inference about means
DESCRIPTION
Inference about means. Central Limit Theorem. X 1 ,X 2 ,…,X n independent identically distributed random variables with mean μ and variance σ 2 , then as n goes to infinity the sample mean Xbar n = ( X 1 +X 2 +…+X n )/n - PowerPoint PPT PresentationTRANSCRIPT
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 T-distribution, 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?
Histogram
0123456
9.5
9.6
9.7
9.8
9.9 10
10.1
10.2
Mor
e
Bin
Frequency
Frequency
• 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