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