Confidence Interval for the Difference of Two Proportions Section 8.3 Confidence Interval for the Difference of Two Proportions

Previously (Section 8.1) we found confidence interval estimates for the proportion of successes in a single population. However, in life, many decisions are based on comparisons.

So we need a method to estimate the size of the difference between the proportion of successes in one population and the proportion of successes in a different population.

A recent poll of 29,700 U. S. households found that 63% owned a pet A recent poll of 29,700 U.S. households found that 63% owned a pet. The percentage in 1994 was 56%. What are the two populations?

A recent poll of 29,700 U. S. households found that 63% owned a pet A recent poll of 29,700 U.S. households found that 63% owned a pet. The percentage in 1994 was 56%. What are the two populations? (1) households now (2) households in the U.S. in 1994

A recent poll of 29,700 U. S. households found that 63% owned a pet A recent poll of 29,700 U.S. households found that 63% owned a pet. The percentage in 1994 was 56%. What was the change in the percentage of U.S. households that own a pet?

A recent poll of 29,700 U. S. households found that 63% owned a pet A recent poll of 29,700 U.S. households found that 63% owned a pet. The percentage in 1994 was 56%. p or p? What was the change in the percentage of U.S. households that own a pet?

A recent poll of 29,700 U. S. households found that 63% owned a pet A recent poll of 29,700 U.S. households found that 63% owned a pet. The percentage in 1994 was 56%. p or p? What was the change in the percentage of U.S. households that own a pet? sample

Recall Variation in Sampling A recent poll of 29,700 U.S. households found that 63% owned a pet. The percentage in 1994 was 56%. What was the change in the percentage of U.S. households that own a pet? 7% seems obvious; however, remember 7% is the difference of two sample percentages and not likely to be equal to the population percentages.

A recent poll of 29,700 U. S. households found that 63% owned a pet A recent poll of 29,700 U.S. households found that 63% owned a pet. The percentage in 1994 was 56%. What was the change in the percentage of U.S. households that own a pet? 7% margin of error

A recent poll of 29,700 U. S. households found that 63% owned a pet A recent poll of 29,700 U.S. households found that 63% owned a pet. The percentage in 1994 was 56%. What was the change in the percentage of U.S. households that own a pet? 7% margin of error Thus, we need a confidence interval for the difference of the two proportions

Formula Confidence interval for a proportion is:

Formula Confidence interval for a proportion is: Confidence interval for the difference of two proportions follows same pattern:

CI for Difference of 2 Proportions

CI for Difference of 2 Proportions

CI for Difference of 2 Proportions Note: the sample sizes do not have to be equal

CI for Difference of 2 Proportions Which sample proportion should be and which ?

CI for Difference of 2 Proportions Which sample proportion should be and which ? It does not matter as long as you define which is which.

Before we start doing computations for a confidence interval based on this formula, what must we do first?

Before we start doing computations for a confidence interval based on this formula, what must we do first? Check the conditions to ensure it is appropriate to use this method for determining the confidence interval.

Before we start doing computations for a confidence interval based on this formula, what must we do first? Check the conditions to ensure it is appropriate to use this method for determining the confidence interval. Three conditions must be met.

Three Conditions (1) Two samples are taken randomly and independently from two populations

Three Conditions (1) Two samples are taken randomly and independently from two populations (2) are all at least 5.

Three Conditions (1) Two samples are taken randomly and independently from two populations (2) are all at least 5. (3) Each population is at least 10 times as large as its sample size --Explain why it is reasonable to assume this is true; do not just show 10 times the sample size

A recent poll of 29,700 U.S. households selected at random found that 63% owned a pet. A 1994 random survey taken by the same organization found that 56% of 6,786 U.S households owned a pet.

A recent poll of 29,700 U.S. households selected at random found that 63% owned a pet. A 1994 random survey taken by the same organization found that 56% of 6,786 U.S households owned a pet. Find and interpret a 95% confidence interval for the difference between the proportion of U.S. households that own a pet now and the proportion of U.S. households that owned a pet in 1994.

Conditions A recent poll of 29,700 U.S. households selected at random found that 63% owned a pet. A 1994 random survey taken by the same organization found that 56% of 6,786 U.S households owned a pet. (1) Problem states the two samples were taken randomly. Also, they were taken independently from the population of U.S. households in two different years.

Conditions A recent poll of 29,700 U.S. households selected at random found that 63% owned a pet. A 1994 random survey taken by the same organization found that 56% of 6,786 U.S households owned a pet. Each is at least 5. (You must show all 4 calculations)

Conditions A recent poll of 29,700 U.S. households selected at random found that 63% owned a pet. A 1994 random survey taken by the same organization found that 56% of 6,786 U.S households owned a pet. (2) The number of U.S. households each year is larger than 10 times 29,700 = 297,000. (Do not just show 10(29,700) = 297000!)

Do Computations

Do Computations (0.057, 0.083)

Do Computations with Calculator STAT

Do Computations with Calculator Enter each sample size and number of successes, not proportion, in each sample

Do Computations with Calculator Enter each sample size and number of successes, not proportion, in each sample

Write Interpretation in Context (0.057, 0.083) I’m 95% confident that the difference in the two rates of pet ownership between now and 1994 is between 0.057 and 0.083.

Write Interpretation in Context (0.057, 0.083) I’m 95% confident that the difference in the two rates of pet ownership between now and 1994 is between 0.057 and 0.083. This means it is plausible that the difference in the percentage of households that own pets now and the percentage in 1994 is 5.7%. It is also plausible the difference is 8.3%.

Write Interpretation in Context (0.057, 0.083) Note that a difference of 0 is not in the confidence interval.

Write Interpretation in Context (0.057, 0.083) Note that a difference of 0 is not in the confidence interval. Thus, you are convinced that there was a change in the percentage of households that own a pet.

Page 521, D41

Dog: 45%, cat: 34%, both 20% (a) 41% Remember not to double count.

Dog: 45%, cat: 34%, both 20% (a) 41% Remember not to double count. 100 – (45 + 34 – 20) = 41%

Page 521, D41 (b)

Page 521, D41 (b) Not independent events here. P(own a dog and own a cat) = 0.20. P(own a dog) P(own a cat) = (0.45)(0.34) = 0.153

Page 521, D41 (b) Not independent events here. P(own a dog and own a cat) = 0.20. P(own a dog) P(own a cat) = (0.45)(0.34) = 0.153 Because the Multiplication Rule for Independent Events ( P(A and B) = P(A) P(B) ) does not hold, the two events are not independent.

Page 521, D41 (c)

Page 521, D41 (c) No. The two percentages, 45% and 34%, did not come from independent events. Therefore, conditions are not met.

Questions?