Determining Sample Size Slide 1 E = z α / 2 ˆ ˆ p q n (solve for n by algebra) n = ( zα α / 2) 2 p ˆ qˆ E 2
Sample Size for Estimating Proportion p When an estimate of ˆp is known: Slide 2 n = ˆ ˆ ( ) 2 p q zα α / 2 E 2 Formula 6-2 When no estimate of p is known: n = ( ) 2 0.25 zα α / 2 E 2 Formula 6-3
Slide 3 Example: Suppose a sociologist wants to determine the current percentage of U.S. households using e-mail. How many households must be surveyed in order to be 95% confident that the sample percentage is in error by no more than four percentage points? a) Use this result from an earlier study: In 1997, 16.9% of U.S. households used e-mail (based on data from The World Almanac and Book of Facts). b) Assume that we have no prior information suggesting a possible value of p. ˆ 1- α=95%=.95, α=0.05 z =z α/2.025 =1.96. E=4%=0.04 (a) phat=16.9%=.169 (b) Use phat=0.5
Slide 4 a) Use this result from an earlier study: In 1997, 16.9% of U.S. households used e-mail (based on data from The World Almanac and Book of Facts). n = [z a/2 ] 2 p q E 2 ˆ ˆ = [1.96] 2 (0.169)(0.831) 0.04 2 = 337.194 = 338 households To be 95% confident that our sample percentage is within four percentage points of the true percentage for all households, we should randomly select and survey 338 households.
Slide 5 b) Assume that we have no prior information suggesting a possible value of p. ˆ n = [z a/2 ] 2 0.25 E 2 = (1.96) 2 (0.25) 0.04 2 = 600.25 = 601 households With no prior information, we need a larger sample to achieve the same results with 95% confidence and an error of no more than 4%.
Finding the Point Estimate and E from a Confidence Interval Slide 6 Point estimate of p: ˆ p ˆ = (upper confidence limit) + (lower confidence limit) 2 Margin of Error: E = (upper confidence limit) (lower confidence limit) 2
Section 6-3 Estimating a Population Mean: Assumptions: σ Known Slide 7 1. The sample is a simple random sample. 2. The value of the population standard deviation σ is known. 3. Either or both of these conditions is satisfied: The population is normally distributed or n > 30.
Point estimate! Population mean: µ (unknown) Slide 8! Point Estimate is a single value (or point) used to approximate a population parameter. Example: The sample mean x is the best point estimate of the population mean µ.
Slide 9 Example: A study found the body temperatures of 106 healthy adults. The sample mean was 98.2 degrees and the sample standard deviation was 0.62 degrees. Find the point estimate of the population mean µ of all body temperatures. Because the sample mean x is the best point estimate of the population mean µ, we conclude that the best point estimate of the population mean µ of all body temperatures is 98.20 o F.
Interval Estimate Level of Confidence Slide 10! The confidence level =1 - α,!where α is the complement of the confidence level.! For a 0.95(95%) confidence level, α = 0.05. For a 0.99(99%) confidence level, α = 0.01.
Slide 11 Margin of Error is the maximum likely difference observed between sample mean x and population mean µ, and is denoted by E. E = z α/2 σ n Standard Error of the sample mean
Confidence Interval (or Interval Estimate) for Population Mean µ when σ is known Slide 12 x E < µ < x + E x + E (x E, x + E)
n = 106 x = 98.20 o s = 0.62 o Slide 13 Example: A study found the body temperatures of 106 healthy adults. The sample mean was 98.2 degrees and the sample standard deviation was 0.62 degrees. Find the margin of error E and the 95% confidence interval for µ. E = z α/ 2 σ = 1.96 0.62 = 0.12 n 106 α = 0.05 α /2 = 0.025 z α/ 2 = 1.96 x E < µ < x + E 98.08 o < µ < 98.32 o 98.20 o 0.12 < µ < 98.20 o + 0.12 98.08 o < µ < 98.32 o
Sample Size for Estimating Mean µ Slide 14 n = (z α/2 ) σ E 2
Round-Off Rule for Sample Size n Slide 15 When finding the sample size n, if the use of Formula 6-5 does not result in a whole number, always increase the value of n to the next larger whole number.
Finding the Sample Size n Slide 16 when σ is unknown 1. Use the range rule of thumb (see Section 2-5) to estimate the standard deviation as follows: σ range/4. 2. Conduct a pilot study by starting the sampling process. Based on the first collection of at least 31 randomly selected sample values, calculate the sample standard deviation s and use it in place of σ. 3. Estimate the value of σ by using the results of some other study that was done earlier.
Slide 17 Example: Assume that we want to estimate the mean IQ score for the population of statistics professors. How many statistics professors must be randomly selected for IQ tests if we want 95% confidence that the sample mean is within 2 IQ points of the population mean? Assume that σ = 15, as is found in the general population. α = 0.05 α /2 = 0.025 z α/ 2 = 1.96 E = 2 σ = 15 n = 1.96 15 2 = 216.09 = 217 2 With a simple random sample of only 217 statistics professors, we will be 95% confident that the sample mean will be within 2 points of the true population mean µ.
6-4: σ Not Known Assumptions Slide 18 1) The sample is a simple random sample. 2) Either the sample is from a normally distributed population, or n > 30. Use Student t distribution
Student t Distribution Slide 19 If the distribution of a population is essentially normal, then the distribution of t = x - µ s n! is essentially a Student t Distribution for all samples of size n, and is used to find critical values denoted by t α/2.
Degrees of Freedom (df ) Slide 20 corresponds to the number of sample values that can vary after certain restrictions have been imposed on all data values df = n 1 in this section.
Margin of Error E for Estimate of µ Slide 21 Based on an Unknown σ and a Small Simple Random Sample from a Normally Distributed Population E = t α / 2 s n where t α / 2 has n 1 degrees of freedom.
Student t Distributions for n = 3 and n = 12 Slide 22
Important Properties of the Student t Distribution Slide 23 1. The Student t distribution is different for different sample sizes (see Figure 6-5 for the cases n = 3 and n = 12). 2. The Student t distribution has the same general symmetric bell shape as the normal distribution but it reflects the greater variability (with wider distributions) that is expected with small samples. 3. The Student t distribution has a mean of t = 0 (just as the standard normal distribution has a mean of z = 0). 4. The standard deviation of the Student t distribution varies with the sample size and is greater than 1 (unlike the standard normal distribution, which has a σ = 1). 5. As the sample size n gets larger, the Student t distribution gets closer to the normal distribution.
Confidence Interval for Slide 24 the Estimate of E Based on an Unknown σ and a Small Simple Random Sample from a Normally Distributed Population x E < µ < x + E where E = t α /2 s n t α /2 found in Table A-3
Procedure for Constructing a Confidence Interval for µ when σ is not known 1. Verify that the required assumptions are met. Slide 25 2. Using n 1 degrees of freedom, refer to Table A- 3 and find the critical value t α/ 2 that corresponds to the desired degree of confidence. 3. Evaluate the margin of error E = t α/ 2 s / n. 4. Find the values of x - E and x + E. Substitute those values in the general format for the confidence interval: x E < µ < x + E 5. Round the resulting confidence interval limits. α/
n = 106 x = 98.20 o s = 0.62 o Slide 26 Example: A study found the body temperatures of 106 healthy adults. The sample mean was 98.2 degrees and the sample standard deviation was 0.62 degrees. Find the margin of error E and the 95% confidence interval for µ. E = t α/ 2 s = 1.984 0.62 = 0.1195 n 106 α = 0.05 α /2 = 0.025 t α/ 2 = 1.984 x E < µ < x + E 98.20 o 0.1195 < µ < 98.20 o + 0.1195 98.08 o < µ < 98.32 o The interval is the same here as in Section 6-2, but in some other cases, the difference would be much greater.
Using the Normal and t Distribution Slide 27 Figure 6-6