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Biostatistics Statistical Estimation Dr Remya.G

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Page 1: Statistical Estimation

BiostatisticsStatistical Estimation

Dr Remya.G

Page 2: Statistical Estimation

Statistics

\][ STATISTICAL ANALYSIS

DESCRIPTIVE INFERENTIALL

NUMERICAL GRAPHICAL

UNIVERIATE ?MULTIVARIATE

?

Estimation Hypothesis testing

Point estimate

Interval estimate Univariate

Multivariate

Page 3: Statistical Estimation
Page 4: Statistical Estimation

Inferential statistics

The part of statistics that allows researchers to generalize their findings to a larger population beyond data from the sample collected.

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Two ways to make inference

–Estimation of parameters * Point Estimation * Intervals Estimation–Hypothesis Testing

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Basic terminology

• Parameter –the numbers that describe the charactreistics of the population(mean, sd, varience etc)

• Statistic- The numbers that describe characteristics of scores in the sample (mean, variance, s.d., correlation coefficient,etc .)

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ParameterStatistic

Mean:

Standard deviation:

Proportion:

s

X _

__

_

from sample

from entire population

p

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Basic Logic

• Information from samples is used to estimate information about the population.

• Statistics are used to estimate parameters.

POPULATION

SAMPLE

PARAMETER

STATISTIC

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Estimation

The process by which one makes inferences about a population, based on information obtained from a sample.

Point estimateInterval estimate

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Point estimate

• Point estimates are single points that estimates parameter directly which serve as a "best guess" or "best estimate" of an unknown population parameter

• sample proportion pˆ (“p hat”) is the point estimate of p

• sample mean x (“x bar”) is the point estimate of μ

• sample standard deviation s is the point estimate of σ

Page 11: Statistical Estimation

Problem

• iIn a health survey of 55 school boys,it was found that the mean hemoglobin level was 10.2 g per 100 ml with a standard deviation of 2.1.Estimate the mean hemoglobin level of the population of such school boys.

Point estimate of the population mean is 10.2

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Disadvantages of point estimatesPoint estimate do not provide

information about sample to sample variability

How precise is x as an estimate of μ

How much can we expect x vary from μ

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Sampling distribution of the mean

X X X

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Sampling Distribution

• Sampling Distribution: A theoretical distribution that shows the frequency of occurrence of values of some statistic computed for all possible samples of size N drawn from some population.

• Sampling Distribution of the Mean: A theoretical distribution of the frequency of occurrence of values of the mean computed for all possible samples of size N from a population

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sampling distribution as N- increases Mean

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Central Limit Theorem

States that the sampling distribution of means, for samples of 30 or more:– Is normally distributed (regardless of the shape of the

population from which the samples were drawn)– Has a mean equal to the population mean, “mu” regardless

of the shape population or of the size of the sample– Has a standard deviation--the standard error of the mean--

equal to the population standard deviation divided by the square root of the sample size

Square root law

Page 17: Statistical Estimation
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Confidence interval

CI is the probability that the interval computed from the sample data includes the population parameter of interest

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FACTORS AFFECTING CONFIDENCE INTERVAL Distribution of Means and Standard Error of the Means

umu

+2sem-2sem +1sem-1sem-3sem +3sem

Population mean

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Confidence interval

Page 22: Statistical Estimation

Confidence limits

• The α (“alpha”) level represents the “lack of confidence”• (1−α)100% represent the confidence level of a

confidence interval

• Confidence interval =

• z1-α/2 instead of z1-α in this formula is because the random error (imprecision) is split between right and left tail

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Z values for different confidence level

Area under the curve

Page 24: Statistical Estimation

Z table 2 tailed

Area

und

er th

e cu

rve

Second decimal places

1.96=1.9+0.06

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Process for Constructing Confidence Intervals

• Compute the sample statistic (e.g. a mean)• Compute the standard error of the mean• Make a decision about level of confidence that is

desired (usually 95% or 99%)• Find tabled value for 95% or 99% confidence

interval• Multiply standard error of the mean by the tabled

value• Form interval by adding and subtracting

calculated value to and from the mean

Page 26: Statistical Estimation

Problems

• iIn a health survey of 55 school boys,it was found that the mean hemoglobin level was 10.2 g per 100 ml with a standard deviation of 2.1.Estimate the mean hemoglobin level of the population of such school boys.

Page 27: Statistical Estimation

Problems

• iIn a health survey of 55 school boys,it was found that the mean hemoglobin level was 10.2 g per 100 ml with a standard deviation of 2.1.Estimate the mean hemoglobin level of the population of such school boys.

X =10.2 s=2.1SE= =0.28395% CI= 10.2-1.96 x 0.283 to 10.2+ 1.96 x 0.283

=9.6 to 10.7599% CI= 9.47 to 10.93

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Problem

• In a survey on hearing level of schoolchildren with normal hearing it was found that in the frequency 500 cycles per second,62 children tested in the sound proof room had a mean hearing threshold of 15.5 db with a standard deviation of 6.5.Another 76 comparable children who were tested in the field had a mean threshold of 20 db with a standard deviation of 7.1.what is the 95% confidence interval for the difference in mean.

Page 29: Statistical Estimation

Here 2 independent samples,sound proof room tested and field tested sample given

The confidence interval of difference in means =difference in means +/_ 1.96 SE of difference in means

sqrt [ s21 / n1 + s2

2 / n2 ]

= 4.5-1.96x1.17 to 4.5+1.96x1.17= 2.21 to 6.79

SE of difference in means = Pooled SD x sqrt [1/ n1 + 1 / n2 ]

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Page 31: Statistical Estimation

Problemm

• In an otological examination of school children out of 146 children examined 21 were found to have otological abnormalities,Find the 99% confidence interval for the proportion of children with otological abnormalities.

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Answer

• p=21 x 100/146 = 14.4%• q= 85.3• 99% CI= p +/_2.57 SE of proportion• SE of proportion = √pq/n

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Problem

• Find the best estimate of the mean and 95% CI of the mean using the data

Sl no Protein value

1 6

2 7

3 8

4 6

5 8

6 7

7 6

8 7

9 8

10 6

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• Best estimate is the mean of sample= 6.9

• Interval estimate -95% CI= x +/- t0.05 SE of x

t0.05 is found from t table with df= 9

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• In case If 2 independent sample is given with sample size less than 30 and difference in CI to be found

• CI=difference in means +/_ t0.05 SE of difference in means

t0.05 found from the t table with df = n1+n2-2 SE of difference in means = use n-1 in the

equation for pooled sd

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T table

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An assumption about the population parameter.

I assume the mean SBP of population is 120 mmHg

What is a Hypothesis?

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How can u test this hypothesis ??????

Hypothesis testing – Next Class

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Thank you