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 has a distribution approximate normal with mean μ and variance Var((X 1 +X 2 +…+X n )/n)= σ 2 /n

Confidence Interval For n large, the confidence interval for the mean is: Xbar n ± z α/2 se where se=s/sqrt(n)

Test of Hypothesis Assumptions: independence & n>60 H 0 : μ= μ 0 H a : μ>μ 0 H a : μ<μ 0 H a : μ≠μ 0 Test statistic: (xbar- μ 0 )/se 0 Where se 0 =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?

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 H 0 : μ= 10H a : μ<10 n=20, df=19, xbar=9.845 s= t= /(.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 Xbar n ± t 19,α/2 * se where se=s/sqrt(n) 95% confidence interval: ± * = (9.75, 9.94) t 19,.025 = 2.093

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