Chapter 6 Estimation and Sample Sizesmath.fau.edu/qian/course/sta2023/L14.pdf · Chapter 6...
Transcript of Chapter 6 Estimation and Sample Sizesmath.fau.edu/qian/course/sta2023/L14.pdf · Chapter 6...
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Chapter 6 Estimation and Sample SizesThis chapter presents the beginning of inferential statistics.
! The two major applications of inferential statistics
! Estimate the value of a population parameter! Test some claim (or hypothesis) about a population.
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! Estimation of Population parameters: proportions, means, and variances.
! Point estimate: Population proportion (p). (today)! Interval estimate: Confidence Interval for p.
! Calculate sample sizes needed to estimate those parameters.
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Point Estimate
p = population proportion
p =ˆ xn sample proportion
(pronounced ‘p-hat’)
of x successes in a sample of size n.Unbiased estimate (best estimate)
q = 1 - p = sample proportion of failures in a sample size of n
ˆ ˆ
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Example: Photo-Cop Survey Responses
" 829 adult Minnesotans were surveyed, and 51% of them are opposed to the use of the photo-cop for issuing traffic tickets. Using these survey results, find the best estimate of the proportion of all adult Minnesotans opposed to photo-cop use.
" Best point estimate=sample proportion=51%.
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Confidence Interval
! Why?: point estimate is not reliable under re-sampling.
! A confidence interval (CI): a range (or an interval) of values used to estimate the true population parameter.
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! αααα: between 0 and 1! A confidence level: 1 - αααα or 100(1- αααα)%. E.g. 95%.
This is the proportion of times that the confidence interval actually does contain the population parameter, assuming that the estimation process is repeated a large number of times.
! Other names: degree of confidence or the confidence coefficient.
Confidence Level
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The Critical Value zα/α/α/α/2
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Finding zα/2 for 95% Degree of Confidence
α/α/α/α/2 = 2.5% = .025
αααα = 5%
Critical value
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! The sampling distribution of sample proportion can be approximated by a normal distribution if np≥≥≥≥5 andnq ≥≥≥≥5 : phat is approximately N(p, pq/n), q=1-p.
p
Sampling Distribution of p̂
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Margin of Error, EWhen data from a simple random sample are used
to estimate a population proportion p, the margin of error, denoted by E, is the maximum likely (with probability 1 – αααα) difference between the observed proportion p and the true value of the
population proportion p.ˆ
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Margin of Error of p
E = α / α / α / α / 2 2 2 2 z p q̂ˆn
^
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100(1-α)% Confidence Interval for Population Proportion
p – E < < + Ep̂pˆ
where
zα / α / α / α / 2 2 2 2 E = nˆ ˆp q
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p – E < < + E
p + E
ˆ ˆ
ˆ
p p
(p – E, p + E)ˆ ˆ
Confidence Interval for Population Proportion
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Round-Off Rule for Confidence Interval Estimates of p
Round the confidence interval limits to
three significant digits
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Procedure for Constructing a Confidence Interval for p
! 1. Verify that the required assumptions are satisfied.Both np>=5 and nq>=5 are satisfied.
! 2. Refer to Table A-2 and find the critical value zαααα/2 that corresponds to the desired confidence level.
! 3. Evaluate the margin of error E = zαααα/2 ˆˆp qn
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! 4. Using the calculated margin of error, E and the value of the sample proportion, p, find the values of p – E and p + E. Substitute those values in the general format for the confidence interval:
ˆˆ
ˆ
p – E < p < p + Eˆ ˆ
!5. Round the resulting confidence interval limits to three significant digits.
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Example:In the Chapter Problem, we noted that 829 adult Minnesotans were surveyed, and 51% of them are opposed to the use of the photo-cop for issuing traffic tickets. Use these survey results.
a) Find the margin of error E that corresponds to a 95% confidence level.
b) Find the 95% confidence interval estimate of the population proportion p.
c) Based on the results, can we safely conclude that the majority of adult Minnesotans oppose use the the photo-cop?
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a) Find the margin of error E that corresponds to a 95% confidence level
First, we check for assumptions. We note that np = 422.79 ≥≥≥≥ 5, and nq = 406.21 ≥≥≥≥ 5.
Next, we calculate the margin of error. We have found that p = 0.51, q = 1 – 0.51 = 0.49, zα/α/α/α/2 = 1.96, and n = 829.
E = 1.96
E = 0.03403
829
(0.51)(0.49)
ˆ ˆ
ˆˆ
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b) Find the 95% confidence interval for the population proportion p.
We substitute our values from Part a to obtain
95% Confidence Interval for p is:
P - E < p < p +E0.51 – 0.03403 < p < 0.51 + 0.03403,
0.476 < p < 0.544
ˆ ˆ
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c) Based on the results, can we safely conclude that the majority of adult Minnesotans oppose use of the photo-cop?
Based on the survey results, we are 95% confident that the limits of 47.6% and 54.4% contain the true percentage of adult Minnesotans opposed to the photo-cop. The percentage of opposed adult Minnesotans is likely to be any value between 47.6% and 54.4%. However, a majority requires a percentage greater than 50%, so we cannot safely conclude that the majority is opposed (because the entire confidence interval is not greater than 50%).
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Determining Sample Size
α / α / α / α / 2222zE =p qnˆ ˆ
(solve for n by algebra)
( )2 ˆp qα / α / α / α / 2222zn = ˆE2
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Sample Size for Estimating Proportion p
When an estimate of p is known: ˆ
ˆ( )2 p qn = ˆE2
α / 2 α / 2 α / 2 α / 2 z
When no estimate of p is known:
( )2 0.25n =E2
α / α / α / α / 2 2 2 2 z
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Example: Suppose a sociologist wants to determine the current percentage of U.S. households using e-mail. How many households must be surveyed in order to be 95% confident that the sample percentage is in error by no more than four percentage points?
a) Use this result from an earlier study: In 1997, 16.9% of U.S. households used e-mail (based on data from The World Almanacand Book of Facts).
b) Assume that we have no prior information suggesting a possiblevalue of p.
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a) Use this result from an earlier study: In 1997, 16.9% of U.S. households used e-mail (based on data from The World Almanacand Book of Facts).
n = [za/2 ]2 p qE2
ˆˆ
0.042
= [1.96]2 (0.169)(0.831)
= 337.194= 338 households
To be 95% confident that our sample percentage is within four percentage points of the true percentage for all households, we should randomly select and survey 338 households.
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b) Assume that we have no prior information suggesting a possiblevalue of p.
0.042
n = [za/2 ]2 • 0.25E2
= (1.96)2 (0.25)
= 600.25 = 601 households
With no prior information, we need a larger sample to achieve the same results with 95% confidence and an error of no more than 4%.
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Finding the Point Estimate and E from a
Confidence Interval
Margin of Error:
E = (upper confidence limit) — (lower confidence limit)
2
Point estimate of p:
p = (upper confidence limit) + (lower confidence limit)
2ˆ
ˆ