Estimating Population Means (Large Samples) Copyright © 2008 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. HAWKES LEARNING SYSTEMS math courseware specialists Section 8.2 Estimating Population Means (Large Samples)

The size of the sample is at least 30 (n ≥ 30). HAWKES LEARNING SYSTEMS math courseware specialists Confidence Intervals 8.2 Estimating Population Means (Large Samples) Criteria for estimating the population mean for large samples: All possible samples of a given size have an equal probability of being chosen. The size of the sample is at least 30 (n ≥ 30). The population’s standard deviation is unknown. When all of the above conditions are met, then the distribution used to calculate the margin of error for the population mean is the Student t-distribution. However, when n ≥ 30, the critical values for the t-distribution are almost identical to the critical values for the normal distribution at corresponding levels of confidence. Therefore, we can use the normal distribution (z-score) to approximate the t-distribution.

Find the critical value for a 95% confidence interval. HAWKES LEARNING SYSTEMS math courseware specialists Confidence Intervals 8.2 Estimating Population Means (Large Samples) Find the critical value: Find the critical value for a 95% confidence interval. Solution: To find the critical value, we first need to find the values for –z0.95 and z0.95. Since 0.95 is the area between –z0.95 and z0.95, there will be 0.05 in the tails, or 0.025 in one tail.

Critical z-Values for Confidence Intervals HAWKES LEARNING SYSTEMS math courseware specialists Confidence Intervals 8.2 Estimating Population Means (Large Samples) Critical Value, zc: Critical z-Values for Confidence Intervals Level of Confidence, c zc 0.80 1.28 0.85 1.44 0.90 1.645 0.95 1.96 0.98 2.33 0.99 2.575

zc = the critical z-value s = the sample standard deviation HAWKES LEARNING SYSTEMS math courseware specialists Confidence Intervals 8.2 Estimating Population Means (Large Samples) Margin of Error, E, for Large Samples: zc = the critical z-value s = the sample standard deviation n = the sample size When calculating the margin of error, round to one more decimal place than the original data, or the same number of places as the standard deviation.

HAWKES LEARNING SYSTEMS math courseware specialists Confidence Intervals 8.2 Estimating Population Means (Large Samples) Find the margin of error: Find the margin of error for a 99% confidence interval, given a sample of size 100 with a sample standard deviation of 15.50. Solution: n = 100, s = 15.50, c = 0.99 z0.99 = 2.575

HAWKES LEARNING SYSTEMS math courseware specialists Confidence Intervals 8.2 Estimating Population Means (Large Samples) Construct a confidence interval: A survey of 85 homeowners finds that they spend on average $67 a month on home maintenance with a standard deviation of $14. Find the 95% confidence interval for the mean amount spent on home maintenance by all homeowners. Solution: c = 0.95, n = 85, s = 14, = 67 z0.95 = 1.96 67 – 2.98 <  < 67 + 2.98 $64.02 <  < $69.98 ($64.02, $69.98)

zc = the critical z-value = the population standard deviation HAWKES LEARNING SYSTEMS math courseware specialists Confidence Intervals 8.2 Estimating Population Means (Large Samples) Finding the Minimum Sample Size for Means: To find the minimum sample size necessary to estimate an average, use the following formula: zc = the critical z-value = the population standard deviation E = the margin of error When calculating the sample size, round to up to the next whole number.

You will need a minimum sample size of 71. 2.575 HAWKES LEARNING SYSTEMS math courseware specialists Confidence Intervals 8.2 Estimating Population Means (Large Samples) Find the minimum sample size: Determine the minimum sample size needed if you wish to be 99% confident that the sample mean is within two units of the population mean, given that  = 6.5. Assume that the population is normally distributed. Solution: c = 0.99,  = 6.5, E = 2 z0.99 = You will need a minimum sample size of 71. 2.575

Z(0.98)=2.3263 n = ((2.3263 * 1.2) / 0.13) 2 n= 461.1

Z(0.99)=2.5758 n = ((2.5758 * 10.6) / 0.36) 2 n= 5752.3

E = 2.5758 x (2.2 / SQRT(1143)) E = 0.1676 4.3 – 0.1676 < μ < 4.3 + 0.1676

E = 1.6449 x (5.8 / SQRT(465)) E = 0.4424 28.3 – 0.4424 < μ < 28.3 + 0.4424