.3% / 43

16 So what does this mean? The confidence level is saying if we repeat this process several different times we would expect to include the population proportion % of the time. For example, the producers of M&Ms say that 24% of the Milk Chocolate M&Ms are blue. There was a study to see if that was true. A sample of 75 milk chocolate M&Ms were randomly selected and then the process was repeated 20 times. The following graph is the percent we got that were blue M&Ms and the confidence intervals for a 95% confidence / 43

17 How many times should the true parameter be included in the confidence interval? / 43

18 What is the average cell phone bill per month? A survey taken in 2010 polled 400 randomly chosen cell phone users. They answered the question: "What is your average monthly cell phone bill?" The following are the characteristics of the sample: The sample mean is x = $71. Assume the population standard deviation to be σ = $20. The sample size is n = / 43

19 What is the population mean monthly cell phone bill? From the sample of 400 people we found that the sample mean was $71. We can use this number to estimate the population mean monthly cell phone bill. However, if we take another random sample of 400 cell phone users, would we get the same sample mean? / 43

20 Confidence Interval To answer the previous questions we create what is called a confidence interval, a range of values to estimate the true parameter. That is, from a sample we get a point estimate. Then we say how confident we are that the true parameter is within the range of values in the interval. Today we will look at estimating the population mean, µ. To find this estimation we will use the sampling distributions of X / 43

21 From Probability to Confidence Let X be a random variable with unknown mean µ and standard deviation σ. Suppose we are looking at the sample means X for a sample of size n. Assume X N(µ, σ n ). Let P( z α/2 X µ σ/ n z α/2) = 1 α / 43

22 / 43

23 The confidence interval The 1 α confidence interval for µ, given that we know the population standard deviation is: x ± z α/2 ( σ n ) / 43

24 Assumptions for Estimating the Population Mean 1. The sample has to be as a result of a simple random sample (SRS). 2. The distribution of the population has to be Normal. By the Central Limit Theorem if our sample size is larger than 30 then the sample means have a Normal distribution / 43

25 Confidence Level C = 100(1 α)% The confidence interval is associated with a degree of confidence. This degree of confidence is the percentage of times that the confidence interval actually does contain the parameter. Most common choices of level of confidence: 80%, 90%, 95% and 99%. The confidence level is predetermined which will be given in the problem. If not assume C = 95% / 43

26 Critical value The critical value depends on the confidence level 1 α. The critical value (CV) is the cut off point where 1 α is the middle area of the density curve. -CV CV / 43

29 Critical Value for Means We are assuming we know the population standard deviation. For means if the population standard deviation is known, the critical value is z. where the area under the Normal curve is between z α/2 and +z α/2 is the confidence level C = 1 α. This can be found using the z-table or qnorm((1 + C)/2) in R. The following table is the common confidence levels with their z-score C 80% 90% 95% 99% z α/2 z 0.10 = 1.28 z 0.05 = z = 1.96 z = / 43

30 Margin of Error The margin of error is m = critical value standard error For means (given the population standard deviation is known), the margin of error is: m = z α/2 ( σ n ) What is the margin of error for the mean monthly cell phone bill? / 43

31 Rcode > 71+c(-1,1)*qnorm(1.95/2)*20/sqrt(400) [1] / 43

32 Gas Prices The following is a random sample of the price of regular unleaded gasoline from 10 gas stations in the area Suppose that the population standard deviation of gasoline prices is σ = 0.1. Give a 97% confidence interval for the mean gasoline prices / 43

33 R code gas=c(1.85,1.99,1.94,2.25,2.39,2.19,2.19,1.95,2.18,2.09) > mean(gas)+c(-1,1)*qnorm(1.97/2)*0.1/sqrt(10) [1] / 43

34 Mean Amount of Coffee Dispensed A coffee machine dispenses coffee into paper cups. Here are the amounts measured in a random sample of 20 cups. 9.9, 9.7, 10.0, 10.1, 9.9, 9.6, 9.8, 9.8, 10.0, 9.5, 9.7, 10.1, 9.9, 9.6, 10.2, 9.8, 10.0, 9.9, 9.5, 9.9 Determine a 90% confidence interval for the mean amount of coffee dispensed from this machine / 43

Determining the Spread of a Distribution

Determining the Spread of a Distribution 1.3-1.5 Cathy Poliak, Ph.D. cathy@math.uh.edu Department of Mathematics University of Houston Lecture 3-2311 Lecture 3-2311 1 / 58 Outline 1 Describing Quantitative