Is Yawning Contagious video

10

34= .29

4

16= .25

𝑃 𝑦𝑎𝑤𝑛 𝑠𝑒𝑒𝑑

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 2

(no yawn seed)

Treatment 1

(yawn seed)

Compare

Yawning

What is this? 𝐻0

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.

𝑝 1 𝑝1

𝑝1

𝑝1 1 − 𝑝1

𝑛1

N 𝑝1, 𝜎𝑝 1

𝑝1

𝜎𝑝 1 =

𝑝 2 𝑝2

𝑝2

𝑝2 1 − 𝑝2

𝑛2

N 𝑝2, 𝜎𝑝 2

𝑝2

𝜎𝑝 2 =

Check: 𝑛1𝑝1 ≥ 10

𝑛1 1 − 𝑝1 ≥ 10

Check: 𝑛2𝑝2 ≥ 10

𝑛2 1 − 𝑝2 ≥ 10

𝑝 1 − 𝑝 2 𝑝1 − 𝑝2

𝑝1 − 𝑝2

𝑝1 1 − 𝑝1

𝑛1+

𝑝2 1 − 𝑝2

𝑛2

N 𝑝1 − 𝑝2, 𝜎𝑝 1−𝑝 2

𝑝1 − 𝑝2

𝜎𝑝 1−𝑝 2 =

Check: 𝑛1𝑝1 ≥ 10 𝑛2𝑝2 ≥ 10

𝑛1 1 − 𝑝1 ≥ 10 𝑛2 1 − 𝑝2 ≥ 10 &

𝑝 1 − 𝑝 2 𝑝1 − 𝑝2

𝑝1 − 𝑝2

𝑝1 1 − 𝑝1

𝑛1+

𝑝2 1 − 𝑝2

𝑛2

N 𝑝1 − 𝑝2, 𝜎𝑝 1−𝑝 2

𝑝1 − 𝑝2

𝜎𝑝 1−𝑝 2 =

Check: 𝑛1𝑝1 ≥ 10 𝑛2𝑝2 ≥ 10

𝑛1 1 − 𝑝1 ≥ 10 𝑛2 1 − 𝑝2 ≥ 10 &

We want to estimate the true difference 𝑝1 − 𝑝2 at a 95%

confidence level.

𝑝 1 =114

150= .76 𝑝 2 =

62

100= .62 𝑝 1 − 𝑝 2 = .14

𝑝1 → true proportion of ECRCHS students who do math HW

𝑝2 → true proportion of Taft students who do math HW

Two-sample 𝑧 interval for 𝑝1 − 𝑝2

Random:

Normal:

Independent:

“random sample of 150 students from ECRCHS” and

“random sample of 100 Taft students”

𝑛1𝑝 1 ≥ 10

𝑛1 1 − 𝑝 1 ≥ 10

→ 150 .76 = 114 ≥ 10

→ 150 .24 = 36 ≥ 10

So the sampling distribution of 𝑝 1 − 𝑝 2 is approximately normal.

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.

𝑛2𝑝 2 ≥ 10

𝑛2 1 − 𝑝 2 ≥ 10

→ 100 .62 = 62 ≥ 10

→ 100 .38 = 38 ≥ 10

There are at least 10 150 = 1500 students at ECRCHS and

at least 10 100 = 1000 students at Taft.

𝑝 1 − 𝑝 2 ± 𝑧∗ 𝑝 1 1 − 𝑝 1

𝑛1+

𝑝 2 1 − 𝑝 2𝑛2

Estimate ± Margin of Error

. 76 − .62 ± 1.96

.14 ± 0.117

0.023, 0.257

With calculator:

114

0.95

0.02286, 0.25714

STAT TESTS 2-PropZInt (B)

𝑝 1 =114

150= .76 𝑝 2 =

62

100= .62 𝑝 1 − 𝑝 2 = .14

x1:

n1:

x2:

n2:

C-Level:

.76 .24

150+

.62 .38

100

Standard Error

150

62

100

We are 95% confident that the interval from 0.023 to 0.257

captures the true difference in proportions 𝑝1 − 𝑝2 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.

We want to estimate the true difference 𝑝1 − 𝑝2 at a 90%

confidence level.

𝑝 1 = .63 𝑝 2 = .68 𝑝 1 − 𝑝 2 = −.05

𝑝1 → true proportion of teens who say they go online every day

𝑝2 → true proportion of adults who say they go online every day

Two-sample 𝑧 interval for 𝑝1 − 𝑝2

Random:

Normal:

Independent:

“random sample of 800 teens and a separate sample

of 2253 adults”

𝑛1𝑝 1 ≥ 10

𝑛1 1 − 𝑝 1 ≥ 10

→ 800 .63 = 504 ≥ 10

→ 800 .37 = 296 ≥ 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.

𝑛2𝑝 2 ≥ 10

𝑛2 1 − 𝑝 2 ≥ 10

→ 2253 .68 = 1532 ≥ 10

→ 2253 .32 = 721 ≥ 10

There are at least 10 800 = 8000 U.S. teens and

at least 10 2253 = 22530 U.S. adults.

𝑝 1 − 𝑝 2 ± 𝑧∗ 𝑝 1 1 − 𝑝 1

𝑛1+

𝑝 2 1 − 𝑝 2𝑛2

Estimate ± Margin of Error

. 63 − .68 ± 1.645

−0.05 ± 0.0324

−0.0824,−0.0176

With calculator:

504

0.9

−0.0824,−0.0176

STAT TESTS 2-PropZInt (B)

x1:

n1:

x2:

n2:

C-Level:

.63 .37

800+

.68 .32

2253

Standard Error

800

1532

2253

𝑝 1 = .63 𝑝 2 = .68 𝑝 1 − 𝑝 2 = −.05

We are 90% confident that the interval from −.0824 to

− .0176 captures the true difference in proportions

𝑝1 − 𝑝2 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:

𝐻0:

𝐻𝑎:

𝑝1 − 𝑝2 = 0

𝑝1 → proportion of AP Calc seniors who are going to college

𝛼 = 0.05

𝑝 1 =98

110= 0.89

𝑝2 → proportion of Pre-Calc seniors who are going to college

𝑝1 − 𝑝2 > 0

𝑝1 = 𝑝2 OR

𝑝1 > 𝑝2 OR 𝑝 2 =

68

80= 0.85

𝑝 1 − 𝑝 2 = .89 − .85 = .04

Plan: Two sample 𝑧 test for 𝑝1 − 𝑝2

Random:

Normal:

Independent:

“SRS of 80 students currently taking Pre-Calc” and

“SRS of 110 students currently taking AP Calculus”

𝑛1𝑝 1 ≥ 10

𝑛1 1 − 𝑝 1 ≥ 10

→ 110 .89 = 98 ≥ 10

→ 110 .11 = 12 ≥ 10

So the sampling distribution of 𝑝 1 − 𝑝 2 is approximately normal.

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.

𝑛2𝑝 2 ≥ 10

𝑛2 1 − 𝑝 2 ≥ 10

→ 80 .85 = 68 ≥ 10

→ 80 .15 = 12 ≥ 10

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, ______

0

𝜎𝑝 1−𝑝 2 =𝑝1 1 − 𝑝1

𝑛1+

𝑝2 1 − 𝑝2

𝑛2

Normally, we would use

this formula for the

standard deviation:

However, since we’re assuming 𝐻0: 𝑝1 = 𝑝2 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

=count of successes in both samples

count of individuals in both samples=

𝑥1 + 𝑥2

𝑛1 + 𝑛2 𝑝 𝑐

We will use this in place of both 𝑝1 and 𝑝2

Do: Sampling Distribution of 𝑝 1 − 𝑝 2

N 0, ______

0

𝜎𝑝 1−𝑝 2 =𝑝 1 1 − 𝑝 1

𝑛1+

𝑝 2 1 − 𝑝 2𝑛2

=𝑝 𝑐 1 − 𝑝 𝑐

𝑛1+

𝑝 𝑐 1 − 𝑝 𝑐𝑛2

.049

0.04

𝑧 =𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 − 𝑝𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟

𝑠𝑡. 𝑑𝑒𝑣. 𝑜𝑓 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 =

𝑝 1 − 𝑝 2 − 𝑝1 − 𝑝2

𝜎𝑝 1−𝑝 2

=.04 − 0

0.049= .82

𝑎𝑟𝑒𝑎 = .2061

𝑝-value normalcdf .04, 99999, 0, .049 = 0.2071

𝑝 𝑐 =𝑥1 + 𝑥2

𝑛1 + 𝑛2=

98 + 68

110 + 80

=166

190

= .874

=.874 .126

110+

.874 .126

80= .049

𝑝-value

𝑝 1 − 𝑝 2 = .04

With calculator:

98

STAT TESTS 2-PropZTest (6)

x1:

n1:

x2:

n2:

p1: ≠p2 <p2 >p2

110

68

80

𝑧 = 0.838

𝑝 = 0.201

𝑝 1 = 0.89

𝑝 2 = 0.85

𝑝 = .874

𝑛1 = 110

𝑛2 = 80

𝑝-value

pooled

proportion

Conclude:

Assuming 𝐻0 is true 𝑝1 = 𝑝2 , 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 𝐻0 and is not statistically significant at 𝛼 = .05 level

Therefore, we fail to reject 𝐻0 and cannot conclude

that seniors taking AP Calc are more likely to attend college next

.2061 > .05 .

year than seniors taking Pre-Calc.

Observational study

no treatments

imposed

No

senior math students in California What population can we target?

State:

𝐻0:

𝐻𝑎:

𝑝1 − 𝑝2 = 0

𝑝1 → true proportion who quit smoking to get money

𝛼 = 0.05

𝑝 1 = 0.15

𝑝2 → true proportion who quit smoking traditional way

𝑝1 − 𝑝2 > 0

𝑝1 = 𝑝2 OR

𝑝1 > 𝑝2 OR

𝑝 2 = 0.05

𝑝 1 − 𝑝 2 = .15 − .05 = .10

Plan: Two sample 𝑧 test for 𝑝1 − 𝑝2

Random:

Normal:

Independent:

“subjects were randomly assigned to treatments”

𝑛1𝑝 1 ≥ 10

𝑛1 1 − 𝑝 1 ≥ 10

→ 439 .15 = 66 ≥ 10

→ 439 .85 = 373 ≥ 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.

𝑛2𝑝 2 ≥ 10

𝑛2 1 − 𝑝 2 ≥ 10

→ 439 .05 = 22 ≥ 10

→ 439 .95 = 417 ≥ 10

Individual observations in each group should also be independent:

knowing whether one subject quits smoking gives no information

whether another subject does.

𝑛1 = 𝑛2 =878

2

= 439

No 10% condition since there was no sampling.

Do: Sampling Distribution of 𝑝 1 − 𝑝 2

N 0, ________

0

𝜎𝑝 1−𝑝 2 =𝑝 𝑐 1 − 𝑝 𝑐

𝑛1+

𝑝 𝑐 1 − 𝑝 𝑐𝑛2

=.10 .90

439+

.10 .90

439=. 0202

.0202

0.10

𝑧 =𝑝 1 − 𝑝 2 − 𝑝1 − 𝑝2

𝜎𝑝 1−𝑝 2

=.15 − .05 − 0

0.0202= 4.95

𝑎𝑟𝑒𝑎 ≈ 0

𝑝-value

normalcdf .10, 99999, 0, .0202 = 0

𝑝 𝑐 =𝑥1 + 𝑥2

𝑛1 + 𝑛2=

66 + 22

878

=88

878

= .10

𝑝-value

𝑝 1 − 𝑝 2 = .10

With calculator:

66

STAT TESTS 2-PropZTest (6)

x1:

n1:

x2:

n2:

p1: ≠p2 <p2 >p2

439

22

439

𝑧 = 4.945

𝑝 = 0

𝑝 1 = 0.15

𝑝 2 = 0.05

𝑝 = .10

𝑛1 = 439

𝑛2 = 439

𝑝-value

pooled

proportion

Conclude:

Assuming 𝐻0 is true 𝑝1 = 𝑝2 , 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 𝐻0 and is statistically significant at 𝛼 = .05 level

Therefore, we reject 𝐻0 and can conclude that financial

incentive helps people quit smoking.

0 < .05 .

Experiment

treatments were

imposed

Yes

Not a random sample, so conclusion only

holds for the people in the experiment