Estimation with Random Samples: Point, Interval, and Confidence Intervals, Study notes of Business Statistics

Download Estimation with Random Samples: Point, Interval, and Confidence Intervals and more Study notes Business Statistics in PDF only on Docsity! 1 Estimation Using Random Samples Stat 201 Prof. Yanni Papadakis Key Terms § Point Estimate § Mean § Std. Deviation § Proportion § Interval Estimate § Mean, s known § Mean, s unknown § t-distribution § Proportion § Determination of Sample Size 2 Unbiased Point Estimates Population Sample Parameter Statistic Formula • Mean, µ • Variance, σ2 • Proportion, π x x = xi∑n 1– 2)–(22 n xixss ∑= p p = x successesn trials Example: Estimating m, s § Bob has been weighing himself once a week for the past month. His measurements are: 190.5 189.0 195.5 187.0 § Estimate his average weight § Estimate the std. deviation of his measurements § His long term average weight is X~N(190lb,5lb). What is the probability his average weight in 4 measurements is the one calculated above or higher? 5 Confidence Intervals: Overview Strategy: Provide a region inside which we expect with great confidence real parameters to be (confidence level is usually 95% or 90%) e.g. forecast proportion in 52±2% § For mean m § s known (use Normal Distribution) § s unknown (use t-Distribution) § For proportion p (use Normal Approximation) CI for m, s known § Confidence level ( C ) § Usually: C=95% § Otherwise clearly stated, say C=90% § Calculate: § Sample mean § Std. Error = s / sqrt(n) § Critical z* (typically z*=1.96) § Margin of Error = z* Std.Error § CI = Sample Mean +/- Margin of Error 6 CI for m, s unknown § Confidence level ( C ) § Usually: C=95% § Otherwise clearly stated, say C=90% § Valid if: § Underlying distribution NORMAL § Sample size n>30 (CLT) § Calculate: § Sample mean § Sample std. deviation (s) § Std. Error = s / sqrt(n) § Degrees of Freedom, df = n-1 § Critical t*=t ( a=(1-C)/2 , df ) § Margin of Error = t* Std.Error § CI = Sample Mean +/- Margin of Error CI for the Mean: Quality Control § Long-term std. deviation of rod diameters is s=0.053in. A sample taken for quality control purposes had size n=30 and sample average equal to 1.4in. § Calculate a CI for the true mean at level=95% § Calculate a CI for the true mean at level=90% 7 CI for the Mean: Quality Control { } 419.1 381.1 019.4.1)95(. 96.1z 0.00968in 30 053. 4.1 * = = ±= = === = UCL LCL CI n x x σ σ CI for the Mean: Quality Control { } 416.1 384.1 .0164.1)95(. 645.1z 0.00968in 30 053. 4.1 * = = ±= = === = UCL LCL CI n x x σ σ 10 CI for the Mean: Example ( ) { } 53.56 28.25 14.1462.39)95(. 25.39110,025.t 35.4 10 75.13 42.39 0.99)(CMean for Interval Confidence * = = ±= ==−=== ==== = UCL LCL CI dft n s sx x α Example: Supermarket Shoppers § A marketing consultant observed 50 consecutive shoppers at a supermarket. One variable of interest was how much each shopper spent in the store. The sample statistics were: mean=$27.40, st.dev=$22.03. § Calculate the 95% CI for the mean shopper expenditure under similar conditions. § The supermarket needs an estimate with margin of error at most equal to $4. What should the sample size be in order to meet this requirement? 11 Example: Supermarket Shoppers ( ) { } 116.5 4 03.22 96.1 03.22 96.1t4 trequiremen width CI 33.6614.21 26.640.27)95(. 01.249150,025.t 12.3 50 03.22 40.27 0.95)(CMean for Interval Confidence 2 * * =     ≥⇔≅≥ == ±= ==−=== ==== = n nn s UCLLCL CI dft n s sx x α CI: Population Proportion ( ) ( ) 576.2960.1645.1 %99%95%90 :z : scores-z Critical 1zpCI :Thus 1, :met are sassumption approx. Normalwhen Recall * * C n pp n ppp~N −±=       −π 12 Proportion CI: Example § In a USA Today/CNN poll, 1406 adults were randomly selected from across the US. In response to the question: “Do you agree that the current system discourages the best candidates from running for president?” 22% responded “strongly agree.” § Calculate 95% CI for proportion “strongly agree.” Proportion CI: Example ( ) ( ) [ ] 0.5 p for Error ofMargin Calculate 242.0,198.0 1406 22.0122.0 96.122.0 1zpCI * = = − ±= −±= n pp 15 Required Sample Size: Example ( )( ) ( )( ) ( )( ) 752integer Min 7.751 03.0/645.15.015.0 90.0 1068integer Min 1.106703.0/96.15.015.0 /1 0.5 useunknown 2 2 2* = = −≥ = = =−= −≥ n n C n ezppn p When Population is Finite ( ) ( ) ( ) ( ) N pp ze pp N nN n pp zp − + − ≥       − −− ± 1 / 1 n :ionDeterminat Size Sample 1 1 :CI 2* * 16 Finite Population Example § FAA lists 8719 pilots holding commercial helicopter certificates. It wants to calculate the proportion of pilots planning to switch to another job in the next three years. § Design a study to estimate this proportion with margin of error 4% at C=0.95. § Is this result different to assuming infinite population? Finite Population Example ( ) ( ) ( ) ( ) ( ) ( ) 562ninteger Min 6.561 8719 5.015.0 96.1/04.0 5.015.0 1 / 1 unknown) (p 0.5p Use 2 2* = = − + − = − + − ≥ = N pp ze pp n 17 Exercise § The ABA survey of community banks also asked about the loan-to-deposit ratio (LTDR), a bank’s total loans as percent of total deposits. The mean LTDR for the 110 banks in a random sample is 76.7 and the standard deviation s=12.3. § The sample size is large enough to assume population s=s. Calculate the 95% CI for LTDR.