Understanding Point Estimates, Confidence Intervals, and Hypothesis Testing
Test 4
Point estimate – single number computed from a sample, which serves as a guess for the parameter.
Estimator – statistic of interest and is therefore a random variable.
Estimate – simply a specific value of an estimator.
Confidence interval – Interval of values constructed so that, with a specified degree of confidence, the value of the population parameter lies within it.
Confidence coefficient – Probability that the CI includes the population parameter in repeated samples.
Confidence level – The confidence coefficient expressed as a percentage.
Confidence interval example when SD is known
Proportion Test Example (If p<= a: Reject null, If p>a: dont)
We need a 95% CI for μ. The population is normal and SD is known. n=2500, a=0.01, x=935
x̄ = 155.7; SD = 18, n = 15
p-x/n=935/2500=0.3740
1 – a = 0.95 -> a = 0.05 -> a/2 = 0.025
Ho: p=0.36, Ha: p>0.36, TS: Z=(P-po)/(SQRT(po(1-po)/n))
P(Z≥z0.025)=1-0.025=0.975
RR: Z>=za=z0.01=2.3263
z0.025=1.96
Ho: p=0.36, Ha: p>0.36, TS: Z=(P-po)/(SQRT(po(1-po)/n)
x ± za/2*SD/sqrt(n)
Another example (n=400, x=204, po=0.58, a=0.05)
=x ± z0.025*SD/sqrt(n)
P=x/n=204/400-0.51
=(146.59,164.81)
Ho: p=0.58
Bound of Error of Estimation Example
Ha: p != 0.58
SD=15, B=5
RR: |Z|>=za/2=z0.025=1.96
1-a=0.95->a/2=0.025->z0.025=1.96
((SD* za/2)/B)^2 = 34.57 = n>=35 (round up)
normcdf(-1e99,-2.84,1,0) = 0.0023
Confidence interval example when SD is unknown
p=2(0.0023)=0.0046(<=0.05=a) we reject
x̄ = 980.2; SD = 27.6, n = 11
1-a=0.95->a=0.05->a/2=0.025
Find t critical value
ta/2,n-1=t0.025,10=2.2281
n=15.95%
x ± ta/2,n-1*s/sqrt(n)
CI for p and inference (p=x/n)
invT(a/2,n-1)
We need a 95% CI for p. If it asks if P is normal test if its greater
n=1200 than 5. (n)(P)>=5 and (n)(1-P)>=5
1-a=0.95->a=0.05->a/2=0.025
za/2=z0.025=1.96
p± za/2*SQRT(p(1-p)/n)
Error of Estimation
B=za/2*SQRT(p(1-p)/n)
n=p(1-p)(za/2/B)^2
Hypothesis Test Errors
Reject Ho Do not reject Ho Ho: u = uo
Truth Ho Type 1 error Correct Ha: u > uo, u < uo, or u != uo
Ha Correct Type II error TS: Z=(X-uo)/(SD/sqrt(n))
RR: Z>=za, Z<=za, or |Z|>=za/2
P value example (u=670;n=20;a=25)
Ho: u=670,Ha: u<670,TS: Z=(x-uo)/(a/sqrt(n)) = -1.771 Random
p=P(Z<=-1.771)=P(Z<=-1.77)=0.0384 Sx = population variance. Make sure to ^2 it
because p=0.0384<=0.05 we reject Ho. Find critical value. Ex. z0.10 InvNorm(0.10,0,1) <– make positive
two sided example
p/2=P(Z>=1.34) = 0.0901 = p=2(0.0901)=p=0.1802