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Understandable Statistics Eighth Edition By Brase and Brase Prepared by: Lynn Smith Gloucester County College Edited by: Jeff, Yann, Julie, and Olivia Chapter 8: Estimation Section 8.2 Estimating μ When σ Is Unknown Focus Points • Learn about degrees of freedom and Student’s t distribution. • Find critical values using degrees of freedom and confidence level. • Compute confidence interval for μ when σ is unknown. What does this information tell you? Statistics Quote “The death of one man is a tragedy. The death of millions is a statistic.” —Joe Stalin What if it is impossible or impractical to use a large sample? Apply the Student’s t distribution. Properties of a Student’s t Distribution 1. The distribution is symmetric about the mean 0. 2. The distribution depends on the degrees of freedom, d.f. (d.f. = n – 1 for μ confidence intervals). 3. The distribution is bell-shaped, but has thicker tails than the standard normal distribution. 4. As the degrees of freedom increase, the t distribution approaches the standard normal distribution. Student’s t Variable x t s n The shape of the t distribution depends only only the sample size, n, if the basic variable x has a normal distribution. When using the t distribution, we will assume that the x distribution is normal. Table 6 in Appendix II gives values of the variable t corresponding to the number of degrees of freedom (d.f.) Student’s t Distribution Critical Values (Table Excerpt) Degrees of Freedom d.f. = n – 1 where n = sample size The t Distribution has a Shape Similar to that of the the Normal Distribution A Normal distribution A “t” distribution The Student’s t Distribution Approaches the Normal Curve as the Degrees of Freedom Increase Find the critical value tc for a 95% confidence interval if n = 8. d.f.=7 c ' '' d.f. ... 6 7 8 ...0.90 ... 0.95 ... 0.98 ... 0.99 ... ...1.943 ...1.895 ...1.860 2.447 2.365 2.306 3.143 2.998 2.896 3.707 3.499 3.355 Convention for Using a Student’s t Distribution • If the degrees of freedom d.f. you need are not in the table, use the closest d.f. in the table that is smaller. This procedure results in a critical value tc that is more conservative in the sense that it is larger. The resulting confidence interval will be longer and have a probability that is slightly higher than c. Confidence Interval for the Mean of Small Samples (n < 30) from Normal Populations xE xE where x Sample Mean s E tc n c = confidence level (0 < c < 1) tc = critical value for confidence level c, and degrees of freedom = n - 1 The mean weight of eight fish caught in a local lake is 15.7 ounces with a standard deviation of 2.3 ounces. Construct a 90% confidence interval for the mean weight of the population of fish in the lake. Mean = 15.7 ounces Standard deviation = 2.3 ounces • n = 8, so d.f. = n – 1 = 7 • For c = 0.90, Table 6 in Appendix II gives t0.90 = 1.895. s 2.3 E tc 1.895 1.54 n 8 Mean = 15.7 ounces Standard deviation = 2.3 ounces. E = 1.54 The 90% confidence interval is: xE xE 15.7 - 1.54 < < 15.7 + 1.54 14.16 < < 17.24 The 90% Confidence Interval: 14.16 < < 17.24 We are 90% sure that the true mean weight of the fish in the lake is between 14.16 and 17.24 ounces. Summary: Confidence Intervals for the Mean • Assume that you have a random sample of size n from an x distribution and that you have computed x-bar and s. A confidence interval for μ is x E x E where E is the margin of error • How do you find E? It depends on how much you know about the x distribution. Situation I (most common) • You don't know the population standard deviation σ. In this situation you • use the t distribution with margin of error s E tc n • with d.f. = n – 1 • Guidelines: If n is less than 30, x should have a distribution that is mound-shaped and approximately symmetric. It's even better if the x distribution is normal. If n is 30 or more, the central limit theorem (Chapter 7) implies these restrictions can be relaxed. Situation II (almost never happens!) • You actually know the population value of σ. In addition, you know that x has a normal distribution. If you don't know that the x distribution is normal, then your sample size n must be 30 or larger. In this situation, you use the standard normal z distribution with margin of error E zc n Which distribution should you use for x? Examine Problem Statement σ is known σ is unknown Use normal distribution with margin of error Use Student’s t distribution with margin of error E zc n s E tc n d.f. = n – 1 Calculator Instructions CONFIDENCE INTERVALS FOR A POPULATION MEAN The TI-83 Plus and TI-84 Plus fully support confidence intervals. To access the confidence interval choices, press Stat and select TESTS. The confidence interval choices are found in items 7 through B. Example (σ is unknown) • A random sample of 16 wolf dens showed the number of pups in each to be 5, 8, 7, 5, 3, 4, 3, 9, 5, 8, 5, 6, 5, 6, 4, and 7. • Find a 90% confidence interval for the population mean number of pups in such dens. Example (σ is unknown) • In this case we have raw data, so enter the data in list using the EDIT option of the Stat key. Since σ is unknown, we use the t distribution. Under Tests from the STAT menu, select item 8:TInterval. Since we have raw data, select the DATA option for Input. The data is in list and occurs with frequency 1. Enter 0.90 for the C-Level. Example (σ is unknown) • Highlight Calculate and press Enter. The result is the interval from 4.84 pups to 6.41 pups. Section 8.2, Problem 11 Diagnostic Tests: Total Calcium Over the past several months, an adult patient has been treated for tetany (severe muscle spasms). This condition is associated with an average total calcium level below 6 mg/dl (Reference: Manual of Laboratory and Diagnostic Tests, F. Fischbach). Recently, the patient's total calcium tests gave the following readings (in mg/dl). 9.3 8.8 10.1 8.9 9.4 9.8 10.0 9.9 11.2 12.1 (a) Use a calculator to verify that x-bar = 9.95 and s = 1.02. (b) Find a 99.9% confidence interval for the population mean of total calcium in this patient's blood. (c) Based on your results in part (b), do you think this patient still has a calcium deficiency? Explain. Solution Section 8.2, Problem 17 Finance: P/E Ratio The price of a share of stock divided by the company's estimated future earnings per share is called the PIE ratio. High PIE ratios usually indicate “growth” stocks or maybe stocks that are simply overpriced. Low P/E ratios indicate “value” stocks or bargain stocks. A random sample of 51 of the largest companies in the United States gave the following P/E ratios (Reference: Forbes). 11 35 19 13 15 21 40 18 60 72 9 20 29 53 16 26 21 14 21 27 10 12 47 14 33 14 18 17 20 19 13 25 23 27 5 16 8 49 44 20 27 8 19 12 31 67 51 26 19 18 32 (a) Use a calculator with mean and sample standard deviation keys to verify that x-bar = 25.2 and s = 15.5. Section 8.2, Problem 17 (b) Find a 90% confidence interval for the P/E population mean μ of all large U.S. companies. (c) Find a 99% confidence interval for the P/E population mean μ of all large U.S. companies. (d) Bank One (now merged with J. P. Morgan) had a P/E of 12, AT&T Wireless had a P/E of 72, and Disney had a P/E of 24. Examine the confidence intervals in parts (b) and (c). How would you describe these stocks at this time? (e) In previous problems, we assumed the x distribution was normal or approximately normal. Do we need to make such an assumption in this problem? Why or why not? Hint: See the central limit theorem in Section 7.2. Solution