1. Consolidation & Review
Inference

2. Sample Proportions Column of zeros and ones
: estimated proportion or fraction equals # of successes/ # of observations = k/n
Where k= Binomial[mean=np, variance=p*(1-p)n]
E[ } = E[k/n] =E[(1/n)*k] = 1/n E[k] =( 1/n)np=p
Var[ ] = Var[k/n] = (1/n)2 Var k =(1/n)2 np(1-p)
Var[ ] = p(1-p)/n
is a point estimate of p
The estimate of the Var of is *( )/n

3. Sample Proportions example from Lab 3
one of the ten columns of 50 observations of ones and zeros with the smallest proportion of 0.32
Var = 0.32*0.68/50 = with square root, i.e standard deviation of
A 95 % confidence interval for an estimate of p from this sample is:
Prob [-1.96≤(0.32-p)/0.066≤1.96]=0.95
Prob[-1.96*0.066≤(0.32-p)≤1.96*0.066]=0.95
Prob[-0.13≤(0.32-p)≤0.13]=0.95
Prob[0.13≥(p-0.32)≥-0.13]=0.95
Prob[0.45≥p≥ 0.19]=0.95

5. Note:
This 95 % confidence interval does not include p=0.5, the population parameter chosen for the simulation, illustrating that 5 % of the time the 95% confidence interval will not include the true value!
The Prob [-1.96≤(0.32-p)/0.066≤1.96]=0.95
Is the same as Prob[-1.96≤z≤1.96]=0.95, where,
( - p)/ = z = (0.32 – p)/0.066, in this example
We can use the normal distribution approximation to the binomial in this example since n*p = 50*0.32 ≥ 5 and
n*(1-p)≥5

6. a Z value
of 1.96 leads
to an area of
0.475, leaving
0.025 in the
Upper tail

7. Interval Estimation
The conventional approach is to choose a probability for the interval such as 95% or 99%

8. So z values
of and
1.96 leave
2.5% in each
tail

9. 1.96
-1.96
2.5%
2.5%

10. Application of Sample Proportions

12. Oct 13
Rasmussen Poll

14. Field Poll Margin of Error
Variance of = *( )/n = 0.47* 0.53/599 =
= √ =
So two standard deviations is about r 4.1%, i.e the margin of error is plus or minus 4.1 percentage points

15. Inferring the unknown population mean from a sample mean
Example from Lab 3: simulate the population as uniform, with random variable x, 0≤x≤1, and density f(x) =1
Note:
Note the expected value of x, E[x]=
Var[x] = E[x-E(x)]2 = E[x – E(x)]2 =E{x2 -2xE[x]+E[x]2}
Var[x] = {E(x2) – 2E[x]*E[x] + E[x]2 } = E[x2] –[Ex]2
E[x2] = = 1/3
Var[x] = E[x2] – E[x]2 = 1/3 –[1/2]2 = 1/3-1/4 = (4 -3)/12 =1/12
X~ Uniform(mean=1/2, Variance=1/12)

16. In lab 3 we drew a random sample of size 50 from this uniform distribution and calculated the sample mean:
From the central limit theorem, we know, and we saw in lab 3, that the sample mean is distributed normally

17. Central tendency and dispersion of sample mean
Where μ is the population mean. In the simulation from Lab 3 using the uniform distribution, we knew that μ = 0.5
Where σ2 is the variance of x. In the simulation from Lab 3 using the
uniform distribution, we knew the σ2 =1/12.

18. Hypothesis testing Example from Lab 3 for sample proportions
Step one: formulate the hypotheses
Null hypothesis, H0: p = 0.5
Alternative hypothesis, HA : p<0.5
Step two: Identify a test statistic
Where the value for p is from the null hypothesis, so z= ( )/0.066 = 0.18/0.066 = 2.73
If the null hypothesis were true, what is the probability of getting a test statistic of this size

19. Hypothesis Testing: 4 Steps
Formulate all the hypotheses
Identify a test statistic
If the null hypothesis were true, what is the probability of getting a test statistic this large?
Compare this probability to a chosen critical level of significance, e.g. 5% 19

20. 20 a Z value Of 2.73 leads to an area of 0.4968, leaving 0.0032 in the
Upper tail, and
Hence
In the lower tail.
If you choose a
risk level of .05,
i.e. α = 0.05 for
The probability
A type I error,
Then reject H0 20

21. 0.0032
-2.73

22. 0.050
-1.645