Statistical Inference Statistical Inference

Lab ThreeLab Three

Bernoulli to Normal Through BinomialBernoulli to Normal Through Binomial

One flip Fair coin

Heads

Tails

Random Variable: k, # of headsp=0.5

1-p=0.5

For n flips, the prob. of k heads is [n!/k!(n-k)!]pk (1-p)n-k

For a sample of size n, the sample proportion of heads is p-hat = k/n, where p-hat is distributed binomially with mean p and variance p*(1-p)/n

For np>=5 n(1-p)>=5, Approximate Distribution of p-hatWith the normal distribution, p-hat~N[p, p*(1-p)/n]

Simulate a random sample of n=50Simulate a random sample of n=50

Simulate Ten Sample ProportionsSimulate Ten Sample Proportions

95% confidence Interval On First 95% confidence Interval On First sample proportion of 0.44sample proportion of 0.44

p

p

p

pervalconfidenceei

pob

pob

pob

nppwhere

ppob

ˆ

ˆ

ˆ

*96.1ˆint%95..

95.0]30.058.0[Pr

95.0]14.044.014.0[Pr

95.0]07071.0*96.1)44.0(07071.0*96.1[Pr

07071.050/5.0*5.0/)1(*

95.0]96.1/)ˆ(96.1[Pr

Distribution of p-hatDistribution of p-hatNormal Approximation to Binomial, n=50, p=0.5

0

1

2

3

4

5

6

0 0.2 0.4 0.6 0.8 1 1.2

sample proportion

den

sity

0.44

95% confidence interval

Around p-hat=0.44

Uniform to the NormalUniform to the Normal

Random variable x is distributed uniformly, on the number line from zero to one

x~U[0.5, 1/12]=U[,2]

density

x0 1

1

Uniform VariableUniform Variable

12/14/13/1)(

4/1)3/()2/1()()()(

])(*2[][)(

*)()()(

2/1)2/(*1*)(*

1)(,10,~

10

31

0

2222

222

0 0

0*

10

1

0

1

0

2

* *

*

xVar

xdxxfxExExxVar

ExxExxEExxExVar

xxdxdxxfxF

xdxxdxxfxEx

xfxUx

x xx

F(x)

x0 1

1

SimulationSimulation

Hypothesis test about population Hypothesis test about population mean from sample mean of 0.45mean from sample mean of 0.45

nnxVarnxVarnxVar

nnnExnxEnxE

nxx

n n

i

n

i

nn

i

n

i

n

i

/)/1()()/1()/1()(

)/1()/1()/1()/1()(

/

2

1 1

222

1

2

111

1

From central limit theorem )/,(~ 2 nNx

Distribution of Sample Mean of a Sample of 50 Random Draws of a Uniformly Distributed Variable

0

2

4

6

8

10

12

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Sample Mean

Den

sity

0.45

Is the sample mean of 0.45 significantly below expected value of 0.5?

)00167.0/,5.0(~ 2 nNx

CDF of Sample of 50 Random Draws from a Uniform Distribution

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1 1.2

Sample Mean

Pro

abab

ility

0.50.45

Hypothesis TestHypothesis Test

5.0:

5.0:0

aH

HStep #1: Formulate hypotheses

Step #2: Choose test statistic

23.10408.0/05.0

)50/29.0/()50.045.0(//()(/)(

x

xx

z

nxxz

Hypothesis TestHypothesis TestStep #3: Formulate probability statement. How low would z have to be to be significant?

Step #4: What level of significance should we choose?For example: 5%, i.e. 95% of time z would be above that value

Density Function for the Standardized Normal Variate

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

-5 -4 -3 -2 -1 0 1 2 3 4 5

Standard Deviations

Den

sity

2]1/)0[(2/1*]2/1[)( zezf

-1.645

0.050

Hypothesis testHypothesis test

• To be significantly below 0.5, a sample mean of 0.45, which corresponds to a z statistic of -1.23, would have to be even smaller, i.e. a z= -1.645 or even more negative. (The sample mean would have to be 0.421 or less)

• Therefore, accept null that population mean is 0.5

Sample means ordered by increasing size: smallest is 0.424,Not significantly below 0.5