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Sumukh Deshpanden
LecturerCollege of Applied Medical
Sciences
Statistics = Skills for life.
BIOSTATISTICS (BST 211)
Lecture 7
Confidence Intervals About m
RevisionSample
parameters are ESTIMATES of
population parameters
Sample parameters → Population parameters… (INFERENTIAL STATS)
Parameter
Mean m
SD S s
Sample
Population
Examplepediatric paracetamol syrup claims to contain 120ml.Measurement of 49 bottles revealed an average of 116ml.
Does this mean the bottle claim is wrong?
How sure can you be when you say the bottle contains 120ml?
116ml? X ml?
Summary
Sample mean is an ESTIMATE of population mean
Point estimate means
ONE single value.. One
number!What is written on the bottle should be m, the TRUE POPULATION MEAN
What you measured is….Sample mean, xbar
Point Estimate
A point estimate of a parameter is the value of a statistic that estimates the value of the parameter.
s is the best POINT ESTIMATE of s
But, what if the xbar
misses m? or s misses
s?
Confidence Interval Estimate
A confidence interval estimate of a parameter consists of an interval of numbers along with a probability that the interval contains the unknown parameter.
Confidence Level
Confidence
Interval
Confidence
LEVEL
Interval width (Error)
Higher confidence level → wider interval → less precise estimate
Confidence Intervals About m
s known
Back to Example1Measurement of 49 bottles of syrup revealed an average of 116ml. If the bottles are normally distributed and SD is known to be 7ml, estimate the true mean with 95% confidence?.
What do we know/Don’t know?
Sample: n = 49 Xbar =
116ml
Population: m = ? s = 7ml
We want to calculate the LOWER and UPPER limits
Solution page1Population SD, s = 7mlSample mean SD is SEM =So, SEM = 7/49 = 1mlWe want a 95% CI, so error is only 5%, a = 0.05Normal Distribution = symmetry… 0.025 on each sideFind z corresponding to P(z) = 0.025 or (1 - 0.025)
Solution page2
Solution Page3
We are 95%confident
that m is between 114.04 and 117.96 ml
Given this data, we can say that the claim of 120ml on syrup bottle
is 95% FALSE!!!
Summary of Process If the data follows
N and s is known: Have a sample
of size n and mean xbar
Have level of confidence CI%, then a = (100 – CI)
Interpretation of Confidence Interval
(1 – a)100% Confidence Interval: if you take many random samples of size (n≥30), from a normally distributed population of known SD = s, then (1 – a)100% of the intervals will contain m.
Example 2
Weight of watermelons from Bagaa follows a normal distribution with SD = 0.5 kg. 36 melons were weighed and averaged 8 kg. Estimate the true population mean weight with 90% CI? Also 99% CI?
What do we know/Don’t know?
Sample: n = 36 Xbar = 8 kg
Population: m = ? s = 0.5 kg
Required levels of confidence 90% and 99%. then a = (100 – CI)= 0.1 and 0.01respectively.
Summary of Process If the data follows
N and s is known: Have a sample
of size n and mean xbar
Have level of confidence CI%, then a = (100 – CI)
Solution page4Population SD, s = 0.5 kgSample mean SD is SEM =So, SEM = 0.5/36 = 0.083 kgWe want a 90% CI, so error is 10%, a = 0.1Normal Distribution = symmetry… 0.05 on each sideFind z corresponding to P(z) = 0.05 or (1 - 0.05)
Solution page5
Solution Page3
We are 90%confident
that m is between 8.14 and 7.86 kg
Now you work out the 99% confidence interval?
We are 99%confident
that m is between 8.19 and 7.81 kg
Summary of Lecture 1 Point Estimate
(single value) Interval Estimate SEM =
a = (100 – CL)
Confidence Interval Sample size n>30 Normal Distr. Population s known
Find za/2
Limits:
Summary of Lecture 2
True mea
n
sample
mean
How confiden
t?
Population SD
sample
size
Margin of ErrorEstimate