Download confidence interval

SUMMARY
• Z-distribution
• Central limit theorem
Sweet demonstration of the sampling
distribution of the mean
Sweet data
𝑛 = 20
𝜇 = 5.05
𝜎 = 2.31
R-code – sampling distribution exact
data.set <c(6,4,5,3,10,3,5,3,6,5,4,8,7,2,8,5,8,5,4,0)
mean(data.set)
sd(data.set)*sqrt(19/20) #standard deviation
(sd(data.set)*sqrt(19/20))/sqrt(20) sample_size<-5
samps <- combn(data.set, sample_size)
xbars <- colMeans(samps)
barplot(table(xbars))
Sampling distribution – exact
𝜇𝑥 = 𝑀 = ? ?
𝜎𝑥 = 𝑆𝐸 = ? ?
𝑀 = 𝜇 = 5.05
𝜎
SE =
= 0.52
20
R-code (sampling distribution simulated)
data.set <c(6,4,5,3,10,3,5,3,6,5,4,8,7,2,8,5,8,5,4,0)
sample_size<-3
number_of_samples<-20
samples <replicate(number_of_samples,sample(data.set,
sample_size, replace=T)); out<-colMeans(samples);
mean(out); sd(out)
barplot(table(out))
Sampling distribution – simulated
𝑀 = 5.05
SE = 0.52
Sampling distribution – simulated
𝑀 = 5.05
SE = 0.52
ESTIMATION
Statistical inference
If we can’t conduct a census, we
collect data from the sample of a
population.
Goal: make conclusions about
that population
Demonstration problem
• You sample 36 apples from your farm’s harvest of over
200 000 apples. The mean weight of the sample is 112
grams (with a 40 gram sample standard deviation).
• What is the probability that the mean weight of all 200 000
apples is within 100 and 124 grams?
What is the question?
• We would like to know the probability that the population
mean is within 12 of the sample mean.
𝑃 𝜇 is within 12 of 𝑥
• But this is the same thing as
𝑃 𝑥 is within 12 of 𝜇
• But this is the same thing as
𝑃 𝑥 is within 12 of 𝑀
• So, if I am able to say how many standard deviations
away from 𝑀 I am, I can use the Z-table to figure out the
probability.
Slight complication
• There is one caveat, can you see it?
• We don’t know a standard deviation of a sampling
distribution (standard error). We only know it equals to
𝜎 6, but 𝜎 is uknown.
• What we’re going to do is to estimate 𝜎. Best thing we can
use is a sample standard deviation 𝑠, that equals to 40.
• 𝑆𝐸 =
40
6
≈ 6.67. This is our best estimate of a standard
error.
• Now you finish the example. What is the probability that
population mean lies within 12 of the sample if the SE
equals to 6.67?
• 92.82%
This is neat!
• You sample 36 apples from your farm’s harvest of over
200 000 apples. The mean weight of the sample is 112
grams (with a 40 gram sample standard deviation). What
is the probability that the population mean weight of all
200 000 apples is within 100 and 124 grams?
• We started with very little information (we know just the
sample statistics), but we can infere that
with the probability of 92.82% a population mean lies
within 12 of our sample mean!
Point vs. interval estimate
• You sample 36 apples from your farm’s harvest of over
200 000 apples. The mean weight of the sample is 112
grams (with a 40 gram sample standard deviation).
• Goal: estimate a population mean
1. A population mean is estimated as a sample mean. i.e.
we say a population mean equals to 112 g. This is
called a point estimate (bodový odhad).
2. However, we can do better. We can estimate, that our
true population mean will lie with the 95% confidence
within an interval of (interval estimate).
𝑠
𝑥 ± 1.96 ×
𝑛
Confidence interval
• This type of result is called a confidence interval
(interval spolehlivosti, konfidenční interval).
𝑠
𝑥±𝑍×
𝑛
critical value
kritická hodnota
margin of error
možná odchylka
• The number of stadandard errors you want to
add/subtract depends on the confidence level (e.g. 95%)
(hladina spolehlivosti).
Confidence level
• The desired level of confidence is set by the researcher
(not determined by data).
• If you want to be 95% confident with your results, you add/subtract
1.96 standard errors (empirical rule says about 2 standard errors).
• 95% interval spolehlivosti
Confidence level Z-value
80
1.28
90
1.64
95
1.96
98
2.33
99
2.58
80%
90%
1.28
1.64
95%
99%
1.96
2.58
Small sample size confidence intervals
• 7 patient’s blood pressure have been measured after
having been given a new drug for 3 months. They had
blood pressure increases of 1.5, 2.9, 0.9, 3.9, 3.2, 2.1 and
1.9. Construct a 95% confidence interval for the true
expected blood pressure increase for all patients in a
population.
CLT consequence
• Change in a blood pressure is a biological process. It’s
going to be a sum of thousands or millions of microscopic
processes.
• Generally, if we think about biological/physical process,
they can be viewed as being affected by a large number
of random subprocesses with individually small effects.
• The sum of all these random components creates a
random variable that converges to a normal
distribution regardless of the underlying distribution of
processes causing the small effects.
• Thus, the Central Limit Theorem explains the ubiquity of
the famous "Normal distribution" in the measurements
domain.
• We will assume that our population distribution is normal,
with 𝜇 and 𝜎.
• We don’t know anything about this distribution but we
have a sample. Let’s figure out everything you can figure
out about this sample:
• 𝑥 = 2.34, 𝑠 = 1.04
• We’ve been estimating the true population standard
deviation with our sample standard deviation
𝜎 ≈ 𝑠 = 1.04
• However, we are estimating our standard deviation with 𝑛
of only 7! This is probably goint to be not so good
estimate.
• In general, if 𝒏 < 𝟑𝟎 this is considered a bad estimate.
William Sealy Gosset aka Student
• 1876-1937
• an employee of Guinness
brewery
• 1908 papers addressed the
brewer's concern with small
samples
• "The probable error of a mean".
Biometrika 6 (1): 1–25. March 1908.
• Probable error of a correlation
coefficient". Biometrika 6 (2/3): 302–
310. September 1908.
Student t-distribution
• Instead of assuming a sampling distribution is normal we
will use a Student t-distribution.
• It gives a better estimate of your confidence interval if you
have a small sample size.
• It looks very similar to a normal distribution, but it has
fatter tails to indicate the higher frequency of outliers
which come with a small data set.
Student t-distribution
Student t-distribution
df – degree of freedom
(stupeň volnosti)
df = 𝑛 − 1
Back to our case
• 𝑥 = 2.34, 𝑠 = 1.04, 𝑛 = 7
• Because a sample size is small, sampling distribution of
the mean won’t be normal. Instead, it will have a Student
t-distribution with d. f. = 6.
• Construct a 95% confidence interval, please
2.34 ± 2.447 ×
1.04
7
= 2.34 ± 0.962
𝑠
for 𝑛 < 30: 𝑥 ± 𝑡𝑛−1 ×
𝑛
neco ×
•
Just to summarize, the margin of error depends on
the confidence level (common is 95%)
2. the sample size 𝑛
1.
•
•
3.
the variability of the data (i.e. on σ)
•
•
•
as the sample size increases, the margin of error decreases
For the bigger sample we have a smaller interval for which
we’re pretty sure the true population lies.
more variability increases the margin of error
Margin of error does not measure anything else
than chance variation.
It doesn’t measure any bias or errors that happen
during the proces.
•
It does not tell anything about the correctness of your
data!!!
𝑠
𝑛