STATISTICAL ANALYSIS - LECTURE 09
Transcript of STATISTICAL ANALYSIS - LECTURE 09
STATISTICAL ANALYSIS -LECTURE 09Dr. Mahmoud Mounir
STATISTICAL INFERENCE
Recap
Statistics
Descriptive Inferential
Estimation
Point Estimation
ยต = เดฅ๐ and p = เท๐
Interval
(Lower Limit, Upper Limit)
Hypothesis
Statistical Inference
โข The statistical inference is the process of making
judgment about a population based on the
properties of a random sample from the
population.
Estimators
Estimator (Sample Statistic)
Population Parameter
เดฅ๐ ESTIMATES ยต
S2 ESTIMATES ฯ2
S ESTIMATES ฯ
เท๐ ESTIMATES p
เดฅ๐๐- เดฅ๐๐ ESTIMATES ยต1 - ยต2
เท๐๐ โ เท๐๐ ESTIMATES p1 โ p2
Central Limit TheoremSampling Distribution of Sample Mean = Distribution of เดฅ๐
Sampling Distribution of sample Proprtion = Distribution of เท๐
Central Limit TheoremDifference Between Two Sample Means
Difference Between Two Sample Proportions
Estimation
Figure: The steps required to draw the statistical inference (estimation) of
an unknown parameter.
Recap ๐. ๐๐ ๐. ๐๐
โ๐๐ +๐๐P (Z < -Za) = P (Z > +Za)
P (Z < -Za) = 0.05
From the Z- Table, we can find the value of โZa
Recap ๐. ๐๐๐ ๐. ๐๐๐
โ๐๐ +๐๐P (Z < -Za) = P (Z > +Za)
P (Z < -Za) = 0.025
From the Z- Table, we can find the value of โZa
Recap ๐. ๐๐๐ ๐. ๐๐๐
โ๐๐ +๐๐P (Z < -Za) = P (Z > +Za)
P (Z < -Za) = 0.005
From the Z- Table, we can find the value of โZa
Standard Z-Score (Zฮฑ)
Standard Z-Score (Zฮฑ)
Standard Z-Score (Zฮฑ)
Confidence Interval (CI)
โข Is a range of values that likely would contain an
unknown population parameter.
โข This means that if we used the same sampling
method to select different samples and computed
an interval estimate for each sample, we would
expect the true population parameter to fall
within the interval estimates 95% of the time.
Confidence Interval (CI)
Population Parameter = Estimator ยฑ Margin of Error (d)
Where:
Margin of Error (d) = Critical Value of Z * Standard Error
โ [ ],Population Parameterโ
[Estimator โ Margin of Error (d) , Estimator + Margin of Error (d)]
Standard Error for Different Sample Statistics
Estimator (Sample Statistic)
Standard Error
เดฅ๐
เท๐
เดฅ๐๐- เดฅ๐๐
เท๐๐ โ เท๐๐
Confidence Interval (CI)
(1-ฮฑ) Confidence Level
ฮฑ Confidence Coefficient
โ๐(๐โ ฮค๐ถ ๐) ๐(๐โ ฮค๐ถ ๐)
Confidence Interval (CI)
If (1-ฮฑ) = 95% or 0.95
(1-ฮฑ) = 0.95เต๐ถ ๐ = ๐. ๐๐๐ เต๐ถ ๐ = ๐. ๐๐๐
Critical Value for Z:
We wish to estimate the average number of heart
beats per minute for a certain population. The
average number of heart beats per minute for a
sample of 49 subjects was found to be 90. Assume
that these 49 patients constitute a random sample,
and that the population is normally distributed with
a standard deviation of 10. Construct 90, 95, 99
percent confidence intervals for the population
mean.
Example (1)
Solution๐ = 49 าง๐ฅ = 90 ๐ = 10
90% confidence interval
๐ โ ๐ = ๐. ๐ ๐ = ๐. ๐๐
๐=. ๐๐
๐ โ๐
๐=. ๐๐
๐ณ๐โ๐/๐ = ๐ฉ ๐ณ < ๐/๐
= ๐ฉ ๐ณ < ๐. ๐๐ = ๐. ๐๐๐
๐ = เดค๐ฑ ยฑ ๐ณ๐โ๐/๐๐
๐ง
๐ = ๐๐ ยฑ ๐. ๐๐๐๐๐
๐๐= ๐๐ ยฑ ๐. ๐๐๐ โ ๐๐. ๐๐, ๐๐. ๐๐
Confidence level 1 โ ๐ผ
Confidence coefficient ๐ผ
1 โ ๐ผ/2 โ ๐ ๐ง1โ๐ผ/2 = ๐ ๐ง < ๐ผ/2
Solution๐ = 49 าง๐ฅ = 90 ๐ = 10
95% confidence interval
๐ โ ๐ = ๐. ๐๐ ๐ = ๐. ๐๐๐
๐=. ๐๐๐
๐ โ๐
๐=. ๐๐๐
๐ณ๐โ๐/๐ = ๐ฉ ๐ณ < ๐/๐
= ๐ฉ ๐ณ < ๐. ๐๐๐ = ๐. ๐๐๐
๐ = เดค๐ฑ ยฑ ๐ณ๐โ๐/๐๐
๐ง
๐ = ๐๐ ยฑ ๐. ๐๐๐๐๐
๐๐= ๐๐ ยฑ ๐. ๐
๐ โ ๐๐. ๐, ๐๐. ๐
Confidence level 1 โ ๐ผ
Confidence coefficient ๐ผ
1 โ ๐ผ/2 โ ๐ ๐ง1โ๐ผ/2 = ๐ ๐ง < ๐ผ/2
Solution๐ = 49 าง๐ฅ = 90 ๐ = 10
Confidence level 1 โ ๐ผ
Confidence coefficient ๐ผ
1 โ ๐ผ/2 โ ๐ ๐ง1โ๐ผ/2 = ๐ ๐ง < ๐ผ/2
99% confidence interval
๐ โ ๐ = ๐. ๐๐ ๐ = ๐. ๐๐๐
๐=. ๐๐๐
๐ โ๐
๐=. ๐๐๐
๐ณ๐โ๐/๐ = ๐ฉ ๐ณ < ๐/๐
= ๐ฉ ๐ณ < ๐. ๐๐๐ = ๐. ๐๐๐
๐ = เดค๐ฑ ยฑ ๐ณ๐โ๐/๐๐
๐ง
๐ = ๐๐ ยฑ ๐. ๐๐๐๐๐
๐๐= ๐๐ ยฑ ๐. ๐๐
๐ โ ๐๐. ๐, ๐๐. ๐
In a simple random sample of machine parts, 18
out of 225 were found to have been damaged in
shipment. Establish a 95% confidence interval
estimate for the proportion of machine parts that
are damaged in shipment.
Suppose there are 50,000 parts in the entire
shipment. How can we translate from the
proportions to actual numbers?
Example (2)
Solution๐ = ๐๐๐ ๐ = ๐๐ เท๐ =
๐๐
๐๐๐= ๐. ๐๐
๐ โ ๐ = ๐. ๐๐ ๐ = ๐. ๐๐๐
๐=. ๐๐๐
๐ โ๐
๐=. ๐๐๐ ๐ณ๐โ๐/๐ = ๐ฉ ๐ณ < ๐/๐ = ๐ฉ ๐ณ < ๐. ๐๐๐ = ๐. ๐๐๐
๐ฉ = เท๐ฉ ยฑ ๐ณ๐โ๐/๐เท๐ฉ ๐ โ เท๐ฉ
๐ง
๐ฉ = ๐. ๐๐ ยฑ ๐. ๐๐๐๐. ๐๐ ๐. ๐๐
๐๐๐
= ๐. ๐๐ ยฑ ๐. ๐๐๐ ๐. ๐๐๐๐๐= ๐. ๐๐ ยฑ ๐. ๐๐๐
๐ โ ๐. ๐๐๐, ๐. ๐๐๐
Thus, we are 95% certain that
the population of machine parts
damaged in shipment is between ๐. ๐๐๐ ๐๐๐ ๐. ๐๐๐
SolutionIf there are 50,000 parts in the entire shipment N=50,000 then
0.045 50000 = 22500.115 50000 = 5750
So, we can be 95% confident that there are between 2250 and
5750 defective parts in the whole shipment