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COMPARING TWO POPULATION (OR TREATMENT) PROPORTIONS 1 COMPARING TWO POPULATION PROPORTIONS USING INDEPENDENT SAMPLES EXAMPLE: The article “Foraging Behavior in the Indian False Vampire Bat” reported that 36 of 193 female bats in flight spent more than 5 minutes in the air before locating food. For male bats, 64 of 168 exceeded 5 minutes when locating food. Is there sufficient evidence to indicate that the proportion of flights taking longer than 5 minutes differs for the two sexes? Note: two independent samples and the interest is in comparing the proportions for the two genders Notation: Population Population Sample Proportion Size π1 n1 1 π2 n2 2 Sample Proportion π̂1 π̂ 2 To compare 2 population proportions we usually consider the size of the difference π1 − π 2 : COMPARING TWO POPULATION (OR TREATMENT) PROPORTIONS 2 π1 − π 2 = 0 ⇒ π1 = π 2 π1 − π 2 > 0 ⇒ π1 > π 2 π1 − π 2 < 0 ⇒ π1 < π 2 Our sampling estimate of this difference is the difference in the sample proportions πˆ1 − πˆ 2 when the two samples are independent of one another. The estimator of π1 − π 2 is πˆ1 − πˆ 2 . Sampling Distribution of πˆ1 − πˆ 2 when the two samples are independently and randomly taken: 1) the mean of the distribution is µπˆ1 −πˆ 2 = π 1 − π 2 (that is, πˆ1 − πˆ 2 is unbiased) 2) the standard deviation of the distribution is π (1 − π 1 ) π 2 (1 − π 2 ) σ πˆ1 −πˆ 2 = 1 + n1 n2 3) the shape of the distribution is approximately normal (a bell curve) if both n1 and n1 are large. The sample sizes are usually considered to be sufficiently large if the following statement is true: COMPARING TWO POPULATION (OR TREATMENT) PROPORTIONS 3 n1πˆ1 ≥ 5, n1 (1 − πˆ1 ) ≥ 5 n2πˆ 2 ≥ 5, n2 (1 − πˆ 2 ) ≥ 5 This, in words, says that you should have enough observations from each population so that at least 5 observations should be successes and at least 5 should be failures. Hence, at a minimum, n1 ≥ 10 and n2 ≥ 10 (but only if π 1 = π 2 = 0.5 ). As we’ll see, the estimator of σ πˆ1 −πˆ 2 = π 1 (1 − π 1 ) π 2 (1 − π 2 ) n1 + n2 depends on whether we are constructing a confidence interval or performing a test of the difference in the two population proportions. COMPARING TWO POPULATION (OR TREATMENT) PROPORTIONS 4 Large Sample Test of the Difference in Two Population Proportions Based on Two Independent Samples H0: π1 − π 2 = 0 Null hypothesis: Alternative Hypothesis is one of three: a) b) c) HA: π1 − π 2 > 0 HA: π1 − π 2 < 0 HA: π1 − π 2 ≠ 0 Test Statistic: z= (πˆ1 − πˆ 2 ) ⎛1 1⎞ + ⎟⎟ ⎝ n1 n2 ⎠ where πˆ C (1 − πˆ C )⎜⎜ n1πˆ1 + n2πˆ 2 n1 + n2 total # successes in both samples = total sample size πˆ C = P-value: depends on the alternative hypothesis: a) P-value = Pr( Z > z) b) P-value = Pr( Z < z) c) P-value = 2 Pr( Z < - |z| ) COMPARING TWO POPULATION (OR TREATMENT) PROPORTIONS 5 Decision Rule: reject Ho if P-value ≤ α Assumptions: 1. n1 and n2 are large enough for the sample proportions to be approximately normally distributed 2. the sampling was random and not more than 5% of the population. 3. the two samples are independently taken EXAMPLE Bats: Sample Statistics: Popula tion 1= female 2= male Sample Size n1= 193 #Suc- Sample Proportion cesses 36 36 πˆ1 = = .1865 193 64 n2 = 168 64 πˆ 2 = = .3809 168 Hypotheses: Assumptions: Ho: π1 − π 2 = 0 HA: π1 − π 2 ≠ 0 n1πˆ1 ≥ 5, n1 (1 − πˆ1 ) ≥ 5 n2πˆ 2 ≥ 5, n2 (1 − πˆ 2 ) ≥ 5 have COMPARING TWO POPULATION (OR TREATMENT) PROPORTIONS 6 been met. And we have 2 random samples. Significance level: let’s use α = 0.05 Test Statistic: first we need the common proportion πˆ C = Then, z= n1πˆ1 + n2πˆ 2 36 + 64 = = .277 n1 + n2 193 + 168 (πˆ1 − πˆ 2 ) ⎛1 1⎞ + ⎟⎟ ⎝ n1 n2 ⎠ πˆ C (1 − πˆ C )⎜⎜ = (.1865 − .3809) = −4.12 1 ⎞ ⎛ 1 + .277(1 − .277)⎜ ⎟ 193 168 ⎝ ⎠ P-value: = 2 Pr(Z< -|z|) = 2 Pr(Z<-4.12) <0.0001 ≈ 0+ Conclusions: We reject the null hypothesis since pvalue <0.0001 <<<< α=0.05. There is strong evidence based on these samples, that the population proportion of female false vampire bats who take COMPARING TWO POPULATION (OR TREATMENT) PROPORTIONS 7 longer than 5 minutes searching for food is different from the proportion for male bats. EXAMPLE Old Faithful, the geyser at Yellowstone National Park, is known to have two distinct types of eruptions: long-duration (> 3 minutes) and short duration (< 3 min). If the types of eruptions are equally likely at all times of the day, then the proportion of long duration eruptions occurring during the day should be the same as the proportion at night. A geologist hypothesized that the length of duration was affected by solar heating during the day and hence, the proportion of daytime long duration eruptions should be higher than the night-time proportion. Two samples were taken in August over several days and nights. The geologist observed 53% long duration eruptions during the day (out of 35 eruptions) and 49% (out of 41 eruptions) at night. Is there sufficient evidence to support the scientist’s claim? Use a significance level of 0.025. Hypotheses: Ho: π1 − π 2 = 0 HA: π1 − π 2 > 0 (population 1 is the daytime eruptions and 2, the night time) COMPARING TWO POPULATION (OR TREATMENT) PROPORTIONS Assumptions: n1πˆ1 ≥ 5, n1 (1 − πˆ1 ) ≥ 5 n2πˆ 2 ≥ 5, n2 (1 − πˆ 2 ) ≥ 5 8 ? random samples? Significance level: α = 0.025 Test Statistic: first we need the common proportion Then, z = (πˆ1 − πˆ 2 ) ⎛1 1⎞ + ⎟⎟ ⎝ n1 n2 ⎠ πˆ C (1 − πˆ C )⎜⎜ P-value: COMPARING TWO POPULATION (OR TREATMENT) PROPORTIONS 9 Conclusions: Large Sample Confidence Interval Estimation of The Difference Between Two Proportions Based on Independent Samples: Interval Estimator: (πˆ1 − πˆ 2 ) ± zα × 2 πˆ1 (1 − πˆ1 ) πˆ 2 (1 − πˆ 2 ) n1 + n2 where the z critical value is based on the confidence level (1 – α) desired Assumptions: 1. n1 and n2 are large enough for the sample proportions to be approximately normally distributed 2. the sampling was random 3. the two samples are independently taken Note that the estimator of SE(πˆ1 − πˆ 2 ) is different than the one used in hypothesis testing! COMPARING TWO POPULATION (OR TREATMENT) PROPORTIONS 10 EXAMPLE for the bats let’s use a 90% C.I. to estimate the difference in proportions of time spent searching for food between males and females. Now, the z critical value for 90% is 1.645. So, a 90% C.I. is (πˆ1 − πˆ 2 ) ± 1.645 πˆ1 (1 − πˆ1 ) πˆ 2 (1 − πˆ 2 ) n1 = (.187 − .381) ± 1.645 + n2 .187(1 − .187) .381(1 − .381) + 193 168 = −.194 ± 1.645(.0468) = −.194 ± .077 = (−.271, − .117) Hence, with 90% confidence, the population proportion of female false vampire bats that spend more than 5 minutes locating food is between 11.7% and 27.1% lower than the population proportion of male bats which spend more than 5 minutes locating food. (We could reverse that and say that the proportion of males spending more than 5 minutes is between 11.7 and 27.1% higher than the proportion of females.)