Shahzad Bashir. 1
Chapter 9 Estimation & Confidence Interval Interval Estimation for Population Mean: σ Known Interval Estimation for Population Mean: σ Unknown Determining the Sample Size 2
A point estimate is a single value (statistic) used to estimate a population value (parameter). An Interval Estimatestates the range within which a population parameter probably lies. A confidence intervalis a range of values within which the population parameter is expected to occur. The two confidence intervals that are used extensively are the 95%and the 99%. 3
Interpretation of Interval Estimation For a 95%confidence interval about 95%of the similarly constructed intervals will contain the parameter being estimated. For the 99%confidence interval, 99%of the sample means for a specified sample size will lie within 2.58standard deviations of the hypothesized population mean. 95%of the sample means for a specified sample size will lie within 1.96standard deviations of the hypothesized population mean. 4
Margin of Error and the Interval Estimate A point estimator cannot be expected to provide the exact value of the population parameter. An interval estimate can be computed by adding and subtracting a margin of error to the point estimate. Point Estimate +/ Margin of Error The purpose of an interval estimate is to provide information about how close the point estimate is to the value of the parameter. 5
Margin of Error and the Interval Estimate (Continued) The general form of an interval estimate of a population mean is x ± Margin of Error Point Estimate 6
Interval Estimation of a Population Mean: σ Known (Continued) In order to develop an interval estimate of a population mean,, the margin of error must be computed using either: the population standard deviation σ,, or the sample standard deviation s σ is rarely known exactly, but often a good estimate can be obtained based on historical data or other information. We refer to such cases as the σ knowncase. 7
Interval Estimation of a Population Mean: σ Known (Continued) There is a 1 α probability that the value of a sample mean will provide a margin of error of z or less. Sampling distribution of x α/2 1 - α of all α/2 x values α/2 σ x z α/2 σ x µ z α/2 σ x x 8
Interval Estimate of a Population Mean: σ Known (Continued) interval does not include µ Sampling distribution of x α/2 1 - α of all x values α/2 x z α/2 σ [------------------------- -------------------------] x µ z σ x x x α/2 [------------------------- -------------------------] [------------------------- -------------------------] x interval includes µ 9
Interval Estimate of a Population Mean: σ Known (Continued) Interval Estimate of µ Point Estimation of Population Mean where: x x σ ± z α n /2 Margin of Error is the sample mean 1 -αα is the confidence coefficient z α/2 is the zvalue providing an area of α/2 in the upper tail of the standard normal probability distribution σ is the population standard deviation n is the sample size 10
Interval Estimate of Population Mean: σ Known Example: Discount Sounds Discount Sounds has 260 retail outlets throughout the United States. The firm is evaluating a potential location for a D new outlet, based in part, on the mean annual income of the individuals in the marketing area of the new location. A sample of size n= = 36was taken; the sample mean income is $31,100. The population standard deviation is estimated to be $4,500, and the confidence coefficient to be used in the interval estimate is 0.95. 11
Interval Estimate of Population Mean: σ Known D The margin of error is: z α/2 σ 4,500 = 1.96 = 1, 470 n 36 Thus, at 95% confidence, the margin of error is $1,470. Note: To find the Z from the table do the following: α/2 =.05/2=.025 and 1-.025 =.975 and from table Z is 1.96 12
Interval Estimate of Population Mean: σ Known D Interval estimate of µ is: x σ ± z α /2 n $31,100 + $1,470 or $29,630 to $32,570 We are 95% confident that the interval contains the population mean. Note that the sample mean was = $31,100. 13
Interval Estimation of a Population Mean: σ Unknown If an estimate of the population standard deviation σ cannot be developed prior to sampling, we use the sample standard deviation sto estimate σ. This is the σ unknown case. In this case, the interval estimate for µis based on the t distribution. 14
t Distribution The t distribution is a family of similar probability distributions. A specific t distribution depends on a parameter known as the degrees of freedom. Degrees of freedom refer to the number of independent pieces of information that go into the computation of s. 15
t Distribution (Continued) A t distribution with more degrees of freedom has less dispersion. As the number of degrees of freedom increases, the difference between the t distribution and the standard normal probability distribution becomes smallerand and smaller. 16
t Distribution (Continued) Standard normal distribution t distribution (20 degrees of freedom) t distribution (10 degrees of freedom) 0 z, t 17
Student s t Table 18
Student s t Table Upper Tail Area df.25.10.05 1 1.000 3.078 6.314 2 0.817 1.886 2.920 3 0.765 1.638 2.353 19
Student s t Table Upper Tail Area df.25.10.05 1 1.000 3.078 6.314 2 0.817 1.886 2.920 3 0.765 1.638 2.353 t values 20
Student s t Table Upper Tail Area α / 2 df.25.10.05 1 1.000 3.078 6.314 2 0.817 1.886 2.920 3 0.765 1.638 2.353 t values 0 α / 2 t 21
Student s t Table Upper Tail Area df.25.10.05 1 1.000 3.078 6.314 α / 2 Assume: n = 3 df = n - 1 = 2 α =.10 α / 2 =.05 2 0.817 1.886 2.920 3 0.765 1.638 2.353 t values 0 α / 2 t 22
Student s t Table Upper Tail Area df.25.10.05 1 1.000 3.078 6.314 α / 2 Assume: n = 3 df = n - 1 = 2 α =.10 α / 2 =.05 2 0.817 1.886 2.920 3 0.765 1.638 2.353 t values 0 α / 2 t 23
Student s t Table Upper Tail Area df.25.10.05 1 1.000 3.078 6.314 α / 2 Assume: n = 3 df = n - 1 = 2 α =.10 α / 2 =.05 2 0.817 1.886 2.920 3 0.765 1.638 2.353 t values 0.05 t 24
Student s t Table Upper Tail Area df.25.10.05 1 1.000 3.078 6.314 α / 2 Assume: n = 3 df = n - 1 = 2 α =.10 α / 2 =.05 2 0.817 1.886 2.920 3 0.765 1.638 2.353.05 t values 0 2.920 t 25
Interval Estimation of a Population Mean: σ Unknown Interval Estimate Point Estimation of Population Mean x ± t α/2 s n Margin of Error where: 1 -α = the confidence coefficient t α/2 =the t value providing an area of α/2α in the upper tail of a t distribution with n-1 1 degrees of freedom s = the sample standard deviation 26
Interval Estimation of a Population Mean: σ Unknown Example: Apartment Rents A reporter for a student newspaper is writing an article on the cost of off-campus housing. A sample of 16 efficiency apartments within a half-mile of campus resulted in a sample mean of $650per month and a sample standard deviation of $55. 27
Interval Estimation of a Population Mean: σ Unknown (Example Continued) Example: Apartment Rents Let us provide a 95%confidence interval estimate of the mean rent per month for the population of efficiency apartments within a half-mile of campus. 28
Interval Estimation of a Population Mean: σ Unknown (Example Continued) At 95% confidence, α= =.05, and α/2 =.025. t.025.025 is based on n 1 1 = 16 1 1 = 15 degrees of freedom. In the t distribution table we see that t.025 = 2.131. Degrees Area in Upper Tail of Freedom.20.100.050.025.010.005 15.866 1.341 1.753 2.131 2.602 2.947 16.865 1.337 1.746 2.120 2.583 2.921 17.863 1.333 1.740 2.110 2.567 2.898 18.862 1.330 1.734 2.101 2.520 2.878 19.861 1.328 1.729 2.093 2.539 2.861....... 29
Interval Estimation of a Population Mean: σ Unknown (Example Continued) Interval Estimate x ± t.025 s n 55 650 ± 2.131 = 650 ± 29.30 16 We are 95%confident that the mean rent per month for the population of efficiency apartments within a half-mile of campus is between $620.70and $679.30. 30
Summary of Interval Estimation Procedures for a Population Mean Yes σ Known Case Can the population standard deviation σ be assumed known? No Use the sample standard deviation s to estimate x ± z Use α/2 σ n σ Unknown Case & Small Sample x ± t Use α/2 s n 31
If σ is unknownand n>30, the standard deviation of the sample, designated by s, is used to approximate the population standard deviation. Interval Estimation Summary X ± z s n If the population standard deviation (σ ) is known or the sample (n) is n 30we use the z distribution. 32
If the population standard deviation (σ) is unknown, and the underlying population is approximately normal, and the sample size is less than 30(n<30) we use the t distribution. Interval Estimation Summary s X ± t n The value oft for a given confidence level depends upon its degrees of freedom. 33
More Examples Confidence Interval Estimate for Mean (σknown)
Thinking Challenge Example You re a Q/C inspector for Gallo. The σfor 2-liter bottles is 0.05liters. A random sample of 100 bottles showedthat sample mean = 1.99liters. What is the 90% confidence interval estimate of the true meanamount amount in 2-liter bottles? To find the Z: α =1-90% =0.1 and α/2 =0.1/2 =0.05 1 0.05 = 0.95 and from table, the Z is: (1.64 + 1.65)/ 2 = 1.645 2 liter 35
Confidence Interval Solution X σ Z µ α / 2 X + Zα / 2 n σ n. 05 199. 1645. µ 199. + 1645. 100. 05 100 1982. µ 1998. We are 90% confident that interval estimate of the true meanamount amount in 2-liter bottles is between 1.98 and 1.99. 36
Confidence Interval of µ (σunknown and n 30) Example The School of Business Dean at CSUwants to estimate the mean number of hours worked per week by business students. A sample of 49students showed a mean of 24hours with a standard deviation of 4 hours. What is the pointestimate of the mean number of hours worked per week by students? The point estimate is 24 hours (sample mean). What is the 95%confidence interval for the average number of hours worked per week by the students? 37
Example & Solution(Continued) Using the formula, we have 24 ±1.96(4/7) or we have 22.88to 25.12. What are the 95% confidence limits? The endpoints of the confidence interval are the confidence limits. The lower confidence limit is 22.88and the upper confidence limit is 25.12. What degree of confidence is being used? The degree of confidence (level of confidence) is 0.95. 38
Example & Solution(Continued) Interpret the findings. If we had time to select 100 samples of size 49 from the population of the number of hours worked per week by business students at CSU and compute the sample means and 95% confidence intervals, the population mean of the number of hours worked by the students per week would be found in about 95 out of the 100 confidence intervals. Either a confidence interval contains the population mean or it does not. In this example, about 5out of the 100confidence intervals would not contain the population mean. 39
More Examples Confidence Interval Estimate for Mean (σunknown, and n<30)
Thinking Challenge Example You re a time study analyst in manufacturing. You ve recorded the following task times (min.): 3.6, 4.2, 4.0, 3.5, 3.8, 3.1. What is the 90% confidence interval estimate of the population mean task time? 41
X X = 3.7 S= = 0.38987 Solution n= = 6, df = n-1 1 = 6-1 = 5 S/ n = 3.8987 / 6 6 = 0.1592 t.05,5 = 2.0150 3.7 -(2.015)(0.1592) µ 3.7 + (2.015)(0.1592) 3.385 µ 4.015 We are 90% confident that the interval estimate of the population mean task time is between 3.4and 4.0 minutes. 42
Confidence Interval Excel Application Given the price of ten (10) houses in one of the subdivisions located in Henry County, use Excel to construct a 95%confidence interval for the population mean. Data: $230,000, $240,000, $310,000, $198,000, $257,000, $345,000, $315,000, $260,000, $198,000, $270,000. 43
Excel Solution--SWStat Data Area 44
Excel Solution SWStat SWStat (Continued) (SWStat Statistics Intervals & Tests) 45
Excel Solution SWStat SWStat (Continued) $227,099 µ $297,502 46
Confidence Interval Excel Application (Another Problem) A random sample of 36 magazine subscribers is taken to estimate the mean age of all subscribers. Use Excelto construct a 90% confidence interval estimate of the mean age of all of this magazine s subscribers. See next slide for the data. 47
The Data Subscriber Age Subscriber Age Subscriber Age 1 39 13 40 25 38 2 27 14 35 26 51 3 38 15 35 27 26 4 33 16 41 28 39 5 40 17 34 29 35 6 35 18 46 30 37 7 51 19 44 31 33 8 36 20 44 32 41 9 47 21 43 33 36 10 28 22 32 34 33 11 33 23 29 35 46 12 35 24 33 36 37 48
Excel Solution--SWStat 49
Excel Solution (Continued) SWStat 50
Sample Size for an Interval Estimate of a Population Mean Let E= = the desired margin of error. We said Eis the amount added to and subtracted from the point estimate to obtain an interval estimate. x s ± t α/2 E = Margin of Error n Interval Estimate of the mean 51
Sample Size for an Interval Estimate of a Population Mean (Continued) Margin of Error E Necessary Sample Size σ = z α /2 n Margin of Error n z = ( α/ 2 ) E 2 2 2 σ 52
Sample Size for an Interval Estimate of a Population Mean--Example Recall that Discount Sounds is evaluating a potential location for a new retail outlet, based in part, on the mean annual income of the individuals in the marketing area of the new location. Suppose that Discount Sounds management team wants an estimate of the population mean such that there is a 0.95probability that the sampling error is $500 or less. How large a sample size is needed to meet the required precision? D 53
Sample Size for an Interval Estimate of a Population Mean--Solution D E = z α σ /2 = 500 n Given At 95% confidence, z.025 = 1.96. Recall that σ = 4,500. n z = ( α/ 2 ) E = 2 2 2 σ 2 2 (1.96) (4, 500) n = = 311.17 = 312 2 (500) A sample of size 312is needed to reach a desired precision of + $500 at 95% confidence. 54
Sample Size for an Interval Estimate of a Population Mean (Continued) There are 3 factors that determine the size of a sample, none of which has any direct relationship to the size of the population. They are: 1- The degree of confidence selected. 2- The maximum allowable error-- --margin of error. 3- The variation of the population. n z = ( α/ 2 ) E 2 2 2 σ 55
Thinking Challenge Sample Size Example 1 A consumer group would like to estimate the meanmonthly monthly electric bill for a single family house in July. Based on similar studies the standard deviation is estimated to be $20.00. A 99%level of confidence is desired, with an accuracy of ± $5.00. How large a sample is required? n=[(2.58)(20)/5] 2 = 106.5024 107 n z = ( α/ 2 ) E 2 2 2 σ 56
Thinking Challenge Sample Size Example 1 (Continued) What sample size is needed to be 90%confident of being correct within ±5?? A pilot study suggested that the standard deviation is 45. 57
Thinking Challenge Sample Size Example 1 (Continued) What sample size is needed to be 90%confident of being correct within ±5?? A pilot study suggested that the standard deviation is 45. Note that in this example boththe the degree of confidence and population standard deviation are changed hence, the sample size is changed too. 58
Thinking Challenge Sample Size Example 2 You work in Human Resources at Merrill Lynch. You plan to survey employees to find their average medical expenses. You want to be 95%confident that the sample mean is within ±$50. A pilot study showed that sample standard deviation was about $400. What sample sizedo you use? 59
Thinking Challenge Sample Size Example 2 (Solution) 60
Interval Estimation of a Population Proportion OPTIONAL READINGS 61
End of Chapter 9 62