Statistical Inference Lab Three. Bernoulli to Normal Through Binomial One flip Fair coin Heads Tails...
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Transcript of Statistical Inference Lab Three. Bernoulli to Normal Through Binomial One flip Fair coin Heads Tails...
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