Comparing Two Population Proportions: Inference and Hypothesis Testing - Prof. Brett D. Hu, Study notes of Data Analysis & Statistical Methods

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21 October Comparing 2 population proportions We want to make inferences about π 1 – π 2 , where π 1 and π 2 come from different populations Take independent samples Find p 1 , the point estimate for π 1 Find p 2 , the point estimate for π 2 So p 1 – p 2 is the point estimate for π 1 - π 2 Sampling Distribution When the samples are independently taken and n is large, π 1 , π 2 not extreme, following hold n 1 π 1 ≥ 10 n 2 π 2 ≥ 10 n 1 (1 – π 1 ) ≥ 10 n 2 (1 – π 2 ) ≥ 10 p 1 – p 2 ~ N (π 1 – π 2 ,

π 1 ( 1 − π 1 )

n 1

+

π 2 ( 1 − π 2 )

n 2

Hypotheses Upper Tail: Is π 1 larger than π 2? H 0 : π 1 - π 2 ≤ 0 HA: π 1 – π 2 > 0 OR H 0 : π 1 ≤ π 2 HA: π 1 > π 2 Lower Tail: Is π 1 smaller than π 2?

H 0 : π 1 – π 2 ≥ 0 HA: π 1 - π 2 < 0 OR H 0 : π 1 ≥ π 2 HA: π 1 < π 2 Two Tail: Is π 1 different from π 2? H 0 : π 1 - π 2 = 0 HA: π 1 - π 2 ≠ 0 OR H 0 : π 1 = π 2 HA: π 1 ≠ π 2 When we perform a hypothesis test, we assume H 0 is true. So here, we are assuming π 1 = π 2 When π 1 = π 2 , we can combine the two populations into one where the true proportion is π = π 1 = π 2 So we can combine our samples into one sample of size n 1 + n 2 , from which we get an estimate

of the overall population proportion (π), denoted pc or p.

Called combined or pooled population That is, pc is the sample proportion from the combined sample of size n 1 + n 2

pc = p^ =

n 1 p 1 + n 2 p 2

n 1 + n 2

=

¿ successes ∈ sample 1 +¿ successes ∈ sample 2

n 1 + n 2

Since we assume π 1 = π 2 = π, we can assume p 1 ≈ p 2 ≈ pc (Theoretically, p 1 would = p 2 , but won’t be in every case. We use pc as an estimate for both) If sampling distribution assumptions hold, then

Ztest = ( p 1 −^ p 2 )−( π 10 − π 20 )

√

pc ( 1 − pc )

n 1

+

pc ( 1 − pc )

n 2

~ N (0, 1)

Use Z-table to find p-values in the normal way. Note: When checking the assumptions, we won’t have hypothesized values for π 1 and π 2 , respectively. We instead use p 1 and p 2 to check the assumptions. Therefore, the following must hold n 1 p 1 ≥ 10 n 2 p 2 ≥ 10 n 1 (1 – n 1 ) ≥ 10 n 2 (1 – n 2 ) ≥ 10 Independent samples Confidence Intervals We aren’t assuming H 0 is true here (as we may not have any hypotheses), so we cannot assume that π 1 = π 2. Therefore, we do not use pc. So if our assumptions hold, the confidence interval for π 1 – π 2 is

CI = p 1 – p 2 ± Zcrit

√

p 1 ( 1 − p 1 )

n 1

+

p 2 ( 1 − p 2 )

n 2