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Business Statistics - QBM117 Revising interval estimation Objective To develop confidence in identifying the correct formula to use when calculating an interval estimate. Identifying the parameter to be estimated In this subject we need only concern ourselves with estimating the population mean , or the population proportion p. Generally it is clear from the question, which parameter we are estimating, if we read it carefully enough. Two confidence interval estimators of There are two different interval estimators of the population mean, and the basis for determining which method is appropriate is quite simple. If is known, the confidence interval estimator of the population mean is x Z / 2 n If is unknown and the population is normally distributed, the confidence interval estimator of the population mean is x t / 2,n 1 s n When d.f > 200 we approximate t by the t / 2 , value. Only one confidence interval estimator of p There are is only one confidence interval estimator of the population proportion so there is no choice. pˆ Z / 2 pˆ qˆ n One condition however must be satisfied before it is appropriate to use this formula both npˆ 5 and nqˆ 5 Exercises relating to interval estimation Example 1 a. Assume that the time it takes to assemble the parts of an electric lamp is normally distributed. Construct a 95% confidence interval for the average assembly time if a random sample of 16 trials has an average assembly time of 400 seconds with a standard deviation of 10 seconds. b. Is it reasonable to suppose that the average assembly time will not exceed 410 seconds? c. How large a sample would be required if we want to estimate the average assembly time to within 2 seconds with 99% confidence? Example 1 Does the question ask us to estimate a mean or a proportion? mean Do we know the population standard deviation, or do we only have the sample standard deviation s? S x t / 2,n 1 s n s 10 n 16 x 400 1 0.95 0.05 / 2 0.025 t..025 025,,15 15 2.131 95%CI ( ) x t00.025,15 15 s n 10 400 (2.131) 16 400 5.33 The average assembly time is between 394.67 secs and 405.33 secs C. s 10 B 2 z / 2 s n B z0.005 2.575 2 (2.575)(10) 2 165.77 166 2 Therefore 166 times should be sampled. Example 2 a. b. c. The daily business at a local restaurant is known to be normally distributed with a standard deviation of $571. Find a 95% confidence interval for the mean daily income, if a random sample of 17 days' receipts shows an average daily income of $9832. Would it be reasonable for the restaurant owner to claim to the taxation department that her average daily income is $9400? How large a sample would be required if we want to estimate the mean daily income to within $50 with 90% confidence? Example 2 Does the question ask us to estimate a mean or a proportion? mean Do we know the population standard deviation, or do we only have the sample standard deviation s? x Z / 2 n 571 n 17 x $9832 1 0.95 0.05 / 2 0.025 Z ..025 025 1.96 95%CI ( ) x Z 0.025 n 571 9832 (1.96) 17 9832 271.4 The average daily income is between $9560.60 and $10 103.40 C. 571 B 50 z / 2 n B z0.05 1.645 2 (1.645)(571) 50 352.9 353 2 Therefore 353 businesses should be sampled. Example 3 a. An interstate savings institution claims that 94% of its depositors have both a savings account and a checking account with the institution. Suppose a random sample of 250 customers shows that 226 have both a saving account and a checking account, whereas the other 24 have either only a savings account or a checking account, not both. Use this sample to construct a 90% confidence interval for the true proportion of depositors who have both types of accounts with this institution. b. What size sample is required to estimate the proportion of depositors with both types of accounts to within 0.1 with 90% confidence? Example 3 Does the question ask us to estimate a mean or a proportion? proportion 226 24 pˆ 0.904 qˆ 0.096 n 250 250 250 1 0.90 0.1 / 2 0.05 Z .05 1.645 npˆ 226 5 nqˆ 24 5 90%CI ( p ) pˆ Z 0.05 pˆ qˆ n (0.904)(0.096) 0.904 (1.645) 250 0.966 0.031 The proportion of depositors with both types of accounts is between 93.5% and 99.7% C. pˆ 0.904 qˆ 0.096 z / 2 pˆ qˆ n B B 0.1 z0.05 1.645 2 (1.645) (0.904)(0.096) 0 . 1 551.5 2 552 Therefore 552 customers should be sampled. Reading for next lecture Chapter 10 Sections 10.1 - 10.2