Inference about means
Central Limit Theorem
X1,X2,…,Xn independent identically
distributed random variables with mean μ
and variance σ2, then as n goes to
infinity the sample mean
Xbarn = (X1+X2+…+Xn)/n
has a distribution approximate normal with
mean μ and variance
Var((X1+X2+…+Xn)/n)= σ2/n
Confidence Interval
For n large,
the confidence interval for the mean is:
Xbarn± zα/2 se
where se=s/sqrt(n)
Test of Hypothesis
• Assumptions: independence & n>60
• H0: μ= μ0
• Ha: μ>μ0
Ha: μ<μ0
Ha: μ≠μ0
• Test statistic: (xbar- μ0)/se0
Where se0 =s/sqrt(n)
• P-value: as before
• conclusions
What if n is small?
• Inferences for small samples can be made
if we know the underlying distribution of
the samples. In a large class of examples,
it is reasonable to assume that the data is
normally distribution. In this case:
T= (xbar- μ)/(s/sqrt(n))
Has a distribution called the student Tdistribution, with
df=n-1
Example
A coffee machine dispenses coffee into
paper cups. You are supposed to get 10
ounces of coffee, but the amount varies
slightly from cup to cup. Here are the
amounts measured in a random sample of
20 cups.
9.9, 9.7, 10, 10.1, 9.9, 9.6, 9.8, 9.8, 10,
9.5, 9.7, 10.1, 9.9, 9.6, 10.2, 9.8, 10, 9.9,
9.5, 9.9
Is there evidence the machine is
shortchanging customers?
6
5
4
3
2
1
0
Bin
More
10.2
10.1
10
9.9
9.8
9.7
9.6
Frequency
9.5
Frequency
Histogram
• Random sample 20< 10% of all cups!
• Distribution looks unimodal and symmetric
so it is reasonable to assume it follows a
normal model.
• Use t-test for means
• No reason to doubt independance
mechanics
•
•
•
•
•
H0: μ= 10
Ha: μ<10
n=20, df=19, xbar=9.845 s= 0.199
t= 9.845-10/(.199/sqrt(20))=-3.49
P-value: P(T<-3.49)=.0012 df=19
Conclusion: small p-value means there is
enough statisticall evidence to conclude
the machine is shortchanging customers.
Confidence interval
Xbarn± t19,α/2 * se
where se=s/sqrt(n)
95% confidence interval:
9.845 ± 2.093 * 0.199 = (9.75, 9.94)
t19,.025= 2.093
HW
CH 23: 1, 3, 4, 5, 7, 9, 11, 13, 15, 25, 28, 34