Lecture 6: Point Estimation and Large Sample Confidence Intervals. Readings: Sections

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1 Lecture 6: Point Estimation and Large Sample Confidence Intervals Readings: Sections Point Estimation Objective of point estimation: use a sample to compute a number that represents in some sense a good guess for the true value of the parameter of interest. Example 1: An automobile manufacturer has developed a new type of bumper, which is supposed to absorb impacts with less damage than previous bumpers. The manufacturer has used this bumper in a sequence of 25 controlled crashes against a wall, each at 10mph, using one of its compact car models. X = the number of crashes that result in no visible damage to the automobile. p = proportion of all such crashes that result in no damage (parameter of interest) X is observed to be x = 15 Estimator of p: X n Estimate of p: x n = = 0.6 A point estimate of a parameter θ is a single number that can be regarded as a sensible value for θ. A point estimate is obtained by selecting a suitable statistic and computing its value from the given sample data. The selected statistic is called the point estimator of θ (usually denoted as ˆθ). Sometimes the choice of point estimator is obvious (e.g., sample mean as an estimator for population mean, etc). In other cases the choice might not be so obvious or there may be a number of potential options. The methods for finding and evaluating estimators are covered in STAT417/517/528. Example 2: Assume that the number of flaws on a newly manufactured item follows a Poisson distribution with parameter λ. A total of 150 items are randomly selected and examined. The number of flaws is recorded as follows: Number of flaws Observed Frequency Find an estimator for λ and compute the estimate using the data. 1

2 Ideally an estimator should be accurate (low bias) and precise (low variability). Bias A point estimator ˆθ is said to be an unbiased estimator of θ if E(ˆθ) = θ. If ˆθ is not unbiased, the difference E(ˆθ) θ is called the bias of ˆθ. Distribution of θ^1 Distribution of θ^2 E(θ^1) = θ E(θ^2) Consistency A point estimator ˆθ is said to be a consistent estimator of θ if the probability that it lies close to θ increases to 1 as the sample size n increases. Sampling Distribution of X n=5 n=10 n=20 µ 2

3 2 Interval Estimation Confidence interval (CI): an interval of plausible values of the parameter being estimated. A confidence interval is always calculated by first selecting a confidence level, which is a measure of the the degree of the reliability of the interval The most commonly used confidence levels are 90%, 95%, and 99%. Notation: Let z α/2 denote the value such that P (Z > z α/2 ) = P (Z < z α/2 ) = α/2, where Z N(0, 1). That is, the area between z α/2 and z α/2 under the standard normal curve is 1 α. Example 3: Find the following z α/2 values: For α = 0.01, z α/2 = For α = 0.05, z α/2 = For α = 0.10, z α/2 = For α = 0.20, z α/2 = 2.1 Large Sample Confidence Interval for a Population Mean µ Idea of Confidence Intervals If X N(µ, σ), then X N(µ X = µ, σ X = σ/ n) 3

4 The 95% CI for Population Mean µ The 95% CI for the population mean µ for a large sample of size n, with sample mean x and sample standard deviation s, is given by: x ± 1.96 s n Rule of thumb: This formula is appropriate when n > 40. This formula is valid regardless of the shape of the population distribution. The 100(1 α)% CI for Population Mean µ The 100(1 α)% CI for the population mean µ for a large sample of size n, with sample mean x and sample standard deviation s, is given by: s x ± z α/2 n Rule of thumb: This formula is appropriate when n > 40. This formula is valid regardless of the shape of the population distribution. Several most commonly used z α/2 s: Confidence Level 90% 95% 99% α z α/ Example 4: Suppose that the seed weight of the princess bean Phaseotus vulgaris has an unknown mean and unknown standard deviation. A random sample of 100 seeds were taken and the average weight is 450mg with a standard deviation of 100mg. a. Give a 90% confidence interval for the mean seed weight of the princess beans. 4

5 Interpretation of CIs We are 90% confident that the mean seed weight of the princess bean is between. and This means: the method that we used to construct the 90% confidence interval captures the true mean µ 90% of the time. The confidence level refers to the METHOD used to construct the interval, rather than to any particular calculated interval. Here is a picture of 50 different 90% CIs for µ, each based on a sample of size µ Note that four (or 8%) of the intervals failed to capture µ. If we repeated this many more times, 10% of the intervals would fail to capture µ and 90% of them would cover µ. Example 4 (cont d): b. Give a 95% confidence interval for the mean seed weight of the princess beans. 5

6 c. Now assume that the sample size is only 50. For the sample sample mean and standard deviation, calculate the 90% and 95% CIs for the mean seed weight of the princess beans. d. Suppose that we plan a larger study on seed weights of princess beans. How many seeds do we have to sample to estimate µ to within ±10mg with 90% confidence? 6

7 Choosing the Sample Size To be 100(1 α)% confident that the sample mean x will be within B of the true population mean µ, the required sample size n is: ( szα/2 ) 2 n =, B where s is the sample standard deviation obtained from prior information about population standard deviation σ dividing the range (difference between the largest and smallest values) of the population by 4 One-sided Confidence Intervals (Confidence Bounds) The 100(1 α)% upper confidence bound for µ is µ < x + z α s n The 100(1 α)% lower confidence bound for µ is µ > x z α s n Confidence Level 90% 95% 99% α z α Large Sample Confidence Interval for a Population Proportion p Recall the sampling distribution of the sample proportion ˆp is ˆp approx p(1 p) N(µˆp = p, σˆp = ), n for large sample size n (np > 5 and n(1 p) > 5). 7

8 The 100(1 α)% CI for the population proportion p for a sample of size n, with sample proportion ˆp, is given by: ˆp + z2 α/2 2n ± z α/2 1 + z2 α/2 n ˆp(1 ˆp) n + z2 α/2 4n 2 Although this formula was based on a large-sample distribution of ˆp, it works quite well even when the sample size n is reasonably small. When n is quite large, the CI reduces to ˆp ± z α/2 ˆp(1 ˆp) n Example 5: To evaluate the policy of routine vaccination for whooping cough, adverse reactions were monitored for 339 infants who received their first injection of the vaccine. Reactions were noted in 69 of the infants. a. Construct a 99% C.I. for the probability of an adverse reaction and interpret it. Choosing the sample size to estimate p To be 100(1 α)% confidence that the sample proportion ˆp will be within B units of the true population proprotion p, the required sample size n is n = ˆp(1 ˆp) ( zα/2 If we had a guess for ˆp (from a preliminary sample or another study), we could use that and determine what the sample size we would need. If we have no guess for ˆp, use ˆp = 0.5 since ˆp(1 ˆp) is maximized at ˆp = 0.5. Example 5 (cont d): B ) 2. 8

9 b. Suppose we wanted to extend our vaccination study to estimate p to within 0.01 with 99% confidence. How many infants do we need to look at? 2.3 Large Sample Confidence Interval for the Difference Between Population Means, µ 1 µ 2 We are often interested in comparing two populations based on a continuous measurement. Compare postmortem serotonin levels in patients who died of heart disease (cases) vs. those who died from other causes (controls). To evaluate impact of light on the growth of plants, one group of seedlings grows in dark conditions, and a second group gets the standard amount of light. Compare heights of plants after a specified time period. Each group has different individuals who may receive different treatments. Responses from each sample are independent of each other. Notations: Population Mean Standard Deviation 1 µ 1 σ 1 2 µ 2 σ 2 Sample Sample size Mean Standard Deviation 1 n 1 x 1 s 1 2 n 2 x 2 s 2 Sampling Distribution of X 1 X 2 µ X1 X 2 = E( X 1 X 2 ) = E( X 1 ) E( X 2 ) = µ 1 µ 2 σ 2 X1 X 2 = V ar( X 1 X 2 ) = V ar( X 1 ) + V ar( X 2 ) = σ2 1 n 1 + σ2 2 n 2 If both population distributions are normal, the sampling distribution of X 1 X 2 is normal. If both sample sizes are large, the sampling distribution of X1 X 2 is approximately normal regardless of the shape of the two population distributions (CLT). 9

10 Therefore when n 1 and n 2 are large, ( X 1 X 2 ) (µ 1 µ 2 ) s 2 1 n 1 + s2 2 n 2 approx. N(0, 1). The 100(1 α)% CI for µ 1 µ 2, the difference between population means, is given by: ( x 1 x 2 ) ± z α/2 s 2 1 n 1 + s2 2 n 2 Rule of thumb: this formula is appropriate when n 1 > 40 and n 2 > 40. Example 6: The lifetime of D batteries from two brands are compared for their lifetime in a flashlight. 50 flashlights were loaded with Brand 1 batteries and 45 with Brand 2 batteries. The lifetime (in hours) of each battery was recorded. Brand n x s What is the 95% confidence interval for the true difference in the mean lifetime of the two brands of batteries? 10