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Chapter 8 Interval Estimation Download this presentation. I. Introduction The previous chapter found a point estimate x of the population parameter . But we know that for every sample, we will never have a point estimate that is exactly equal to the true population parameter. Thus this chapter shows us how interval estimates of and p can be developed to provide information about the precision of that estimate. II. Interval Estimation of In a large sample case, n>=30. We will show how a sampling distribution of x can be used to develop an interval estimate of . A. An example CJW is a mail-order sporting equipment firm that conducts a monthly customer service survey. Their scale is 0-100 where 100 is “Excellent” service. They find that the population =20, x changes every month and is unknown. Most recent survey: x =82, =20, n=100 B. Sampling Error Question: How good is the sample estimate of the population parameter? Sampling Error = x We don’t precisely know the sampling error because we don’t know . But we can draw some probability conclusions about the size of the sampling error. How? The Central Limit Theorem allows us to conclude that the sampling distribution of x can be approximated by a normal probability distribution. n=100, =20 20 x 2 100 x C. Probability Statements From the standard normal probability table (z-table) we know that 95% of all values are within z=1.96* x of the mean. Precision statement: “There is 95% of all sample x-bar values fall here. a 95% probability that x will provide a sampling error of 3.92 or less.” -3.92 +3.92 x D. General Lingo • : probability that the sampling error is larger than the sampling error in the precision statement. • /2: probability in each tail of the distribution • (1-): probability that a sample mean will provide a sampling error less than or equal to the sampling error in the precision statement. • Z/2 is the value of the standard normal random variable corresponding to an area of /2 in the upper tail. A picture should help... In our example (1-) =.95, =.05, /2=.025. (1- ) of all xbar values. /2 /2 x Precision statement: There is a (1-) probability that the value of a sample mean will provide a sampling error of z/2* x or less. E. Interval Estimate: large sample, known. x z /2 n Confidence Level 90% 95% 99% /2 z/2 .10 .05 .01 .05 .025 .005 1.645 1.96 2.576 z/2 is the z-value providing an area of /2 in the upper tail of the standard normal probability distribution. F. Interval Estimate: large sample, unknown. x z /2 s n If the population standard deviation is unknown, use the sample value. Example: Suppose a 50-point exam is given to 36 statistics students. The mean is 39.5 and the standard deviation is 7.77. Construct a 95% confidence interval. Solution With 95% confidence, we find a z.025=1.96. Using the rest of our information, we construct: 39.5 ± 1.96(7.77/6) What does this interval tell us? or: 39.5 ± 2.54 36.96 39.5 42.04