inference about means

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Inference about means

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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 Presentation

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Page 1: Inference about means

Inference about means

Page 2: 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

Page 3: Inference about means

Confidence Interval

For n large,

the confidence interval for the mean is:

Xbarn± zα/2 se

where se=s/sqrt(n)

Page 4: Inference about means

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

Page 5: Inference about means

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

Page 6: Inference about means

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

Page 7: Inference about means

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

Page 8: Inference about means

• 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

Page 9: Inference about means

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.

Page 10: Inference about means

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

Page 11: Inference about means

HW

CH 23: 1, 3, 4, 5, 7, 9, 11, 13, 15, 25, 28, 34