Ch. 9-2 Tests About a Population Proportion
Ch. 9-2 Tests About a Population Proportion
State:
One or two sided? one!
๐ป0:
๐ป๐:
๐ = 0.8
๐ < 0.8
๐ โ true proportion of FT made by Brinkhus
๐ผ = 0.01 ๐ =33
50= 0.66
Plan: One sample ๐ง test for a proportion
Random:
Normal:
Independent:
โThink of these 50 shots as being an SRSโ
๐๐ โฅ 10
๐ 1 โ ๐ โฅ 10
โ 50 .8 = 40 โฅ 10
โ 50 .2 = 10 โฅ 10
So the sampling distribution of ๐ is approximately normal.
Mr. Brinkhus has shot more than 10 50 = 500 free throws over the years.
NOTE: weโre using ๐, not ๐ !
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 ๐
N 0.8, ______
0.8
๐๐ =.8 .2
50= .057
.057
0.66
๐ง =๐ ๐ก๐๐ก๐๐ ๐ก๐๐ โ ๐๐๐๐๐๐๐ก๐๐
๐ ๐ก. ๐๐๐ฃ. ๐๐ ๐ ๐ก๐๐ก๐๐ ๐ก๐๐ =
๐ โ ๐
๐๐ =
0.66 โ 0.8
0.057= โ2.47
๐๐๐๐ = .007
๐-value
normalcdf โ99999, 0.66, 0.8, .057 = 0.007
Assuming ๐ป0 is true ๐ = .8 , there is a .007 probability of obtaining a
๐ value of .66 or lower purely by chance. This provides strong evidence
against ๐ป0 and is statistically significant at ๐ผ = .01 level .007 < .01 .
Therefore, we reject ๐ป0 and can conclude that the true proportion of
free throws made by Mr. Brinkhus is less than 0.8.
Conclude:
1) Interpret ๐-value 2) evidence 3) decision with context
When a problem doesnโt specify ๐ผ, use ๐ผ = 0.05
One or two sided? two!
State: ๐ป0:
๐ป๐:
๐ = 0.23
๐ โ 0.23
๐ โ true proportion of math teachers who are left handed
๐ผ = 0.05 ๐ =28
100= 0.28
Plan: One sample ๐ง test for a proportion
Random:
Normal:
Independent:
โrandom sample of 100 math teachersโ
๐๐ โฅ 10
๐ 1 โ ๐ โฅ 10
โ 100 .23 = 23 โฅ 10
โ 100 .77 = 77 โฅ 10
So the sampling distribution of ๐ 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 ๐
N 0.23, ______
0.23
๐๐ =.23 .77
100= .042
.042
0.18
๐ง =๐ ๐ก๐๐ก๐๐ ๐ก๐๐ โ ๐๐๐๐๐๐๐ก๐๐
๐ ๐ก. ๐๐๐ฃ. ๐๐ ๐ ๐ก๐๐ก๐๐ ๐ก๐๐ =
๐ โ ๐
๐๐ =
0.28 โ 0.23
0.042= 1.19
๐๐๐๐ = .117
๐-value = 2 .117 = .234
normalcdf 0.28, 99999, 0.23, .042 = 0.117
0.28
0.117 .05 .05
Assuming ๐ป0 is true ๐ = .23 , there is a .234 probability of obtaining a
๐ value that is .05 or more away from ๐ purely by chance.
This provides weak evidence against ๐ป0 and is not statistically
and cannot conclude that the true proportion of math teacher who are
left-handed is not 23%.
Conclude:
significant at ๐ผ = 0.05 level (.234 > .05). Therefore, we fail to reject ๐ป0
State: ๐ป0:
๐ป๐:
๐ = 0.3
๐ โ 0.3
๐ โ true proportion of current high school students who have seen the 2002 Spider-Man movie
๐ผ = 0.05 ๐ =175
500= 0.35
Plan: One sample ๐ง test for a proportion
Random:
Normal:
Independent:
โSRS of 500 current high school studentsโ
๐๐ โฅ 10
๐ 1 โ ๐ โฅ 10
โ 500 .3 = 150 โฅ 10
โ 500 .7 = 350 โฅ 10
So the sampling distribution of ๐ 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 ๐
N 0.3, _______
0.3
๐๐ =.3 .7
500= .0205
.0205
0.25
๐ง =๐ โ ๐
๐๐ =
0.35 โ 0.3
0.0205= 2.44
๐๐๐๐ = .0073
๐-value = 2 .0073 = .0146
normalcdf .35, 99999, 0.3, .0205 = 0.0073
0.35
0.0073 .05 .05
Assuming ๐ป0 is true ๐ = .3 , there is a .015 probability of obtaining a
๐ value that is 0.05 or more away from ๐ purely by chance.
This provides strong evidence against ๐ป0 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 ๐ผ = 0.05 level (.015 < .05). Therefore, we reject ๐ป0 and can conclude
1-PropZTest (STATโTESTSโ5)
reject ๐ป0
confidence interval
With calculator:
๐0: ๐ฅ: n: prop: โ ๐0 < ๐0 > ๐0
175
STAT TESTS 1-PropZTest (5)
0.3
500
๐ง = 2.44 ๐ = 0.015 ๐ = 0.35 ๐ = 500
๐-value
We want to estimate the actual proportion, ๐, of high school students who have seen the Spider-Man movie at a 95% confidence level.
One-sample ๐ง interval for proportion
Random:
Normal:
Independent:
(same)
๐๐ โฅ 10
๐ 1 โ ๐ โฅ 10
โ 500 .35 = 175 โฅ 10
โ 500 .65 = 325 โฅ 10
So the sampling distribution of ๐ is approximately normal.
๐ =175
500= .35
(same)
๐ ยฑ ๐งโ
๐ 1 โ ๐
๐
Estimate ยฑ Margin of Error
.35 ยฑ 1.96 (.35) .65
500
.35 ยฑ 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 ๐ = 0.3
STAT โ TESTS โ 1-PropZInt ๐ฅ = 122 ๐ = 500
C-Level: 0.95 0.206, 0.282
notice ๐ = .28 is captured just barely
Make a guess based off of ๐ in the interval. Is ๐-value going to be greater or less than .05? By how much?
.06?
๐ = .28 is captured by the 95% confidence interval, so it is NOT statistically significant at ๐ผ = .05.
STAT โ TESTS โ 1-PropZTest
โ1.79
If ๐ผ = 0.05 is being used, notice ๐ง is less than 1.96 std dev away from ๐, so ๐-value will be higher than ๐ผ.
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, itโs acceptable to use a confidence interval rather than a sig test to address a two-sided alternative hypothesis. HOWEVER, if ๐ป๐ is one-sided, you must do a sig test.