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Estimating Population Values
Business Statistics Estimating Population Values
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Business Statistics Topic Index Probability Distribution
Sampling Distribution Point & Interval Estimates
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Types of Estimates Point Estimate Interval Estimate
A single number used to estimate an unknown population parameter Department Head: “Our current data indicate that this course will have 350 students in the fall” Characteristics Either right or wrong No estimate of reliability Interval Estimate A range of values used to estimate a population parameter Department Head: “I estimate that the true enrollment in this course in this fall will be between 330 & 380 and that it is very likely that the exact enrollment will fall within this interval” Better idea of reliability of estimate Decision making is facilitated; e.g. on the basis of estimate cancel one of the sections & offer an elective instead
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Point and Interval Estimates
A point estimate is a single number, a confidence interval provides additional information about variability Upper Confidence Limit Lower Confidence Limit Point Estimate Width of confidence interval
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Point Estimates Estimate Population Parameters …
with Sample Statistics Mean Proportion Variance Difference
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Confidence Intervals How much uncertainty is associated with a point estimate of a population parameter? An interval estimate provides more information about a population characteristic than does a point estimate Such interval estimates are called confidence intervals
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Confidence Interval Estimate
An interval gives a range of values: Takes into consideration variation in sample statistics from sample to sample Based on observation from 1 sample Gives information about closeness to unknown population parameters Stated in terms of level of confidence Never 100% sure
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Estimation Process Random Sample Population
I am 95% confident that μ is between 40 & 60. Random Sample Population Mean x = 50 (mean, μ, is unknown) Sample
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Point Estimate (Critical Value)(Standard Error)
General Formula The general formula for all confidence intervals is: Point Estimate (Critical Value)(Standard Error)
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Confidence Level Confidence Level
Confidence in which the interval will contain the unknown population parameter A percentage (less than 100%)
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Confidence Level, (1-) Suppose confidence level = 95%
Also written (1 - ) = .95 A relative frequency interpretation: In the long run, 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 No probability involved in a specific interval
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Confidence Intervals Confidence Intervals Population Mean Population
Proportion σ Known σ Unknown
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Confidence Interval for μ (σ Known)
Assumptions Population standard deviation σ is known Population is normally distributed If population is not normal, use large sample Confidence interval estimate for μ
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Finding the Critical Value
Consider a 95% confidence interval: z.025= -1.96 z.025= 1.96 z units: Lower Confidence Limit Upper Confidence Limit x units: Point Estimate Point Estimate
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Common Levels of Confidence
Commonly used confidence levels are 90%, 95%, and 99% Confidence Coefficient, Confidence Level z value, 80% 90% 95% 98% 99% 99.8% 99.9% .80 .90 .95 .98 .99 .998 .999 1.28 1.645 1.96 2.33 2.57 3.08 3.27
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Interval and Level of Confidence
Sampling Distribution of the Mean x Intervals extend from to x1 100(1-)% of intervals constructed contain μ; 100% do not. x2 Confidence Intervals
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Margin of Error Margin of Error (e): the amount added and subtracted to the point estimate to form the confidence interval Example: Margin of error for estimating μ, σ known:
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Factors Affecting Margin of Error
Intervals Extend from X - Z to X + Z x x Data variation, σ : e as σ Sample size, n : e as n Level of confidence, 1 - : e if
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Case Study 4.A 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 .35 ohms. Determine a 95% confidence interval for the true mean resistance of the population.
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Solution – Case Study 4.A Solution:
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 .35 ohms. Solution:
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Interpretation We are 98% confident that the true mean resistance is between and ohms Although the true mean may or may not be in this interval, 98% of intervals formed in this manner will contain the true mean An incorrect interpretation is that there is 98% probability that this interval contains the true population mean. (This interval either does or does not contain the true mean, there is no probability for a single interval)
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Case Study 4.B Police Chief Ackert has recently instituted a crackdown on drug dealers in her city. Since the crackdown began, 750 of the 12,368 drug dealers in the city have been caught. The mean dollar value of drugs found on these 750 dealers is $250,000. The SD of the dollar value of drugs for these 750 dealers is $41,000. Construct a 90% confidence interval for the mean dollar value of drugs possessed by the city’s drug dealers.
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Determining Sample Size
The required sample size can be found to reach a desired margin of error (e) and level of confidence (1 - ) Required sample size, σ known:
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Required Sample Size Example
If = 45, what sample size is needed to be 90% confident of being correct within ± 5? So the required sample size is n = 220 (Always round up)
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If σ is unknown If unknown, σ can be estimated when using the required sample size formula Use a value for σ that is expected to be at least as large as the true σ Select a pilot sample and estimate σ with the sample standard deviation, s
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Confidence Intervals Confidence Intervals Population Mean Population
Proportion σ Known σ Unknown
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Confidence Intervals for the Population Proportion, p
An interval estimate for the population proportion ( p ) can be calculated by adding an allowance for uncertainty to the sample proportion ( p )
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Confidence Intervals for the Population Proportion, p
Recall that the distribution of the sample proportion is approximately normal if the sample size is large, with standard deviation We will estimate this with sample data:
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Confidence interval endpoints
Upper and lower confidence limits for the population proportion are calculated with the formula where z is the standard normal value for the level of confidence desired p is the sample proportion n is the sample size
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Example A random sample of 100 people shows that 25 are left-handed. Form a 95% confidence interval for the true proportion of left-handers. 1. 2. 3.
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Interpretation We are 95% confident that the true percentage of left-handers in the population is between 16.51% and 33.49%. Although this range 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.
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Changing the sample size
Increases in the sample size reduce the width of the confidence interval. Example: If the sample size in the above example is doubled to 200, and if 50 are left-handed in the sample, then the interval is still centered at .25, but the width shrinks to .19 …… .31
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Finding the Required Sample Size for proportion problems
Define the margin of error: Solve for n: p can be estimated with a pilot sample, if necessary (or conservatively use p = .50)
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What sample size...? 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 = .12)
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What sample size...? Solution: For 95% confidence, use Z = 1.96
p = .12, so use this to estimate p So use n = 451
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Determining Sample Size for Mean
What sample size is needed to be 90% confident of being correct within ± 5? A pilot study suggested that the standard deviation is 45. Round Up
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Confidence Intervals Confidence Intervals Population Mean Population
Proportion σ Known σ Unknown
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Confidence Interval for μ (σ Unknown)
If the population standard deviation σ is unknown, we can substitute the sample standard deviation, s This introduces extra uncertainty, since s is variable from sample to sample So we use the t distribution instead of the normal distribution
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Determining Sample Size (Cost)
Too small: Won’t do the job Too Big: Requires too many resources
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Confidence Interval for μ (σ Unknown)
Assumptions Population standard deviation is unknown Population is normally distributed If population is not normal, use large sample Use Student’s t Distribution Confidence Interval Estimate
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Approximation for Large Samples
Since t approaches z as the sample size increases, an approximation is sometimes used when n 30: Technically correct Approximation for large n
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Student’s t Distribution
Standard Normal Bell-Shaped Symmetric ‘Fatter’ Tails t (df = 13) t (df = 5) Z t
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Degrees of Freedom (df )
Number of observations that are free to vary after sample mean has been calculated Example: Mean of 3 numbers is 2 degrees of freedom = n -1 = 3 -1 = 2
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Student’s t Table Let: n = 3 df = n - 1 = 2 = .10 /2 =.05 .05 2 t
Upper Tail Area df .25 .10 .05 1 1.000 3.078 6.314 / 2 = .05 2 0.817 1.886 2.920 3 0.765 1.638 2.353 t 2.920 t Values
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With comparison to the z value
t distribution values With comparison to the z value Confidence t t t z Level (10 d.f.) (20 d.f.) (30 d.f.) ____ Note: t z as n increases
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Example
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Confidence Interval Estimate for Proportion
Assumptions Two categorical outcomes Population follows binomial distribution Normal approximation can be used if and Confidence interval estimate
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Example A random sample of 400 voters showed 32 preferred candidate A. Set up a 95% confidence interval estimate for p.
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Determining Sample Size for Proportion
Out of a population of 1,000, we randomly selected 100, of which 30 were defective. What sample size is needed to be within ± 5% with 90% confidence? Round Up
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Confidence Interval for Population Total Amount
Point estimate Confidence interval estimate
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Confidence Interval for Population Total: Example
An auditor is faced with a population of 1000 vouchers and wants to estimate the total value of the population. A sample of 50 vouchers is selected with average voucher amount of $ , standard deviation of $ Set up the 95% confidence interval estimate of the total amount for the population of vouchers.
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Example Solution The 95% confidence interval for the population total amount of the vouchers is between 1,000,559.15, and 1,152,220.85
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Confidence Interval for Total Difference in the Population
Point estimate Where is the sample average difference Confidence interval estimate Where
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Estimation for Finite Population
Samples are selected without replacement Confidence interval for the mean ( unknown) Confidence interval for proportion
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Sample Size Determination for Finite Population
Samples are selected without replacement When estimating the mean When estimating the proportion
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Case Study LR An apartment manager wants to inform potential renters about how much electricity they can expect to use during August. She randomly selects 61 residents and discovers their average electricity usage in August to be 894 kwh. She believes the variance in usage to be 131 kwh2. Establish an interval estimate for the average August electricity usage so that the apartment manager can be 68.3% certain that the true population mean lies within this interval Repeat part (a) with 99.7% certainty If the price per kwh is $0.12, within what interval can the apartment manager be 68.3% certain that the average August cost for electricity will lie?
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Case Study LR 356 SC 7.3 For a population with a known variance of 185, a sample of 64 individuals leads to 217 as an estimate of the mean Find the standard error of the mean Establish an interval estimate that should include the population mean 68.3% of the time.
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Case Study LR The manager of Cardinal Electric’s light bulb division must estimate the average number of hours that a light bulb made by each light bulb machine will last. A sample of 40 light bulbs was selected from machine A and the average burning time was 1,416 hours. The standard deviation of burning time is known to be 30 hours. Compute the standard error of the mean Construct a 90% confidence interval for the true population mean.
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Case Study LR 365 SC 7.7 In an automotive safety test conducted by the North Carolina Highway Safety Research Center, the average tire pressure in a sample of 62 tires was found to be 24 pounds per square inch, and the standard deviation was 2.1 pounds per square inch. What is the estimated population standard deviation for this population? (There are about a million cars registered in North Carolina.) Calculate the estimated standard error of the mean. Construct a 95% confidence interval for the population mean.
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Case Study LR 369 SC 7.8 When a sample of 70 retail executives was surveyed regarding the poor November performance of the retail industry, 66% believed that decreased sales were due to unseasonably warm temperatures, resulting in consumers’ delayed purchase of cold-weather items. Estimate the standard error of the proportion of retail executives who blame warm weather for low sales. Find the upper & lower confidence limits for this proportion, given a 95% confidence level.
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Case Study LR For a year and a half, sales have been falling consistently in all 1,500 franchises of a fast-food chain. A consulting firm has determined that 31 percent of a sample of 95 indicate clear signs of mismanagement. Construct a 98% confidence interval for this proportion.
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Case Study LR Twelve bank teller were randomly sampled and it was determined they made an average of 3.6 errors per day with a sample standard deviation of .42 error. Construct a 90% confidence interval for the population mean of errors per day. What assumption is implied about the number of errors bank tellers make?
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Case Study LR A local store that specializes in candles and clocks is interested in obtaining an interval estimate for the mean number of customers that enter the store daily. The owners are reasonably sure that the actual standard deviation of the daily number of customers is 15 customers. Help the store out of the fix by determining the sample size it should use in order to develop a 96% confidence interval for the true mean that will have a width of only 8 customers.
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