MATH408: Probability & StatisticsSummer 1999

WEEK 7

Dr. Srinivas R. ChakravarthyProfessor of Mathematics and Statistics

Kettering University(GMI Engineering & Management Institute)

Flint, MI 48504-4898Phone: 810.762.7906

Email: schakrav@kettering.eduHomepage: www.kettering.edu/~schakrav

Test on population variance

•Recall that sample variance is an UMV for the population variance.

•To test Ho: 2 = 2o vs Ho: 2 > 2

o we look at the test statistic :

Decision Rule: Reject Ho if the absolute value of the calculated value of

the above statistic exceeds the table value (2,n-1).

EXAMPLES

EXAMPLES

Inference on Population ProportionTest of hypotheses

•Recall that sample proportion is an UMV for the population proportion

•To test Ho: p = po vs Ho: p po we look at the test statistic

Decision Rule: Reject Ho if the absolute value of the calculated value of

the above statistic exceeds the table value (z/2).

EXAMPLES

Inference on Population ProportionConfidence Interval

EXAMPLES

CHOICE OF SAMPLE SIZE

EXAMPLES

HOMEWORK PROBLEMS

Sections 4.6 through 4.7

43-46, 48-50, 52, 54-56, 59

Decision Making for Two Samples

• So far we talked about making inferences about the population parameter(s) when dealing with only one population at a time.

• Suppose we ask:– Is the new method of assembling a product better than the

existing one?– Is there any difference between workers in two assembly

plant?

– How do we answer these scientifically?

Using statistical methods for two-population case.

Inference for a difference in the means

Under the assumption listed earlier, we have the following result.

Test of hypotheses

EXAMPLES

Confidence Interval

EXAMPLES

• Verify that the pooled standard deviation, sp = 2.7 and the test statistic value is, to = -0.35.

• At 5% level of significance, we do not reject the null hypothesis and conclude that there is no sufficient evidence to say the means of the two catalysts differ.

EXAMPLES

EXAMPLES

Dependent PopulationsPairwise test

• So far, we assumed the populations under study to be independent.

• What happens when we need to compare, say, two assembling methods using the same set of operators?

• Obviously, the populations are not independent.

• So, what do we do?

• Verify that the calculated value of the statistic is 6.05.• Since this value is greater than the table value at 5% level, we

reject the null hypothesis and conclude at 5% level that there is

sufficient evidence to say that Karlsruhe method produces more

strength on the average than the Lehigh method.

Verify that 90% confidence interval for the difference in the means

is given by (-4.79, 7.21).

Inference on the variances of two normal populations

Test on the variances

EXAMPLES

Confidence interval

1,1,2/22

21

22

21

1,1,2/122

21

1212 nnnn fs

sf

s

s

Examples

Large Sample Test for proportions

To test H0: p1 = p2 vs H1: p1 p2, the test statistic is

2

22

1

11

2121

)1()1(

)(ˆˆ

npp

npp

ppPPz

Decision Rule: Reject the null hypothesis if the absolute

value of the calculated value of the above statistic exceeds

the table value z/2

HOMEWORK PROBLEMS

Sections 5.1 through 5.7

1-5, 9, 11, 13, 14, 16, 17, 20, 22, 30-33, 40-42, 46, 48-51, 53, 54