Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Chapter 7 – 2C: Determining the Sample Size required to Estimate A Population Mean µ The Sample Size required to estimate a Population Mean µ is found by the following formula ⎡ zα 2 • estimated σ x ⎤ n=⎢ ⎥ E ⎣ ⎦ 2 where The Confidence Level required for the estimate is 1− α E is the maximum Margin of Error you will allow The Population Standard Deviation σ has been estimated Example 1: Jenny Craig wants to estimate the average weight loss for new customers after the first 90 days of using their product. They estimate that the population standard deviation for this group is 14.3 pounds Find the sample size required that will estimate the population mean µ within 1.2 pounds at a 95% confidence level. E = 1.2 σ = 14.3 α = .05 so α 2 = .025 and z α 2 = 1.96 2 2 ⎡ zα 2 • estimated σ x ⎤ ⎡ 1.96 •14.3 ⎤ n=⎢ ⎥ = ⎢ ⎥⎦ E 1.2 ⎣ ⎣ ⎦ = 545.5 ≈ 546 (rounded up) Statement: If I conduct a random sample of 546 new Jenny Craig customers after the first 90 days of using the product the sample mean I get from the sample will be within 1.2 pounds of the true value of population mean at a 95% confidence level. α = .05 so α 2 = .025 Find +z α 0.0250 2 is an exact table value the negative Z score that has an .025 area to the left is – 1.96 The positive +z α 2 = | − z α 2 | The positive + zα 2 value is |– 1.96| = 1.96 Negative Z Scores Standard Normal (Z) Distribution: Cumulative Area to the LEFT of Z Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 –1.9 0.0287 0.0281 0.0274 0.0268 0.0262 0.0256 0.0250 0.0244 7– 2D Sample Size for Means Page 1 of 3 0.08 0.09 0.0239 0.0233 ©2013 Eitel Example 2: The Department of Educational Research wants to estimate the average numbers of hours a week students spend on homework. They estimate that the population standard deviation for this group is 8.25 hours Find the sample size required that will estimate the population mean µ within .75 hours at a 99% confidence level. E = .75 σ = 8.25 α = .01 so α 2 = .005 and z α 2 = 2.575 2 2 ⎡ zα 2 • estimated σ x ⎤ ⎡ 2.575 • 8.25 ⎤ n=⎢ = ⎥ ⎢⎣ ⎥⎦ = 802.3 ≈ 803 (rounded up) E .75 ⎣ ⎦ Statement: If I conduct a random sample of 803 college students about the time spent doing homework per week the sample mean I get from the sample will be within .75 hours of the true value of population mean at a 99% confidence level. α = .01 so α 2 = .005 Find +z α 2 Find an area in the body of the table that is as close to .005 as possible 0.0050 is an exact table value The cells at the bottom of the table says for an area of .005 use Z = –2.575 The positive +z α 2 = | − z α 2 | The positive + zα 2 value is |– 2.575| = 2.575 Negative Z Scores Standard Normal (Z) Distribution: Cumulative Area to the LEFT of Z Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 –2.5 0.0062 0.0060 0.0059 0.0057 0.0055 0.0054 0.0052 0.0051 Z scores of –3.5 or less use .0001 AREA Z Score 0.0500 –1.645 7– 2D Sample Size for Means Page 2 of 3 AREA 0.08 0.09 0.0049 0.0048 Z Score 0.0050 – 2 . 5 7 5 ©2013 Eitel Example 3: The Department of Transportation wants to estimate the average numbers of miles a week Californians drive. They estimatethat the population standard deviation for this group is 12.4 miles. Find the sample size required that will estimate the population mean µ within 1.5 miles at a 92% confidence level. E = 1.5 σ = 12.4 α = .08 so α 2 = .04 and z α 2 = 1.75 2 2 ⎡ zα 2 • estimated σ x ⎤ ⎡ 1.75 •12.4 ⎤ n=⎢ = ⎥ ⎢⎣ ⎥⎦ E 1.5 ⎣ ⎦ = 209.2 ≈ 210 (rounded up) Statement: If I conduct a random sample of 210 Californian drivers about the miles they drive per week, the sample mean I get from the sample will be within 1.5 hours of the true value of population mean at a 92% confidence level. α = .08 so α 2 = .04 Find +z α 2 Find an area in the body of the table that is as close to .04 as possible 0.0401 is as close to .04 as possible the negative Z score that has an .04 area to the left is – 1.75 The positive +z α 2 = | − z α 2 | The positive + zα 2 value is |– 1.75| = 1.75 Negative Z Scores Standard Normal (Z) Distribution: Cumulative Area to the LEFT of Z Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 –1.7 0.0446 0.0436 0.0427 0.0418 0.0409 0.0401 0.0392 0.0384 7– 2D Sample Size for Means Page 3 of 3 0.08 0.09 0.0375 0.0367 ©2013 Eitel