1. Chapter 9: testing a claim
Ch. 9-2 Tests About a Population Proportion

2. One or two sided?
one!
State:
𝐻 0 :
𝐻 𝑎 :
𝑝=0.8
𝑝→ true proportion of FT made by Brinkhus
𝑝 = =0.66
𝑝<0.8
𝛼=0.01

3. Plan:
One sample 𝑧 test for a proportion
Random:
“Think of these 50 shots as being an SRS”
Normal:
𝑛𝑝≥10
𝑛 1−𝑝 ≥10
→50 .8 =40≥10
→50 .2 =10≥10
NOTE: we’re using 𝑝, not 𝑝 !
So the sampling distribution of 𝑝 is approximately normal.
Independent:
Mr. Brinkhus has shot more than =500 free throws over the years.
You can also write that the observations are already independent. The outcome of one shot does not affect the outcome of another shot.

4. Do:
Sampling Distribution of 𝑝
𝜎 𝑝 = =.057
N 0.8, ______
.057
0.66
0.8
𝑧= 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐−𝑝𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟 𝑠𝑡. 𝑑𝑒𝑣. 𝑜𝑓 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐
= 𝑝 −𝑝 𝜎 𝑝
= 0.66− =−2.47
𝑎𝑟𝑒𝑎=.007
𝑝-value
normalcdf −99999, 0.66, 0.8, .057 =0.007

5. Conclude:
Assuming 𝐻 0 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 𝐻 0 and is statistically significant at 𝛼=.01 level .007<.01 .
Therefore, we reject 𝐻 0 and can conclude that the true proportion of
free throws made by Mr. Brinkhus is less than 0.8.
1) Interpret 𝑝-value
2) evidence
3) decision with context

6. When a problem doesn’t specify 𝛼, use 𝛼=0.05
One or two sided?
two!
𝑝→ true proportion of math teachers
who are left handed
State:
𝐻 0 :
𝐻 𝑎 :
𝑝=0.23
𝑝≠0.23
𝑝 = =0.28
𝛼=0.05
Plan:
One sample 𝑧 test for a proportion
Random:
“random sample of 100 math teachers”
Normal:
𝑛𝑝≥10
𝑛 1−𝑝 ≥10
→ =23≥10
→ =77≥10
So the sampling distribution of 𝑝 is approximately normal.
Independent:
Sampling without replacement so check 10% condition
We can assume there are more than =1000 math teachers in the country.

7. Do:
Sampling Distribution of 𝑝
𝜎 𝑝 = =.042
N 0.23, ______
.042
0.117
.05
.05
0.18
0.23
0.28
𝑧= 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐−𝑝𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟 𝑠𝑡. 𝑑𝑒𝑣. 𝑜𝑓 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐
= 𝑝 −𝑝 𝜎 𝑝
= 0.28− =1.19
𝑎𝑟𝑒𝑎=.117
normalcdf 0.28, 99999, 0.23, .042 =0.117
𝑝-value = =.234

8. Conclude:
Assuming 𝐻 0 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 𝐻 0 and is not statistically
significant at 𝛼=0.05 level (.234>.05).
Therefore, we fail to reject 𝐻 0
and cannot conclude that the true proportion of math teacher who are
left-handed is not 23%.

9. 𝑝→ true proportion of current high school students who have seen the 2002 Spider-Man movie
State:
𝐻 0 :
𝐻 𝑎 :
𝑝=0.3
𝑝≠0.3
𝑝 = =0.35
𝛼=0.05
Plan:
One sample 𝑧 test for a proportion
Random:
“SRS of 500 current high school students”
Normal:
𝑛𝑝≥10
𝑛 1−𝑝 ≥10
→ =150≥10
→ =350≥10
So the sampling distribution of 𝑝 is approximately normal.
Independent:
Sampling without replacement so check 10% condition
We can assume there are more than =5000 current high school students.

10. Do:
Sampling Distribution of 𝑝
𝜎 𝑝 = =.0205
N 0.3, _______
.0205
0.0073
.05
.05
0.25
0.3
0.35
𝑧= 𝑝 −𝑝 𝜎 𝑝
= 0.35− =2.44
𝑎𝑟𝑒𝑎=.0073
normalcdf .35, 99999, 0.3, =0.0073
𝑝-value = =.0146

11. Assuming 𝐻 0 is true 𝑝=.3 , there is a .015 probability of obtaining a
Conclude:
Assuming 𝐻 0 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 𝐻 0 and is statistically significant
at 𝛼=0.05 level (.015<.05).
Therefore, we reject 𝐻 0 and can conclude
that the true proportion of current high school students that have seen
the 2002 Spider-Man movie is not 0.3.
1-PropZTest (STAT→TESTS→5)
reject 𝐻 0
confidence interval
With calculator:
𝑧=2.44
𝑝=0.015
𝑝 =0.35
𝑛=500
STAT  TESTS  1-PropZTest (5)
𝑝-value
𝑝 0 : 𝑥: n:
prop: ≠ 𝑝 0 < 𝑝 0 > 𝑝 0
0.3
175
500

12. 𝑝 = =.35
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:
(same)
Normal:
𝑛 𝑝 ≥10
𝑛 1− 𝑝 ≥10
→ =175≥10
→ =325≥10
So the sampling distribution of 𝑝 is approximately normal.
Independent:
(same)

13. Estimate ± Margin of Error
𝑝 1− 𝑝 𝑛
𝑝 ± 𝑧 ∗
(.35)
.35 ± 1.96
.35 ±0.042
0.308, 0.392
We are 95% confident that the interval from to captures the true proportion of current high school students who have seen the Spider-Man movie.

14. plausible
strong
reject
𝑝=0.3
0.308, 0.392
plausible
weak
fail to reject

15. STAT → TESTS → 1-PropZInt 𝑥=122 𝑛=500 0.206, 0.282 C-Level: 0.95
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.

16. STAT → TESTS → 1-PropZTest
−1.79
0.073
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.
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 𝐻 𝑎 is one-sided, you must do a sig test.