Ch. 9-2 Tests About a Population Proportion

State:

One or two sided? one!

𝐻0:

𝐻𝑎:

𝑝 = 0.8

𝑝 < 0.8

𝑝 → true proportion of FT made by Brinkhus

𝛼 = 0.01 𝑝 =33

50= 0.66

Plan: One sample 𝑧 test for a proportion

Random:

Normal:

Independent:

“Think of these 50 shots as being an SRS”

𝑛𝑝 ≥ 10

𝑛 1 − 𝑝 ≥ 10

→ 50 .8 = 40 ≥ 10

→ 50 .2 = 10 ≥ 10

So the sampling distribution of 𝑝 is approximately normal.

Mr. Brinkhus has shot more than 10 50 = 500 free throws over the years.

NOTE: we’re using 𝑝, not 𝑝 !

You can also write that the observations are already independent. The outcome of one shot does not affect the outcome of another shot.

Do: Sampling Distribution of 𝑝

N 0.8, ______

0.8

𝜎𝑝 =.8 .2

50= .057

.057

0.66

𝑧 =𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 − 𝑝𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟

𝑠𝑡. 𝑑𝑒𝑣. 𝑜𝑓 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 =

𝑝 − 𝑝

𝜎𝑝 =

0.66 − 0.8

0.057= −2.47

𝑎𝑟𝑒𝑎 = .007

𝑝-value

normalcdf −99999, 0.66, 0.8, .057 = 0.007

Assuming 𝐻0 is true 𝑝 = .8 , there is a .007 probability of obtaining a

𝑝 value of .66 or lower purely by chance. This provides strong evidence

against 𝐻0 and is statistically significant at 𝛼 = .01 level .007 < .01 .

Therefore, we reject 𝐻0 and can conclude that the true proportion of

free throws made by Mr. Brinkhus is less than 0.8.

Conclude:

1) Interpret 𝑝-value 2) evidence 3) decision with context

When a problem doesn’t specify 𝛼, use 𝛼 = 0.05

One or two sided? two!

State: 𝐻0:

𝐻𝑎:

𝑝 = 0.23

𝑝 ≠ 0.23

𝑝 → true proportion of math teachers who are left handed

𝛼 = 0.05 𝑝 =28

100= 0.28

Plan: One sample 𝑧 test for a proportion

Random:

Normal:

Independent:

“random sample of 100 math teachers”

𝑛𝑝 ≥ 10

𝑛 1 − 𝑝 ≥ 10

→ 100 .23 = 23 ≥ 10

→ 100 .77 = 77 ≥ 10

So the sampling distribution of 𝑝 is approximately normal.

We can assume there are more than 10 100 = 1000 math teachers in the country.

Sampling without replacement so check 10% condition

Do: Sampling Distribution of 𝑝

N 0.23, ______

0.23

𝜎𝑝 =.23 .77

100= .042

.042

0.18

𝑧 =𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 − 𝑝𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟

𝑠𝑡. 𝑑𝑒𝑣. 𝑜𝑓 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 =

𝑝 − 𝑝

𝜎𝑝 =

0.28 − 0.23

0.042= 1.19

𝑎𝑟𝑒𝑎 = .117

𝑝-value = 2 .117 = .234

normalcdf 0.28, 99999, 0.23, .042 = 0.117

0.28

0.117 .05 .05

Assuming 𝐻0 is true 𝑝 = .23 , there is a .234 probability of obtaining a

𝑝 value that is .05 or more away from 𝑝 purely by chance.

This provides weak evidence against 𝐻0 and is not statistically

and cannot conclude that the true proportion of math teacher who are

left-handed is not 23%.

Conclude:

significant at 𝛼 = 0.05 level (.234 > .05). Therefore, we fail to reject 𝐻0

State: 𝐻0:

𝐻𝑎:

𝑝 = 0.3

𝑝 ≠ 0.3

𝑝 → true proportion of current high school students who have seen the 2002 Spider-Man movie

𝛼 = 0.05 𝑝 =175

500= 0.35

Plan: One sample 𝑧 test for a proportion

Random:

Normal:

Independent:

“SRS of 500 current high school students”

𝑛𝑝 ≥ 10

𝑛 1 − 𝑝 ≥ 10

→ 500 .3 = 150 ≥ 10

→ 500 .7 = 350 ≥ 10

So the sampling distribution of 𝑝 is approximately normal.

We can assume there are more than 10 500 = 5000 current high school students.

Sampling without replacement so check 10% condition

Do: Sampling Distribution of 𝑝

N 0.3, _______

0.3

𝜎𝑝 =.3 .7

500= .0205

.0205

0.25

𝑧 =𝑝 − 𝑝

𝜎𝑝 =

0.35 − 0.3

0.0205= 2.44

𝑎𝑟𝑒𝑎 = .0073

𝑝-value = 2 .0073 = .0146

normalcdf .35, 99999, 0.3, .0205 = 0.0073

0.35

0.0073 .05 .05

Assuming 𝐻0 is true 𝑝 = .3 , there is a .015 probability of obtaining a

𝑝 value that is 0.05 or more away from 𝑝 purely by chance.

This provides strong evidence against 𝐻0 and is statistically significant

that the true proportion of current high school students that have seen

the 2002 Spider-Man movie is not 0.3.

Conclude:

at 𝛼 = 0.05 level (.015 < .05). Therefore, we reject 𝐻0 and can conclude

1-PropZTest (STAT→TESTS→5)

reject 𝐻0

confidence interval

With calculator:

𝑝0: 𝑥: n: prop: ≠ 𝑝0 < 𝑝0 > 𝑝0

175

STAT TESTS 1-PropZTest (5)

0.3

500

𝑧 = 2.44 𝑝 = 0.015 𝑝 = 0.35 𝑛 = 500

𝑝-value

We want to estimate the actual proportion, 𝑝, of high school students who have seen the Spider-Man movie at a 95% confidence level.

One-sample 𝑧 interval for proportion

Random:

Normal:

Independent:

(same)

𝑛𝑝 ≥ 10

𝑛 1 − 𝑝 ≥ 10

→ 500 .35 = 175 ≥ 10

→ 500 .65 = 325 ≥ 10

So the sampling distribution of 𝑝 is approximately normal.

𝑝 =175

500= .35

(same)

𝑝 ± 𝑧∗

𝑝 1 − 𝑝

𝑛

Estimate ± Margin of Error

.35 ± 1.96 (.35) .65

500

.35 ± 0.042

0.308, 0.392

We are 95% confident that the interval from 0.308 to 0.392 captures the true proportion of current high school students who have seen the Spider-Man movie.

plausible strong

reject

plausible weak

fail to reject

0.308, 0.392 𝑝 = 0.3

STAT → TESTS → 1-PropZInt 𝑥 = 122 𝑛 = 500

C-Level: 0.95 0.206, 0.282

notice 𝑝 = .28 is captured just barely

Make a guess based off of 𝑝 in the interval. Is 𝑝-value going to be greater or less than .05? By how much?

.06?

𝑝 = .28 is captured by the 95% confidence interval, so it is NOT statistically significant at 𝛼 = .05.

STAT → TESTS → 1-PropZTest

−1.79

If 𝛼 = 0.05 is being used, notice 𝑧 is less than 1.96 std dev away from 𝑝, so 𝑝-value will be higher than 𝛼.

0.073

However, this only gives us information about one sample. Confidence intervals give more info than significance tests. A confidence interval gives a whole range of plausible values, whereas a sig test concentrates only on the one statistic as a possibility for the population proportion. On the AP Exam, it’s acceptable to use a confidence interval rather than a sig test to address a two-sided alternative hypothesis. HOWEVER, if 𝐻𝑎 is one-sided, you must do a sig test.