The (“Sampling”) Distribution for the Sample Mean*
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Transcript of The (“Sampling”) Distribution for the Sample Mean*
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The (“Sampling”) Distribution for the Sample Mean*
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Distribution of Sample Means
A quantitative population of N units with parameters
mean standard deviation
A random sample of n units from the populationStatistic: The sample mean .X
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Distribution of Sample Means
Statistic: The sample mean .This statistic is an unbiased point estimate (on average correct) of the parameter .
X
X
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20 Times Rule / 5% Rule(same thing)
If the population size (N) is at least 20 times the sample size (n)
N / n 20 or n / N 0.05
then the standard deviation is (essentially)
nX
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Distribution of the Sample MeanGiven
A variable with population that is not Normally distributed with mean and standard deviation .
A random sample of size n.
ResultThe sample mean has approximate Normal distribution with
X nX
When the population size is at least 20 times n.
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ExampleRolls of paper leave a factory with weights that are Normal with mean = 1493 lbs, and standard deviation = 12 lbs.
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Finding probabilitiesWhat is the probability a roll weighs over 1500 lbs?
ANS: 0.2798
(about 28% of rolls exceed 1500 lbs)
5833.012
14931500
Z
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New QuestionA truck transports 8 rolls at a time. The legal weight limit for the truck is 12,000 lbs. What is the probability 8 rolls have total weight exceeding this limit?
Since 12000/8 = 1500, the question could also be phrased:
What is the probability 8 rolls have (sample) mean weight exceeding 1500?
The bad news: The answer is not 0.2798.
The good news: It’s not that tough.
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Distribution of the Sample MeanGiven
A variable with population that is Normally distributed with mean and standard deviation .
A random sample of size n. (N/n 20)
ResultThe sample mean has Normal distribution:
X nX
Review of previous slide
X
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Example - continuedRolls (single rolls) of paper leave a factory with weights that are Normal with mean = 1493 lbs, and standard deviation = 12 lbs.
If n = 8 rolls are randomly selected, what is the probability their sample mean weight exceeds 1500?
The distribution of sample means is Normal.
1493X 243.48
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nX
X
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Finding probabilitiesFind the probability the sample mean is over 1500 lbs.Here we’re using the same mean, but a standard deviation reduced to 4.243. ANS: 0.0495
650.1243.47
243.414931500
Z
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Interpreting the ResultThe probability the sample mean for 8 rolls exceeds 1500 lbs is 0.0495.
For 4.95% of all possible samples of 8 rolls, the sample mean exceeds 1500 lbs.
Equivalent: There is a 0.0495 probability that the total weight will exceed 81500 = 12,000 lbs.
We’re working towards using the sample mean as an estimate of the population mean.
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The Picture
Weight (lbs)153315231513150314931483147314631453
1500
Weights of single rolls.Sample mean weights for samples of 8 rolls.
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The Picture
About 28% of all rolls are > 1500 lbs
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The PictureAbout 5% of all
samples of 8 rolls have mean > 1500 lbs
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ExampleSurvival times have a right skewed distribution with mean = 13 months and standard deviation = 12 months.
What can we say about the distribution of sample mean survival times for samples of n patients?
As n gets larger, the distribution gets closer to Normal.
nnX0.12
13X
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6050403020100
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Single values
SD = 12.0
Sample mean n = 4SD = 6.0
Sample mean n = 16SD = 3.0
Sample mean n = 64SD = 1.5
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Distribution of the Sample MeanGiven
A variable with population that is not Normally distributed with mean and standard deviation .
A random sample of size n.
ResultThe sample mean has approximate Normal distribution with
X nX
Assume the population size is at least 20 times n.
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Distribution of the Sample MeanGiven
A variable with population that is not Normally distributed with mean and standard deviation .
A random sample of size n.
ResultThe sample mean has generally unknown distribution with
X nX
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Distribution of the Sample MeanGiven
A variable with population that is not Normally distributed with mean and standard deviation .
A random sample of size n, where n is sufficiently large.
ResultThe sample mean has approximate Normal distribution with
X nX
Central Limit Theorem (CLT)
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What is “Sufficiently Large?”Your book says “generally n at least 30.”
If the population is fairly symmetric without outliers, considerably less than 30 will do the trick.If the population is highly skewed, or not unimodal, considerably more than 30 may be required.
If the population is Normal then sample size is not a concern: The sample mean is Normal.You may use the “30” rule if you recognize that it’s not that black and white, and that for Normal populations, n = 1 is “sufficiently large.”
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ExampleThe Census Bureau reports the average age at death for female Americans is 79.7 years, with standard deviation 14.5 years.
= 79.7 years = 14.5 yearsWhat can we say about the distribution of sample means for samples of size 7? It has mean It has standard deviation Is the distribution Normal?
7.79X
48.575.14
nX
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ExampleDistribution of longevity: 80 15
Within 1 s.d.:
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ExampleDistribution of longevity: 80 15
If Normal
Within 1 s.d.: (65, 95)
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ExampleDistribution of longevity: 80 15
If Normal
Within 1 s.d.: (65, 95) 68%
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ExampleDistribution of longevity: 80 15
If Normal
Within 1 s.d.: (65, 95) 68%
Within 2 s.d.s: (50, 110) 95%
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ExampleDistribution of longevity: 80 15
If Normal
Within 1 s.d.: (65, 95) 68%
Within 2 s.d.s: (50, 110) 95%
Above 110
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ExampleDistribution of longevity: 80 15
If Normal
Within 1 s.d.: (65, 95) 68%
Within 2 s.d.s: (50, 110) 95%
Above 110 2.5%
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ExampleDistribution of longevity: 80 15
If Normal
Within 1 s.d.: (65, 95) 68%
Within 2 s.d.s: (50, 110) 95%
Above 110 2.5%
1 in 40 ???
No way! The distribution is not Normal.
309075604530
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8
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2
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Age at Death (years) for Women
Perc
ent o
f Wom
enExample
The Normal shouldn’t be used here (why not?)
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ExampleThe Normal shouldn’t be used here (why not?)
The distribution of age at death is not Normal. It is quite left skewed.The sample size is not sufficiently large. (At least 30 by your book, although for this situation your instructor would probably buy into as low as 20.)
The Central Limit Theorem can’t be applied.The sample mean doesn’t have approximate Normal distribution
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ExampleWhat can we say about the distribution of sample means for samples of size 7?
It has mean
It has standard deviation
Is the distribution Normal?
NO!
7.79X
48.575.14
nX
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Example = 79.7 years = 14.5 years
I looked at a few recent obituaries in the Oswego Daily News (online):
79 70 48 99 85 71 45
36.1900.71 SX
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Example
This sample has . A difference of 8.7.
Can we compute a Z score for 71.0? Should we?
Z = (71.0 – 79.7) /5.48 = 8.7/5.48 = –1.59
Why not? This suggests 71.0 (8.7 from 79.7) is somewhat, but not extremely, unusually low. 71.0 is 1.59 standard deviations from 79.7.
7.79X 48.575.14
nX
0.71X
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ExampleShould we use the Table to obtain probabilities from Z scores (such as our Z = –1.59)?
NOIf not, how could we get the probabilityof a result within 8.7 from 79.7?Using either
a huge database of longevities:Simulate many (all possible) samples of size 7. Determine what proportion of samples give a mean at no more than 8.7 from 79.7.
a mathematical model for the longevitiesEither determine the model for sample means using calculus, or approximate it using numerical methods.
Preferred method: Much more compact; faster to work with; essentially identical results.
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ExampleWhat is the distribution of the sample mean of samples of size n = 48?
Even though age at death is left skewed, with n = 48 (large enough) the Central Limit Theorem applies, and the sample mean has approximate Normal distribution.
7.79X 09.248
5.14
nX
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ExampleI looked at 41 more recent obituaries (total of 48) 79 70 48 99 85 71 45 more data 87 75 90 95 51 99 69 71 49 93 80 89 77 72101 69 92 92 86 78 92 89 91 81 74 68 89 92 64 71 50 81 88 42 91 44 51 85 81 92 93
37.1652.77 SX
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100908070605040
MedianMean
90858075
95% Confidence Intervals
Example
Mean
Median
Mode
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ExampleMeans for samples of 48 US longevities: NormalMy sampleThe sample mean is (79.7 – 77.52) = 2.18 from the population mean.What is the probability that a random sample of 48 U.S. women’s deaths gives a sample mean at within 2.18 of 79.7.
2.18 below 79.7 is 77.52.2.18 above 79.7 is 81.88
7.79X52.77X09.2X
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77.52
0.7054
81.8879.7
Normal, Mean=79.7, StDev=2.08
ExampleBelow 77.52 or above 81.88.Z = 2.18/2.08 = 1.04Probability = 0.852 – 0.148 = 0.704
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ExampleFind the probability that a random sample of 48 U.S. women’s deaths gives a sample mean at within 2.18 of 79.7.Probability = 0.704About 30% (that’s almost 1 in 3) of all samples of 48 deaths give a sample mean more than 2.18 from 79.7.
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ExampleGive two explanations that account for the 2.18 year difference between the data on Oswego longevity (which were lower on average) and the U.S. longevity parameter of 79.7.1. Women in Oswego do not live as long on average as they do nationwide. That is:
Oswego< 79.7
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ExampleGive two explanations that account for the 2.18 year difference between the data on Oswego longevity (which were lower on average) and the U.S. longevity parameter of 79.7.
2. Sampling variability (sampling “error”):
Oswego= 79.7
About 30% of all samples of 48 women yield a mean 2.18 or more from 79.7. That isn’t so uncommon. Our data aren’t very inconsistent with the national result.
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Sampling Without Replacement
What to do if the sample size is more than 5% of the population size…
N= population size
n = sample size
N / n 20 n / N ≤ 0.05
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Distribution of Sample MeansThe distribution of the sample mean has
> mean (“unbiased”)
> standard deviation
> shape closer to Normal(but not necessarily Normal)
X
1
N
nNnX
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1110987654321Individual Word Lengths
Dotplot of word length
Each symbol represents up to 2 observations.
Word Lengths – Gettysburg Address
N = 268 words: Mean length = 4.295.
Standard Deviation = 2.123.
Not Normal. Right skewed. Can’t use Table A2.
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Distribution of Sample Means: n = 5Sample means from samples of size n = 5 have
> mean> standard deviation
> shape closer to Normal (but not Normal – a bit right skewed)
295.4X
12685268
5123.2
1
N
nNnX
942.09925.09494.02672639494.0
X
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8.07.67.26.86.46.05.65.24.84.44.03.63.22.82.42.01.6Sample Mean Word Length
Each symbol represents up to 22 observations.
4.295
Distribution of Sample Means : n = 5The mean of this distribution is 4.295.
The standard deviation of this distribution is 0.942.
The shape is close to Normal (but not Normal – there’s right skew).
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Distribution of Sample Means : n = 10Sample means from samples of size n = 10 have
> mean> standard deviation
> shape closer to Normal (but not exactly Normal – a bit right skewed)
295.4X
126810268
10123.2
1
N
nNnX
660.09830.06714.02672586714.0
X
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8.07.67.26.86.46.05.65.24.84.44.03.63.22.82.42.01.6Sample Mean Word Length
Each symbol represents up to 31 observations.
4.295 The mean of this distribution is 4.295.
The standard deviation of this distribution is 0.660.
The shape is quite close to Normal (just a little right skew – not enough to fuss over).
Distribution of Sample Means : n = 10
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n = 5
n = 10
126810268
10123.2
1
N
nNnX
660.09830.06714.02672586714.0
12685268
5123.2
1
N
nNnX
942.09925.09494.02672639494.0
Awful close to 1
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n = 5
n = 10
126810268
10123.2
1
N
nNnX
660.09830.06714.02672586714.0
12685268
5123.2
1
N
nNnX
942.09925.09494.02672639494.0
Almost the same.
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n = 100
1268100268
100123.2
1
N
nNnX
1684.07932.02123.02671682123.0
Not so close to 1
Not almost the same.
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Distribution of the Sample MeanGiven
A variable with population that is distributed with mean and standard deviation .
A random sample of size n. PARAMETERS
Results 1 and 2 STATISTIC
The sample mean has distribution with the same mean and a smaller standard deviation.
X 1
N
nNnX
X
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Distribution of the Sample MeanGiven
A variable with population that is distributed with mean and standard deviation .
A random sample of size n.
Results 3The sample mean has distribution with a shape that is closer to Normal.
X 1
N
nNnX
X