5.3 Interval Estimation

Transcription

1 5.3 Interval Estimation Ulrich Hoensch Wednesday, March 13, 2013

2 Confidence Intervals Definition Let θ be an (unknown) population parameter. A confidence interval with confidence level C is an interval of the form L θ U, where P(L θ U) = C. Remarks. Usually, the confidence interval is symmetric; this means P(θ < L) = P(θ > U) = (1 C)/2. In other words, the confidence interval covers the middle C of the distribution of θ. If θ has a discrete distribution, then it might not be possible to find an interval [L, U] so that P(L θ U) equals C exactly. In this case we choose [L, U] to be the smallest interval so that P(L θ U) C.

3 Example Suppose we have a random sample X 1,..., X n taken from an N(µ, σ 2 ) distribution (or a large sample), where σ 2 is known. Find a 95% confidence interval for µ. 1. We know that Z = ( X µ ) /(σ/ n) has (approximately) an N(0, 1) distribution. 2. We have that P( 1.96 Z 1.96) = 0.95, so P ( 1.96 X µ ) σ/ n 1.96 = Solving this inequality for µ gives P (X 1.96 n σ µ X n σ ) = 0.95, so the CI is [X 1.96(σ/ n), X (σ/ n)].

4 Percentiles for the Standard Normal Distribution Note that the value 1.96 is the 97.5th percentile of a standard normal distribution. In general, if the confidence level is C, we need to find the 1 (1 C)/2 percentile. The 1 α percentile is denoted by z α and can be computed as follows: 1. z α = invnorm(1 α, 0, 1) on a TI 83/84 Plus calculator. 2. In Mathematica: Quantile NormalDistribution 0, 1, 1 Α

5 Large Sample Confidence Interval for a Population Mean The large sample confidence interval for the population mean at the confidence level C is where: [ x zα/2 (s/ n), x + z α/2 (s/ n) ], x is the observed sample mean; s is the observed sample standard deviation; n is the sample size; z α/2 is the 1 (α/2) percentile of a standard normal distribution, and α/2 = (1 C)/2. (x x) 2 Replacing σ by the sample standard deviation s = n 1 is justified since for large n, σ will be well-approximated by s (see the discussion in the next section).

6 Example Fifty vehicles were observed at random for their speeds (in mph) on a highway with speed limit posted as 70 mph, and it was found that their average speed was 73.3 mph, with a standard deviation of 3.2 mph. Construct a 90% confidence interval for the average speed of all vehicles on this highway. Here, n = 50, x = 73.3, s = 3.2, and z α/2 = invnorm(0.95, 0, 1) = 1.645, so the confidence interval is [ (3.2/ 50), (3.2/ ] 50). The quantity 1.645(3.2/ 50) = 0.74 is called the margin of error of the CI. Thus, the CI is 73.3 ± 0.74 or [72.56, 74.04]. This means that we are 90% confident that the average speed of all vehicles lies between 72.6 and 74.0 mph.

7 Minimum Sample Size In the large sample situation, we have that the margin of error for the mean is E = z α/2s. n To obtain the minimal sample size needed to achieve this margin or error, we solve this equation for n: Example ( zα/2 s ) 2 n =. E Find the minimum sample size needed to so that the margin of error in the previous example is 0.5 mph (at C = 0.90, assume s = 3.2). ( ) (1.645)(3.2) 2 n =

8 Large Sample Confidence Interval for a Population Proportion Since for n large, the sample proportion ˆp has an approximate normal distribution with mean µ = p and standard deviation σ = p(1 p)/n (p is the population proportion), the large sample confidence interval for the population proportion at the confidence level C is [ˆp z α/2 ˆp(1 ˆp)/n, ˆp + zα/2 ˆp(1 ˆp)/n ], where: ˆp is the observed sample proportion; n is the sample size; z α/2 = invnorm(1 (α/2), 0, 1) is the 1 (α/2) percentile of a standard normal distribution, and α/2 = (1 C)/2.

9 Example An auto manufacturer gives a bumper to bumper warranty for 3 years or 36,000 miles for its new vehicles. In a random sample of 60 of its vehicles, 20 of them need five or more major repairs within the warranty period. Find the 95% confidence interval for the true proportion of vehicles that need five or more major repairs during the warranty period. Here, ˆp = 20/60 = 1/3, n = 60. The confidence interval is [ (1/3) 1.96 (1/3)(2/3)/60, (1/3) ] (1/3)(2/3)/60 or [0.214, 0.453]. So we are 95% confident that the actual proportion of vehicles lie between 21.4% and 45.3%.

10 Minimum Sample Size To find the minimum sample for size a CI involving proportions, we proceed as follows. If we know that the population proportion will never exceed p < 0.5 or if we know that the population proportion will never be less than p > 0.5, then the minimal sample size needed to achieve the margin or error E is n = ( zα/2 p (1 p ) E ) 2 = (z α/2) 2 p (1 p ) E 2. If no information is available about the population proportion, choose p = 0.5. Thus, the minimal sample size needed to achieve the margin or error E is n = ( zα/2 0.5(1 0.5) E ) 2 = (z α/2) 2 4E 2.

11 Example In the previous example, the auto manufacturer wants to determine the minimum sample size needed to achieve a margin of error of 5% at C = If the population proportion is known to be less than 40%, then n = (1.96)2 0.4(1 0.4) (0.05) If no information is available about the population proportion, then n = (1.96)2 4(0.05)

12 Homework Problems for Section 5.3 (Points) p.309: (2), (2), (2), (2), (2). Homework problems are due at the beginning of the class on Wednesday, March 20, 2013.