Is Yawning Contagious video 10 34 =.29 P yawn seed 4 16 =.25 P yawn no seed.29.25 =.04
No, maybe this occurred purely by chance. 50 subjects Random Assignment Group 1 (34) Group 2 (16) Treatment 1 (yawn seed) Treatment 2 (no yawn seed) Compare Yawning What is this? H 0
Get in your groups.
difference in proportions p-value =
p-value =, not surprising at all. Not statistically significant. We don t have enough evidence to prove that yawning is contagious.
Check: n 1 p 1 10 n 1 1 p 1 10 p 1 p 1 N p 1, σ p1 p 1 σ p1 = p 1 Check: n 2 p 2 10 n 2 1 p 2 10 p 2 p 2 p 1 1 p 1 n 1 N p 2, σ p2 p 2 σ p2 = p 2 p 2 1 p 2 n 2
p 1 p 2 p 1 p 2 p 1 p 2 N p 1 p 2, σ p1 p 2 σ p1 p 2 = p 1 p 2 p 1 1 p 1 n 1 + p 2 1 p 2 n 2 Check: n 1 p 1 10 n 2 p 2 10 & n 1 1 p 1 10 n 2 1 p 2 10
p 1 p 2 p 1 p 2 p 1 p 2 N p 1 p 2, σ p1 p 2 σ p1 p 2 = p 1 p 2 p 1 1 p 1 n 1 + p 2 1 p 2 n 2 Check: n 1 p 1 10 n 2 p 2 10 & n 1 1 p 1 10 n 2 1 p 2 10
p 1 = 114 150 =.76 p 2 = 62 100 =.62 p 1 p 2 =.14 p 1 true proportion of ECRCHS students who do math HW p 2 true proportion of Taft students who do math HW We want to estimate the true difference p 1 p 2 at a 95% confidence level.
Two-sample z interval for p 1 p 2 Random: Normal: random sample of 150 students from ECRCHS and random sample of 100 Taft students n 1 p 1 10 n 1 1 p 1 10 n 2 p 2 10 n 2 1 p 2 10 150.76 = 114 10 150.24 = 36 10 100.62 = 62 10 100.38 = 38 10 So the sampling distribution of p 1 p 2 is approximately normal. Independent: Two things to check: 1) The two samples need to be independent of each other. 2) Individual observations in each sample have to be independent. When sampling without replacement for both samples, must check 10% condition for both. We clearly have two independent samples one from each school. There are at least 10 150 = 1500 students at ECRCHS and at least 10 100 = 1000 students at Taft.
p 1 = 114 150 =.76 p 2 = 62 100 =.62 p 1 p 2 =.14 Estimate ± Margin of Error p 1 p 2 ± z p 1 1 p 1 n 1 + p 2 1 p 2 n 2 Standard Error. 76.62 ± 1.96.76.24 150 +.62.38 100.14 ± 0.117 0.023, 0.257 With calculator: STAT TESTS 2-PropZInt (B) x1: 114 n1: 150 0.02286, 0.25714 x2: 62 n2: 100 C-Level: 0.95
We are 95% confident that the interval from 0.023 to 0.257 captures the true difference in proportions p 1 p 2 of ECRCHS and Taft students who do their math HW. This suggests that 2.3% to 25.7% more students at ECRCHS do math HW than Taft students.
p 1 =.63 p 2 =.68 p 1 p 2 =.05 p 1 true proportion of teens who say they go online every day p 2 true proportion of adults who say they go online every day We want to estimate the true difference p 1 p 2 at a 90% confidence level.
Two-sample z interval for p 1 p 2 Random: Normal: Independent: random sample of 800 teens and a separate sample of 2253 adults n 1 p 1 10 n 1 1 p 1 10 n 2 p 2 10 n 2 1 p 2 10 800.63 = 504 10 800.37 = 296 10 2253.68 = 1532 10 2253.32 = 721 10 So the sampling distribution of p 1 p 2 is approximately normal. We clearly have two independent samples one of teens and one of adults. There are at least 10 800 = 8000 U.S. teens and at least 10 2253 = 22530 U.S. adults.
p 1 =.63 p 2 =.68 p 1 p 2 =.05 Estimate ± Margin of Error p 1 p 2 ± z p 1 1 p 1 n 1 + p 2 1 p 2 n 2 Standard Error. 63.68 ± 1.645.63.37 800 +.68.32 2253 0.05 ± 0.0324 0.0824, 0.0176 With calculator: STAT TESTS 2-PropZInt (B) x1: 504 n1: 800 0.0824, 0.0176 x2: 1532 n2: 2253 C-Level: 0.9
We are 90% confident that the interval from.0824 to.0176 captures the true difference in proportions p 1 p 2 of teens and adults who go online every day. This suggests that 1.76% to 8.24% more adults are online every day than teens.
A LOT of writing!! Don t write too big. State: p 1 proportion of AP Calc seniors who are going to college p 2 proportion of Pre-Calc seniors who are going to college H 0 : p 1 p 2 = 0 p 1 = 98 OR p 1 = p 2 110 = 0.89 H a : p 1 p 2 > 0 α = 0.05 OR p 1 > p 2 p 2 = 68 80 = 0.85 p 1 p 2 =.89.85 =.04
Plan: Two sample z test for p 1 p 2 Random: Normal: SRS of 80 students currently taking Pre-Calc and SRS of 110 students currently taking AP Calculus n 1 p 1 10 n 1 1 p 1 10 n 2 p 2 10 n 2 1 p 2 10 110.89 = 98 10 110.11 = 12 10 80.85 = 68 10 80.15 = 12 10 So the sampling distribution of p 1 p 2 is approximately normal. Independent: Two things to check: 1) The two samples need to be independent of each other. 2) Individual observations in each sample have to be independent. When sampling without replacement for both samples, must check 10% condition for both. We clearly have two independent samples one from each class level. There are at least 10 80 = 800 Pre-Calc students and at least 10 110 = 1100 AP Calculus students in California.
Do: Sampling Distribution of p 1 p 2 N 0, Normally, we would use this formula for the standard deviation: 0 σ p1 p 2 = p 1 1 p 1 n 1 + p 2 1 p 2 n 2 However, since we re assuming H 0 : p 1 = p 2 is true, we ll be using a different value in the formula. The two parameters are the same, so we will call their common value p. In order to estimate p, we can get a more precise estimate when we use more data, so it makes sense to combine the data from the two samples. This pooled (or combined) sample proportion is p c = count of successes in both samples count of individuals in both samples = x 1 + x 2 n 1 + n 2 We will use this in place of both p 1 and p 2
Do: Sampling Distribution of p 1 p 2 N 0,.049 0 0.04 p 1 p 2 =.04 p c = x 1 + x 2 98 + 68 = n 1 + n 2 110 + 80 = 166 190 =.874 σ p1 p 2 = p 1 1 p 1 n 1 + p 2 1 p 2 n 2 = p c 1 p c n 1 + p c 1 p c n 2 =.874.126 110 +.874.126 80 =.049 z = statistic parameter st. dev. of statistic = p 1 p 2 p 1 p 2 σ p1 p 2 =.04 0 0.049 =.82 area =.2061 p-value normalcdf.04, 99999, 0,.049 = 0.2071 p-value
Conclude: Assuming H 0 is true p 1 = p 2, there is a.2061 probability of obtaining a p 1 p 2 value of.04 or more purely by chance. This provides weak evidence against H 0 and is not statistically significant at α =.05 level.2061 >.05. Therefore, we fail to reject H 0 and cannot conclude that seniors taking AP Calc are more likely to attend college next year than seniors taking Pre-Calc. With calculator: STAT TESTS 2-PropZTest (6) x1: 98 n1: 110 x2: 68 n2: 80 p1: p2 <p2 >p2 z = 0.838 p = 0.201 p 1 = 0.89 p 2 = 0.85 p =.874 n 1 = 110 n 2 = 80 Observational study p-value pooled proportion no treatments imposed senior math students in California What population can we target? No
State: p 1 true proportion who quit smoking to get money p 2 true proportion who quit smoking traditional way H 0 : p 1 p 2 = 0 OR p 1 = p 2 p 1 = 0.15 p 2 = 0.05 H a : p 1 p 2 > 0 OR p 1 > p 2 p 1 p 2 =.15.05 =.10 α = 0.05
Plan: Two sample z test for p 1 p 2 Random: Normal: Independent: subjects were randomly assigned to treatments n 1 p 1 10 n 1 1 p 1 10 n 2 p 2 10 n 2 1 p 2 10 439.15 = 66 10 439.85 = 373 10 439.05 = 22 10 439.95 = 417 10 So the sampling distribution of p 1 p 2 is approximately normal. Due to random assignment, these two groups can be viewed as independent. No 10% condition since there was no sampling. n 1 = n 2 = 878 2 = 439 Individual observations in each group should also be independent: knowing whether one subject quits smoking gives no information whether another subject does.
Do: Sampling Distribution of p 1 p 2 σ p1 p 2 = p c 1 p c n 1 0 N 0,.0202 0.10 + p c 1 p c n 2 =.10.90 439 p 1 p 2 =.10 p c = x 1 + x 2 66 + 22 = n 1 + n 2 878 = 88 878 =.10 +.10.90 439 =. 0202 z = p 1 p 2 p 1 p 2 σ p1 p 2 =.15.05 0 0.0202 = 4.95 area 0 p-value normalcdf.10, 99999, 0,.0202 = 0 p-value
Conclude: Assuming H 0 is true p 1 = p 2, there is a 0 probability of obtaining a p 1 p 2 value of.10 or more purely by chance. This provides strong evidence against H 0 and is statistically significant at α =.05 level 0 <.05. Therefore, we reject H 0 and can conclude that financial incentive helps people quit smoking. With calculator: STAT TESTS 2-PropZTest (6) x1: 66 n1: 439 x2: 22 n2: 439 p1: p2 <p2 >p2 z = 4.945 p = 0 p 1 = 0.15 p 2 = 0.05 p =.10 n 1 = 439 n 2 = 439 Experiment p-value pooled proportion treatments were imposed Yes Not a random sample, so conclusion only holds for the people in the experiment