MATH408: Probability & Statistics Summer 1999 WEEK 7 Dr. Srinivas R. Chakravarthy Professor of Mathematics and Statistics Kettering University (GMI Engineering & Management Institute) Flint, MI Phone: Homepage:

Test on population variance Recall that sample variance is an UMV for the population variance. To test H o :  2 =  2 o vs H o :  2 >  2 o we look at the test statistic : Decision Rule: Reject H o if the absolute value of the calculated value of the above statistic exceeds the table value (  2 ,n-1 ).

EXAMPLES

Inference on Population Proportion Test of hypotheses Recall that sample proportion is an UMV for the population proportion To test H o : p = p o vs H o : p  p o we look at the test statistic Decision Rule: Reject H o if the absolute value of the calculated value of the above statistic exceeds the table value (z  /2 ).

EXAMPLES

Inference on Population Proportion Confidence Interval

EXAMPLES

CHOICE OF SAMPLE SIZE EXAMPLES

HOMEWORK PROBLEMS Sections 4.6 through , 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, s p = 2.7 and the test statistic value is, t o = 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

Dependent Populations Pairwise 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 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 Examples

Large Sample Test for proportions To test H 0 : p 1 = p 2 vs H 1 : p 1  p 2, the test statistic is 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 , 9, 11, 13, 14, 16, 17, 20, 22, 30-33, , 46, 48-51, 53, 54