Chapter 8 Confidence Interval Estimation Confidence Interval Example

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. Chapter 8 Confidence Interval Estimation

. Chapter 8 Confidence Interval Estimation

. Confidence Interval Example Cereal fill example n Population has µ = 368 and

. Confidence Interval Example Cereal fill example n Population has µ = 368 and σ = 15. n If you take a sample of size n = 25 you know n n 368 ± 1. 96 * 15 / = (362. 12, 373. 88). 95% of the intervals formed in this manner will contain µ. When you don’t know µ, you use X to estimate µ n n If X = 362. 3 the interval is 362. 3 ± 1. 96 * 15 / = (356. 42, 368. 18) Since 356. 42 ≤ µ ≤ 368. 18 the interval based on this sample makes a correct statement about µ. But what about the intervals from other possible samples of size 25?

. Confidence Interval Example (continued) Sample # X Lower Limit Upper Limit Contain µ?

. Confidence Interval Example (continued) Sample # X Lower Limit Upper Limit Contain µ? 1 362. 30 356. 42 368. 18 Yes 2 369. 50 363. 62 375. 38 Yes 3 360. 00 354. 12 365. 88 No 4 362. 12 356. 24 368. 00 Yes 5 373. 88 368. 00 379. 76 Yes

. General Formula n The general formula for all confidence intervals is: Point Estimate

. General Formula n The general formula for all confidence intervals is: Point Estimate ± (Critical Value)(Standard Error) Where: • Point Estimate is the sample statistic estimating the population parameter of interest • Critical Value is a table value based on the sampling distribution of the point estimate and the desired confidence level • Standard Error is the standard deviation of the point estimate

. Confidence Level, (1 - ) n n n Suppose confidence level = 95%

. Confidence Level, (1 - ) n n n Suppose confidence level = 95% Also written (1 - ) = 0. 95, (so = 0. 05) A relative frequency interpretation: n n (continued) 95% of all the confidence intervals that can be constructed will contain the unknown true parameter A specific interval either will contain or will not contain the true parameter n No probability involved in a specific interval

. Finding the Critical Value, Zα/2 n Consider a 95% confidence interval: Z units:

. Finding the Critical Value, Zα/2 n Consider a 95% confidence interval: Z units: X units: Zα/2 = -1. 96 Lower Confidence Limit 0 Point Estimate Zα/2 = 1. 96 Upper Confidence Limit

. Example n n A sample of 11 circuits from a large normal population

. Example n n A sample of 11 circuits from a large normal population has a mean resistance of 2. 20 ohms. We know from past testing that the population standard deviation is 0. 35 ohms. Determine a 95% confidence interval for the true mean resistance of the population.

. Example (continued) n n A sample of 11 circuits from a large normal

. Example (continued) n n A sample of 11 circuits from a large normal population has a mean resistance of 2. 20 ohms. We know from past testing that the population standard deviation is 0. 35 ohms. Solution:

. Interpretation n n We are 95% confident that the true mean resistance is

. Interpretation n n We are 95% confident that the true mean resistance is between 1. 9932 and 2. 4068 ohms Although the true mean may or may not be in this interval, 95% of intervals formed in this manner will contain the true mean

. Examples: Confidence Interval Estimate for the Mean (σ known) n Example 8. 1,

. Examples: Confidence Interval Estimate for the Mean (σ known) n Example 8. 1, page 277 n n n n Use Confidence function in excel Confidence (alpha, population standard deviation, sample size) Enter = confidence(. 05, . 02, 100) in cell A 1 Then add/subtract from mean to get interval Enter 10. 998 + value in A 1 in A 2 Enter 10. 998 – value in A 1 in A 3 This is your interval

. Confidence Interval for μ (σ Unknown) n Assumptions n n n (continued) Population

. Confidence Interval for μ (σ Unknown) n Assumptions n n n (continued) Population standard deviation is unknown Population is normally distributed If population is not normal, use large sample (n > 30) Use Student’s t Distribution Confidence Interval Estimate: (where tα/2 is the critical value of the t distribution with n -1 degrees of freedom and an area of α/2 in each tail)

. Student’s t Distribution n n The t is a family of distributions The

. Student’s t Distribution n n The t is a family of distributions The tα/2 value depends on degrees of freedom (d. f. ) n Number of observations that are free to vary after sample mean has been calculated d. f. = n - 1

. Degrees of Freedom (df) Idea: Number of observations that are free to vary

. Degrees of Freedom (df) Idea: Number of observations that are free to vary after sample mean has been calculated Example: Suppose the mean of 3 numbers is 8. 0 Let X 1 = 7 Let X 2 = 8 What is X 3? If the mean of these three values is 8. 0, then X 3 must be 9 (i. e. , X 3 is not free to vary) Here, n = 3, so degrees of freedom = n – 1 = 3 – 1 = 2 (2 values can be any numbers, but the third is not free to vary for a given mean)

. Student’s t Table Upper Tail Area df . 10 . 05 . 025

. Student’s t Table Upper Tail Area df . 10 . 05 . 025 1 3. 078 6. 314 12. 706 Let: n = 3 df = n - 1 = 2 = 0. 10 /2 = 0. 05 2 1. 886 2. 920 4. 303 3 1. 638 2. 353 /2 = 0. 05 3. 182 The body of the table contains t values, not probabilities 0 2. 920 t

. Example of t distribution confidence interval A random sample of n = 25

. Example of t distribution confidence interval A random sample of n = 25 has X = 50 and S = 8. Form a 95% confidence interval for μ n d. f. = n – 1 = 24, so The confidence interval is 46. 698 ≤ μ ≤ 53. 302

. Examples: Confidence Interval Estimate for the Mean (σ unknown) n Figure 8. 7,

. Examples: Confidence Interval Estimate for the Mean (σ unknown) n Figure 8. 7, page 282 n n Use TINV function in excel (because we’re using tdistribution) TINV(1 -confidence interval, degrees of freedom) n n Then get interval half width value (i. e. , +/- number) n n Enter = B 3 – B 11 in cell B 14 Take sample mean plus half width to get upper limit n n Enter = B 10*B 8 in cell B 11 Take sample mean minus half width to get lower limit n n Enter = TINV(1 -. 95, B 9) in cell B 10 Enter = B 3 + B 11 in cell B 15 B 14 and B 15 constitute the interval

. Confidence Interval Endpoints n n Upper and lower confidence limits for the population

. Confidence Interval Endpoints n n Upper and lower confidence limits for the population proportion are calculated with the formula where n n Zα/2 is the standard normal value for the level of confidence desired p is the sample proportion n is the sample size Note: must have np > 5 and n(1 -p) > 5

. Example n n A random sample of 100 people shows that 25 are

. Example n n A random sample of 100 people shows that 25 are left-handed. Form a 95% confidence interval for the true proportion of left-handers

. Example (continued) n A random sample of 100 people shows that 25 are

. Example (continued) n A random sample of 100 people shows that 25 are left-handed. Form a 95% confidence interval for the true proportion of left-handers.

. Interpretation n n We are 95% confident that the true percentage of left-handers

. Interpretation n n We are 95% confident that the true percentage of left-handers in the population is between 16. 51% and 33. 49%. Although the interval from 0. 1651 to 0. 3349 may or may not contain the true proportion, 95% of intervals formed from samples of size 100 in this manner will contain the true proportion.

. Examples: Confidence Interval Estimate for the Proportion n Figure 8. 12, page 288

. Examples: Confidence Interval Estimate for the Proportion n Figure 8. 12, page 288 n n Use NORMSINV function in excel to get Z-value NORMSINV(1 -confidence level/2), for two tails n n Get standard error of proportion n n Enter = B 7 – B 10 in cell B 13 Take sample proportion plus half width to get upper limit n n Enter = abs(B 8*B 9) in cell B 10 Take sample proportion minus half width to get lower limit n n Enter = sqrt(B 7*(1 -B 7)/B 2) in cell B 9 Then get interval half width value (i. e. , +/- number) n n Enter = NORMSINV((1 -. 95)/2) in cell B 8 Enter = B 7 + B 10 in cell B 14 B 13 and B 14 constitute the interval

. Determining Sample Size For the Mean Sampling error (margin of error)

. Determining Sample Size For the Mean Sampling error (margin of error)

. Determining Sample Size For the Mean Now solve for n to get (continued)

. Determining Sample Size For the Mean Now solve for n to get (continued)

. Required Sample Size Example If = 45, what sample size is needed to

. Required Sample Size Example If = 45, what sample size is needed to estimate the mean within ± 5 with 90% confidence? So the required sample size is n = 220 (Always round up)

. Determining Sample Size (continued) Determining Sample Size For the Proportion Now solve for

. Determining Sample Size (continued) Determining Sample Size For the Proportion Now solve for n to get

. Examples: Determine Sample Size for mean n Figure 8. 13, page 291 n

. Examples: Determine Sample Size for mean n Figure 8. 13, page 291 n n Use NORMSINV function in excel to get Z-value NORMSINV(alpha/2), for two tails n n Get sample size n n n (Z-value^2*standard deviation)/sampling error^2 Enter = =(B 7^2*B 2^2)/B 3^2) in cell B 8 Then round up to whole number n n Enter = NORMSINV((. 05)/2) in cell B 7 Use ROUNDUP function ROUNDUP(calculated sample size, 0) Enter = roundup(B 8, 0) in cell B 11 This is your sample size needed

. Required Sample Size Example How large a sample would be necessary to estimate

. Required Sample Size Example How large a sample would be necessary to estimate the true proportion defective in a large population within ± 3%, with 95% confidence? (Assume a pilot sample yields p = 0. 12)

. Required Sample Size Example (continued) Solution: For 95% confidence, use Zα/2 = 1.

. Required Sample Size Example (continued) Solution: For 95% confidence, use Zα/2 = 1. 96 e = 0. 03 p = 0. 12, so use this to estimate π So use n = 451

. Examples: Determine Sample Size for mean n Figure 8. 14, page 293 n

. Examples: Determine Sample Size for mean n Figure 8. 14, page 293 n n Use NORMSINV function in excel to get Z-value NORMSINV(1 -confidence level/2), for two tails n n Get sample size n n n (Z-value^2*estimate of true population*(1 -estimate of true population))/sampling error^2 Enter = ((B 7^2*B 2*(1 -B 2))/B 3^2) in cell B 8 Then round up to whole number n n Enter = NORMSINV((1 -. 95)/2) in cell B 7 Use ROUNDUP function ROUNDUP(calculated sample size, 0) Enter = roundup(B 8, 0) in cell B 11 This is your sample size needed