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Chapter 8 Confidence Intervals on and p Confidence Interval Estimation Normal Distribution for the Mean • An interval estimator is a formula that tells us how to use sample data to calculate an interval that estimates a population parameter. • The confidence coefficient is the probability that an interval estimator encloses the population parameter. • The confidence level is the confidence coefficient expressed as a percentage. Confidence Intervals Distribution Revisited Useful Useful Probabilities for Normal Distributions 68% 95% 99% • Confidence intervals assume that the sample means 3 are normally distributed. • A 95% confidence interval represents a range of values within which you are 95% certain that the true population mean exists. – One interpretation is that if 100 different samples were drawn from the same population and 100 intervals were calculated, approximately 95 of them would contain the population mean. 4 1 Chapter 8 Confidence Intervals Sometimes Miss Large-Sample Confidence Interval for x z x 2 z • where 2 is the z value with an area 2 to its right and n • The parameter is the standard deviation of the sampled population and n is the sample size. • When is unknown (most cases) and n is large, the value of can be approximated by the sample standard deviation, s. x 5 Commonly Used Values of z 2 C o n fid e n c e L e v e l 1 0 0 (1 ) 90% 95% 99% .1 0 .0 5 .0 1 2 .0 5 .0 2 5 .0 0 5 z 2 1 .6 4 5 1 .9 6 2 .5 7 5 Example • Suppose you sampled 400 college students to determine the average soft drink consumption of college students. The results are a sample mean of 20.1 and a sample standard deviation of 0.5. What is the 95% CI for the average soft drink consumption for college students? 2 Chapter 8 Interpretation of a Confidence Interval for a Population Mean • We can be 1 0 0 ( 1 ) % that lies between the lower and upper bounds of the confidence interval. • The statement reflects our confidence in the estimation process rather than in the particular interval that is calculated from sample data. Answer 0.2 L,U 3.6 2.57 10 L 3.44,U 3.76 Example • A tire manufacturer is testing a new compound for tread wear. Tires made with the compound are placed on a machine that simulates road wear, and the amount of tread left after the equivalent of 40,000 road miles is recorded. The mean amount of tread left after a test of 10 tires was 3.6 mm. Construct a confidence interval for the population mean amount of tread left after 40,000 miles at the 99% level of confidence. Assume the population distribution is normally distributed with a standard deviation of 0.2 mm. Example • An automobile dealer plans to order enough cars to have a 90-business-day inventory. Over the past 30 business days the dealer has sold an average of 20 cars a day. Construct a Confidence Interval for the population mean number of cars sold per day at the 95% level of confidence. Assume that the distribution of cars sold per day is approximately normal with a standard deviation of 5 cars. 3 Chapter 8 Answer 5 L,U 20 196 . 30 L 18.21,U 2179 . t-statistic • Formula - t x s n • t-statistic is more variable than the zstatistic. The variability depends upon the sample size, n. • The t-statistic has (n-1) degrees of freedom (df). Small-Sample Estimation of a Normal Probability • The sample standard deviation s may provide a poor approximation of the population standard deviation when the sample size is small. • In these cases the we use the t-statistic rather than the z-statistic. Small-Sample Confidence Interval for • Formula - x t 2 s n • The confidence interval using the t-statistic is wider than the corresponding confidence interval using the standard zstatistic. • Curves: 4 Chapter 8 Example • A sample of monthly sales for 20 Circle K convenience stores shows mean sales to be $30,000 with a standard deviation of $16,000: – Calculate a 95% confidence interval for monthly sales using a t-statistic. – Calculate a 95% confidence interval for monthly sales using a z-statistic. Large-Sample Estimation of a Binomial Probability • Often we want to the know the proportion of a population such as the proportion of smokers, proportion of smokers that prefer a specific brand, proportion of viewers that remember a commercial, etc. • How do we estimate p, the proportion of successes in the sample? Answer FIX THE NOTES L,U x t s x 2 16,000 L,U 30,000 2.093 20 L $22,512,U $37,488 L,U x z x 2 16,000 L,U 30,000 1.96 20 L $22,987 ,U $37,012 Estimation of p • One logical answer is to calculate p as follows: p nx – where – x = number of successes – n = number of trials – p is an unbiased estimator of p. • The standard deviation of the sampling distribution of p is p q p n 5 Chapter 8 Large-Sample Confidence Interval for p pˆ z 2 • Where: pˆ x n pˆ qˆ 1 pˆ • When n is large, we can use p to approximate the value of p in the formula for p . Answer Example • A sample of 500 consumers chosen at random shows that 265 are optimistic about the state of the economy: – Calculate a 90% confidence interval to estimate the proportion of all consumers that are optimistic about the state of the economy. Determining Sample Size: Population Mean Estimation x 265 0.53 n 500 qˆ 1 pˆ 1 0.53 0.47 pˆ L, U pˆ z pˆ 2 L, U 0.53 1.645 0.53 * 0.47 500 L 0.493, U 0.567 4 ( z n ) 2 2 2 W 2 • Where: – W is the desired width of the confidence interval. – The population standard deviation must be estimated. – n is rounded up to ensure that the sample size will be sufficient to achieve the desired confidence interval. 6 Chapter 8 Example Determining Sample Size: Binomial Probability • Footballs inflated to a mean pressure of 13.5 pounds: – Due to machine calibration, individual footballs vary in pressure from 13.3 to 13.7 pounds. – What sample size is necessary for a 99% confidence interval that is only 0.05 wide? 4 ( z )2 n 2 2 W 2 ( 4 )( 2 . 575 ) 2 ( 0 . 1 ) 2 106 . 09 ( 0 . 05 ) 2 Example • The probability that a consumer will feel optimistic about the economy is 0.53: – What sample size is necessary for a 90% confidence interval that is only 0.05 wide? 2 W 2 ) 2 ( pq ) 2 W 2 • Where: – W is the desired width of the confidence interval. – The population standard deviation is estimated by p*q. – n is rounded up to ensure that the sample size will be sufficient to achieve the desired confidence interval. Confidence Interval Review • Large-Sample Confidence Interval: x z x 2 • Small-Sample Confidence Interval: 4 ( z ) 2 ( pq ) n 4 ( z n 4 ( 1 . 645 ) 2 ( 0 . 53 )( 0 . 47 ) 1078 . 5 ( 0 . 05 ) 2 s n x t 2 • Large-Sample Binomial Confidence Interval: pˆ z 2 pˆ qˆ n 7 Chapter 8 Commonly Used Values of z Example • Example: 2 C o n fid e n c e L e v e l 1 0 0 (1 ) 90% 95% 99% .1 0 .0 5 .0 1 2 .0 5 .0 2 5 .0 0 5 z 2 1 .6 4 5 1 .9 6 2 .5 7 5 Solution • Large sample size so use z-statistic rather than t-statistic. x z x 2 • Formula: – where z-value is at 10/2=5% level. Z=1.645. – the population standard deviation is estimated by the sample standard deviation. 4 .1 x z x 11 .6 1 .645 ( ) 11 .6 0 .45 (11 .15,12 .05 ) 225 2 – We are 90% confident that the mean number of unoccupied seats per flight (population mean) lies between 11.15 and 12.05 seats. – We want to estimate the mean number of unoccupied seats per flight for a major airline. Specifically, we want to construct a 90% confidence interval for the population mean. – 225 flights sampled. – Sample mean is 11.6 empty seats. – Sample standard deviation is 4.1 seats. Example • Earnings per share example: – We want to estimate the earnings per share of a type of stock so we ask five portfolio analysts what their projections are for the upcoming year. Specifically, we want to construct a 95% confidence interval for the mean projected earnings estimate for all analysts. – 5 analysts sampled. – Sample mean is $2.63 per share. – Sample standard deviation is $0.72 per share. 8 Chapter 8 Solution • Small sample size so use t-statistic rather than zstatistic. s x t • Formula: n 2 – where t-value is at 5/2=2.5% level and there are 5-1=4 degrees of freedom. t=2.776. – the population standard deviation is estimated by the sample standard deviation. s 0 .72 x t 2 .63 2 .776 ( ) 2 .63 0 .89 (1 .74 ,3 .52 ) n 5 2 – We are 95% confident that the mean of all analysts’ earnings per share projections (population mean) for this type of stock is between $1.74 and $3.52. Example • Smokers brand preference example: – Philip Morris wants to determine the proportion of smokers who prefer Marlboro. Specifically, they want to construct a 95% confidence interval for the proportion of smokers in the smoking population that prefer Marlboro. – 1000 smokers interviewed. – X, the number of smokers (out of 1000 sampled) that prefer Marlboro is a binomial random variable. – 313 out of the 1000 smokers prefer Marlboro: pˆ Solution • Binomial distribution so use the following formula: pˆ qˆ pˆ z 2 2 pˆ qˆ 0.313 * 0.687 0.313 1.96 0.313 0.029 (0.284,0.342) n 1000 – We are 95% confident that the proportion of smokers (population proportion) that prefer Marlboro is between 28.4% and 34.2%. qˆ 1 pˆ 1 0.313 0.687 Determining Sample Size • Population Mean Estimation: n – where z-value is at 5/2=2.5% level. Z=1.96. – the population standard deviation is estimated by the sample standard deviation in the formula above. pˆ z 313 0.313 1000 4 ( z n ) 2 2 2 2 W • Binomial Probability: 4 ( z n ) 2 ( pq ) 2 W 2 9 Chapter 8 Population Mean Example • We want to determine the sample size necessary such that a 95% confidence interval for the mean overdue amount for all delinquent accounts is within $5 of the population mean: – A previous sample of delinquent accounts shows a standard deviation of $90. – Note that information concerning mean is NOT necessary. 4( z )2 2 2 2 n 2 W2 (4)(1.96) (90) 1244.68 (10)2 Binomial Probability Example • The probability that a consumer will choose brand X is 0.20. What sample size is necessary for a 95% confidence interval that is only 0.08 wide? 4 ( z ) 2 ( pq ) n 2 W 2 4 (1 . 96 ) 2 ( 0 . 2 )( 0 .8 ) 384 . 16 ( 0 . 08 ) 2 – Use 385 as the sample size. – Use 1245 as the sample size. 10