Section 7-3, 7.4. Estimating a Population Mean

Transcription

1 Section 7-3, 7.4 Slide 1 Estimating a Population Mean Assumptions: 1. The sample is a simple random sample. 2. The population is normally distributed or n > 30.

2 Point estimate Slide 2 Population mean: µ (unknown) Point Estimate: The sample mean: x σ Exact standard error: n Estimated standard error (se): s n

3 (i) 100(1-α)% Confidence Interval for Population Mean µ if σ is known point estimate ± margin of error Slide 3 x ± E (x E, x + E) E z = α / 2 σ n

4 Example: Constructing confidence interval Slide 4 A study found the body temperatures of 106 healthy adults. The sample mean was 98.2 degrees and the standard deviation was 0.62 degrees. Find (a) the point estimate of the population mean μ of all body temperatures degree (b) the margin of error E (c) the 95% confidence interval for µ.

5 Slide 5 Solution: (a) n = 106 x = o σ = 0.62 o (b) α = 0.05 α /2 = z α/ 2 = 1.96 E = z α/ 2 σ = = 0.12 n 106 (c) Recall x E < μ < x + E o 0.12 < μ < o o < μ < o

6 Sample Size for Estimating n = Mean μ (z α/2 ) σ E 2 Slide 6 Sample size formula with 95% confidence level and margin of error E is approximated by if σ is unknown n = 4s E 2 2

7 Round-Off Rule for Sample Size n Slide 7 When finding the sample size n, if it does not result in a whole number, always increase the value of n to the next larger whole number.

8 Finding the Sample Size n Slide 8 when σ is unknown 1. Use the range rule of thumb 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.

9 Example: Slide 9 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 = z = 1.96 E = 2 σ = 15 n = = = 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.

10 (ii) σ is not known Slide 10 Use Student t distribution

11 Important Properties of the Student t Distribution Slide The Student t distribution is different for different sample sizes (see Figure 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.

12 Table B Slide 12

13 Student t -score Slide 13 If the distribution of a population is essentially normal, then the distribution of t = x - µ s is essentially a Student t Distribution for all samples of size n Degrees of Freedom (df)=n-1 n

14 Margin of Error E for Estimating μ Slide 14 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.

15 100(1-α)%Confidence Interval for µ Slide 15 x s ± t ( ); df = a/2 n n -1 t α/2 found in Table B Based on an Unknown σ and a Small Simple Random Sample from a Normally Distributed Population A 95% confidence interval for the population mean µ is: x s ± t ( ); df =.025 n n -1

16 Procedure for Constructing a Confidence Interval for µ when σ is not known Slide Verify that the required assumptions are met. 2. Using n 1 degrees of freedom, refer to Table B: A3 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.

17 The Standard Normal Distribution is Slide 17 the t-distribution with df =

18 Example: Slide 18 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. n = 106 x = o s = 0.62 o α = 0.05 α /2 = t = E = t α/ 2 s = = n 106 x E < μ < x + E o < μ < o o < μ < o The interval is the same here as in Section 6-2, but in some other cases, the difference would be much greater.

19 Using the Normal and Flow Chart t Distribution Slide 19

20 Summary: Sections Slide 20 Point estimates: pˆ x > p > μ Margin of Error: E=(critical value)(standard error) Confidence Interval: (point estimate) ± E Sample size n: a solution of E=(critical value)(standard error)