Section 11.3: Large-Sample Inferences Concerning a Difference Between Two Population or Treatment Proportions

Population or Treatment 1: Proportion of “successes” = π 1 Population or Treatment 2: Proportion of “successes” = π 2 Sample from Population or Treatment 1: Proportion of “successes” = p 1 Sample from Population or Treatment 2: Proportion of “successes” = p 2

Properties of the Sampling Distribution of p 1 – p 2 3. If both n 1 and n 2 are large (n 1 π 1 ≥ 10, n 1 (1-π 1 ) ≥ 10, n 2 π 2 ≥ 10, and n 2 (1-π 2 ) ≥ 10), then each have a sampling distribution that is approximately normal, and their difference also has a sampling distribution that is approximately normal.

Summary of Large-Sample z Tests for H 0 : π 1 – π 2 = 0 Assumptions: 1.The samples are independently chosen random samples, or treatments were assigned at random to individuals or objects (vice versa) 2.Both sample sizes are large n 1 p 1 ≥ 10, n 1 (1-p 1 ) ≥ 10, n 2 p 2 ≥ 10, n 2 (1-p 2 ) ≥ 10

Example Some people seem to believe that you can fix anything with duct tape. Even so, many were skeptical when researchers announced that duct tape may be a more effective and less painful alternative to liquid nitrogen, which doctors routinely use to freeze warts. Patients with warts were randomly assigned to either the duct tape treatment or the more traditional freezing treatment.

100 people were given the liquid nitrogen treatment with a successful wart removal from 60 of them. 104 people were given the duct tape treatment with 88 being successful. Does the data suggest that freezing is less successful that duct tape in removing warts?

Let π 1 represent the true proportion of warts that would be successfully removed by freezing, and let π 2 represent the true proportion of warts that would be successfully removed with the duct tape treatment. We will use.01 for the significance level.

Calculations before we test the hypothesis The nine-step procedure can now be used to perform the hypothesis test:

1.π 1 – π 2 is the difference between the true proportions of warts removed for freezing and the duct tape treatment. 2.H 0 : π 1 – π 2 = 0 (π 1 = π 2 ) 3.H a : π 1 – π 2 < 0 (π 1 < π 2, in which case the proportion of warts removed for freezing is lower than the proportion for duct tape). 4.Significance Level: α =.01

8. P-value: This is a lower-tailed test, so the P-value is the area under the z curve and to the left of the computed z = From appendix table 2, we find the P-value ≈ Conclusion: Because P-value ≤ α, the null hypothesis is rejected at level.01. There is convincing evidence that the proportion of warts successfully removed is lower for freezing than for the duct tape treatment.

Example Researchers at the National Cancer Institute released the results of a study that examined the effect of weed-killing herbicides on house pets. Dogs, some of whom were from homes where the herbicide was used on a regular basis, were examined for the presence of malignant lymphoma. The following data are compatible with summary values given in the report:

GroupSample Size Number with Lymphoma P Exposed Unexposed

We use the given data to estimate the difference between the proportion of exposed dogs that develop lymphoma and the proportion of unexposed dogs that develop lymphoma. Let π 1 denote the proportion of exposed dogs that develop lymphoma, and define π 2 similarly for unexposed dogs.

The sample sizes are large enough for the large sample interval to be valid. A 90% confidence interval.