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Chapter

Chapter. 9. Estimating the Value of a Parameter. Section. 9.2. Estimating a Population Mean. Objectives. Obtain a point estimate for the population mean State properties of Student’s t -distribution Determine t -values Construct and interpret a confidence interval for a population mean

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  1. Chapter 9 Estimating the Value of a Parameter

  2. Section 9.2 Estimating a Population Mean

  3. Objectives • Obtain a point estimate for the population mean • State properties of Student’s t-distribution • Determine t-values • Construct and interpret a confidence interval for a population mean • Find the sample size needed to estimate the population mean within a given margin or error

  4. Objective 1 • Obtain a Point Estimate for the Population Mean

  5. A point estimate is the value of a statistic that estimates the value of a parameter. For example, the sample mean, , is a point estimate of the population mean μ.

  6. Parallel Example 1: Computing a Point Estimate Pennies minted after 1982 are made from 97.5% zinc and 2.5% copper. The following data represent the weights (in grams) of 17 randomly selected pennies minted after 1982. 2.46 2.47 2.49 2.48 2.50 2.44 2.46 2.45 2.49 2.47 2.45 2.46 2.45 2.46 2.47 2.44 2.45 Treat the data as a simple random sample. Estimate the population mean weight of pennies minted after 1982.

  7. Solution The sample mean is The point estimate of μ is 2.464 grams.

  8. Objective 2 • State Properties of Student’s t-Distribution

  9. Student’s t-Distribution Suppose that a simple random sample of size n is taken from a population. If the population from which the sample is drawn follows a normal distribution, the distribution of follows Student’s t-distribution with n – 1 degrees of freedom where is the sample mean and s is the sample standard deviation.

  10. Parallel Example 1: Comparing the Standard Normal Distribution to the t-Distribution Using Simulation • Obtain 1,000 simple random samples of size n=5 from a normal population with μ =50 and σ=10. • Determine the sample mean and sample standard deviation for each of the samples. • Compute and for each sample. • Draw a histogram for both z and t.

  11. Histogram for z

  12. Histogram for t

  13. CONCLUSION: • The histogram for z is symmetric and bell-shaped with the center of the distribution at 0 and virtually all the rectangles between –3 and 3. In other words, z follows a standard normal distribution.

  14. CONCLUSION (continued): • The histogram for t is also symmetric and bell-shaped with the center of the distribution at 0, but the distribution of t has longer tails (i.e., t is more dispersed), so it is unlikely that t follows a standard normal distribution. The additional spread in the distribution of t can be attributed to the fact that we use s to find t instead of σ. Because the sample standard deviation is itself a random variable (rather than a constant such as σ), we have more dispersion in the distribution of t.

  15. Properties of the t-Distribution 1. The t-distribution is different for different degrees of freedom. 2. The t-distribution is centered at 0 and is symmetric about 0. 3. The area under the curve is 1. The area under the curve to the right of 0 equals the area under the curve to the left of 0, which equals 1/2. 4. As t increases or decreases without bound, the graph approaches, but never equals, zero.

  16. Properties of the t-Distribution 5. The area in the tails of the t-distribution is a little greater than the area in the tails of the standard normal distribution, because we are using s as an estimate of σ, thereby introducing further variability into the t- statistic. 6. As the sample size n increases, the density curve of t gets closer to the standard normal density curve. This result occurs because, as the sample size n increases, the values of s get closer to the values of σ, by the Law of Large Numbers.

  17. Objective 3 • Determine t-Values

  18. Parallel Example 2: Finding t-values Find the t-value such that the area under the t-distribution to the right of the t-value is 0.2 assuming 10 degrees of freedom. That is, find t0.20 with 10 degrees of freedom.

  19. Solution The figure to the left shows the graph of the t-distribution with 10 degrees of freedom. The unknown value of t is labeled, and the area under the curve to the right of t is shaded. The value of t0.20with 10 degrees of freedom is 0.879.

  20. Objective 4 • Construct and Interpret a Confidence Interval for the Population Mean

  21. Constructing a (1– α)100%Confidence Interval for μ Provided • Sample data come from a simple random sample or randomized experiment • Sample size is small relative to the population size(n ≤ 0.05N) • The data come from a population that is normally distributed, or the sample size is large A (1–α)·100% confidence interval for μ is given by Lower Upper bound: bound: where is the critical value with n – 1 df.

  22. Constructing a (1– α)100%Confidence Interval for μ A (1–α)·100% confidence interval for μ is given by Lower Upper bound: bound: where is the critical value with n – 1 df.

  23. Parallel Example 2: Using Simulation to Demonstrate the Idea of a Confidence Interval We will use Minitab to simulate obtaining 30 simple random samples of size n=8 from a population that is normally distributed with μ=50 and σ=10. Construct a 95% confidence interval for each sample. How many of the samples result in intervals that contain μ=50 ?

  24. Sample Mean 95.0% CI C1 47.07 ( 40.14, 54.00) C2 49.33 ( 42.40, 56.26) C3 50.62 ( 43.69, 57.54) C4 47.91 ( 40.98, 54.84) C5 44.31 ( 37.38, 51.24) C6 51.50 ( 44.57, 58.43) C7 52.47 ( 45.54, 59.40) C9 43.49 ( 36.56, 50.42) C10 55.45 ( 48.52, 62.38) C11 50.08 ( 43.15, 57.01) C12 56.37 ( 49.44, 63.30) C13 49.05 ( 42.12, 55.98) C14 47.34 ( 40.41, 54.27) C15 50.33 ( 43.40, 57.25) C8 59.62 ( 52.69, 66.54)

  25. SAMPLE MEAN 95% CI C16 44.81 ( 37.88, 51.74) C17 51.05 ( 44.12, 57.98) C18 43.91 ( 36.98, 50.84) C19 46.50 ( 39.57, 53.43) C20 49.79 ( 42.86, 56.72) C21 48.75 ( 41.82, 55.68) C22 51.27 ( 44.34, 58.20) C23 47.80 ( 40.87, 54.73) C24 56.60 ( 49.67, 63.52) C25 47.70 ( 40.77, 54.63) C26 51.58 ( 44.65, 58.51) C27 47.37 ( 40.44, 54.30) C29 46.89 ( 39.96, 53.82) C30 51.92 ( 44.99, 58.85) C28 61.42 ( 54.49, 68.35)

  26. Note that 28 out of 30, or 93%, of the confidence intervals contain the population mean μ=50. In general, for a 95% confidence interval, any sample mean that lies within 1.96 standard errors of the population mean will result in a confidence interval that contains μ. Whether a confidence interval contains μ depends solely on the sample mean, .

  27. Parallel Example 3: Constructing a Confidence Interval Construct a 99% confidence interval about the population mean weight (in grams) of pennies minted after 1982. Assume μ = 0.02 grams. 2.46 2.47 2.49 2.48 2.50 2.44 2.46 2.45 2.49 2.47 2.45 2.46 2.45 2.46 2.47 2.44 2.45

  28. Weight (in grams) of Pennies

  29. Lower bound: = 2.464 – 1.746 = 2.464 – 0.008 = 2.456 • Upper bound: = 2.464 + 2.575 = 2.464 + 0.008 = 2.472 We are 99% confident that the mean weight of pennies minted after 1982 is between 2.456 and 2.472 grams.

  30. Objective 5 • Find the Sample Size Needed to Estimate the Population Mean within a Given Margin of Error

  31. Determining the Sample Size n The sample size required to estimate the population mean, µ, with a level of confidence (1– α)·100% with a specified margin of error, E, is given by where n is rounded up to the nearest whole number.

  32. Parallel Example 7: Determining the Sample Size Back to the pennies. How large a sample would be required to estimate the mean weight of a penny manufactured after 1982 within 0.005 grams with 99% confidence? Assume  = 0.02.

  33. s = 0.02 • E = 0.005 Rounding up, we find n = 107.

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