1. Pull 2 samples of 20 pennies and record both averages (2 dots).
We are going to discuss 9.2 today.

2. Chapter 9 Testing a Claim Section 9.2 Tests About a Population
Proportion

3. Tests About a Population Proportion
STATE and CHECK the Random, 10%, and Large Counts conditions for performing a significance test about a population proportion.
CALCULATE the standardized test statistic and P-value for a test about a population proportion.
PERFORM a significance test about a population proportion.

4. Performing a Significance Test About p
Conditions for Performing a Significance Test
About a Proportion
• Random: The data come from a random sample from the population of interest.
Independent/10%: When sampling without replacement, n < 0.10N.
• Normal/Large Counts: Both np0 and n(1 – p0) are at least 10.

5. Performing a Significance Test About p
Conditions for Performing a Significance Test
About a Proportion
• Random: The data come from a random sample from the population of interest.
Independent/10%: When sampling without replacement, n < 0.10N.
• Normal/Large Counts: Both np0 and n(1 – p0) are at least 10.
Because we assume H0 is true when performing a significance test, we use the parameter value specified by the null hypothesis (denoted p0) when checking the Large Counts condition.

6. Performing a Significance Test About p
Conditions for Performing a Significance Test
About a Proportion
• Random: The data come from a random sample from the population of interest.
Independent/10%: When sampling without replacement, n < 0.10N.
• Normal/Large Counts: Both np0 and n(1 – p0) are at least 10.
Because we assume H0 is true when performing a significance test, we use the parameter value specified by the null hypothesis (denoted p0) when checking the Large Counts condition.
For a test, use the null value p0, not the sample proportion 𝑝 to check the Large Counts condition. That is, the expected count of successes np0 and of failures n(1 – p0) are both at least 10.

7. Performing a Significance Test About p
Problem: According to the U.S. Census Bureau, the proportion of students in high school who have a part-time job is An administrator at a local high school suspects that the proportion of students at her school who have a part-time job is less than the national figure. She would like to carry out a test at the α = 0.05 significance level of
H0: p = 0.25
Ha: p < 0.25
where p = the true proportion of all students at the school who have a part-time job. The administrator selects a random sample of 200 students from the school and finds that 39 of them have a part-time job. Check if the conditions for performing the significance test are met.
ArtWell/Shutterstock.com

8. Performing a Significance Test About p
Problem: She would like to carry out a test at the α = 0.05 significance level of
H0: p = 0.25
Ha: p < 0.25
where p = the true proportion of all students at the school who have a part-time job. The administrator selects a random sample of 200 students from the school and finds that 39 of them have a part-time job. Check if the conditions for performing the significance test are met.

9. Performing a Significance Test About p
Problem: She would like to carry out a test at the α = 0.05 significance level of
H0: p = 0.25
Ha: p < 0.25
where p = the true proportion of all students at the school who have a part-time job. The administrator selects a random sample of 200 students from the school and finds that 39 of them have a part-time job. Check if the conditions for performing the significance test are met.
• Random? Random sample of 200 students from the school. ✓
º Independent/10%: 200 is less than 10% of students at a large high school. ✓
• Normal/Large Counts? np0 = 200(0.25) = 50 ≥ 10 and n(1 – p0) = 200(1 – 0.25) = 150 ≥ 10 ✓

10. Performing a Significance Test About p
A basketball player claims to be an 80% free-throw shooter. You think the player is exaggerating.
H0: p = 0.25
Ha: p < 0.25

11. Performing a Significance Test About p
A basketball player claims to be an 80% free-throw shooter. You think the player is exaggerating.
H0: p = 0.25
Ha: p < 0.25
If H0 is true (if p = 0.80), then what values of 𝑝 should we expect in a random sample of 50 shots?

12. Performing a Significance Test About p
A basketball player claims to be an 80% free-throw shooter. You think the player is exaggerating.
H0: p = 0.25
Ha: p < 0.25
If H0 is true (if p = 0.80), then what values of 𝑝 should we expect in a random sample of 50 shots?
Shape: The sample size is 50 so we can say the sampling distribution of 𝑝 will be approximately Normal.

13. Performing a Significance Test About p
A basketball player claims to be an 80% free-throw shooter. You think the player is exaggerating.
H0: p = 0.25
Ha: p < 0.25
If H0 is true (if p = 0.80), then what values of 𝑝 should we expect in a random sample of 50 shots?
Shape: The sample size is 50 so we can say the sampling distribution of 𝑝 will be approximately Normal.
Center: For random samples, 𝜇 𝑝 = 𝑝 0 =0.80

14. Performing a Significance Test About p
A basketball player claims to be an 80% free-throw shooter. You think the player is exaggerating.
H0: p = 0.25
Ha: p < 0.25
If H0 is true (if p = 0.80), then what values of 𝑝 should we expect in a random sample of 50 shots?
Shape: The sample size is 50 so we can say the sampling distribution of 𝑝 will be approximately Normal.
Center: For random samples, 𝜇 𝑝 = 𝑝 0 =0.80
Variability: 𝜎 𝑝 = 𝑝 0 1− 𝑝 0 𝑛 = − =0.0566

15. Performing a Significance Test About p
A basketball player claims to be an 80% free-throw shooter. You think the player is exaggerating.
H0: p = 0.25
Ha: p < 0.25
If H0 is true (if p = 0.80), then what values of 𝑝 should we expect in a random sample of 50 shots?

16. Performing a Significance Test About p
A basketball player claims to be an 80% free-throw shooter. You think the player is exaggerating.
H0: p = 0.25
Ha: p < 0.25
If H0 is true (if p = 0.80), then what values of 𝑝 should we expect in a random sample of 50 shots?
The player makes 32 out of 50 shots.
𝑝 = =0.64

17. Performing a Significance Test About p
A basketball player claims to be an 80% free-throw shooter. You think the player is exaggerating.
H0: p = 0.25
Ha: p < 0.25
If H0 is true (if p = 0.80), then what values of 𝑝 should we expect in a random sample of 50 shots?
𝑧= 𝑝 − 𝑝 𝑝 0 1− 𝑝 0 𝑛 = 0.64− =−2.83

18. Performing a Significance Test About p
A basketball player claims to be an 80% free-throw shooter. You think the player is exaggerating.
H0: p = 0.25
Ha: p < 0.25
If H0 is true (if p = 0.80), then what values of 𝑝 should we expect in a random sample of 50 shots?
𝑧= 𝑝 − 𝑝 𝑝 0 1− 𝑝 0 𝑛 = 0.64− =−2.83
The value z = –2.83 is called the standardized test statistic.

19. Performing a Significance Test About p
A standardized test statistic measures how far a sample statistic is from what we would expect if the null hypothesis H0 were true, in standard deviation units. That is,
𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑𝑖𝑧𝑒𝑑 𝑡𝑒𝑠𝑡 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐= 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 −𝑝𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐

20. Performing a Significance Test About p
A standardized test statistic measures how far a sample statistic is from what we would expect if the null hypothesis H0 were true, in standard deviation units. That is,
𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑𝑖𝑧𝑒𝑑 𝑡𝑒𝑠𝑡 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐= 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 −𝑝𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐

21. Performing a Significance Test About p
A standardized test statistic measures how far a sample statistic is from what we would expect if the null hypothesis H0 were true, in standard deviation units. That is,
𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑𝑖𝑧𝑒𝑑 𝑡𝑒𝑠𝑡 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐= 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 −𝑝𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐
We can find the P-value using Table A or technology.

22. Performing a Significance Test About p
A standardized test statistic measures how far a sample statistic is from what we would expect if the null hypothesis H0 were true, in standard deviation units. That is,
𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑𝑖𝑧𝑒𝑑 𝑡𝑒𝑠𝑡 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐= 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 −𝑝𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐
Table A gives
P(z ≤ –2.83) =

23. Performing a Significance Test About p
A standardized test statistic measures how far a sample statistic is from what we would expect if the null hypothesis H0 were true, in standard deviation units. That is,
𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑𝑖𝑧𝑒𝑑 𝑡𝑒𝑠𝑡 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐= 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 −𝑝𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐
Table A gives
P(z ≤ –2.83) =
normalcdf(lower: –1000, upper:22.83, mean:0, SD:1) =

24. Performing a Significance Test About p
Problem: In the preceding example, an administrator at a local high school decided to perform a test at the α = significance level of
H0: p = 0.25
Ha: p < 0.25
where p = the true proportion of all students at the school who have a part-time job. The administrator selects a random sample of 200 students from the school and finds that 39 of them have a part-time job. We already confirmed that the conditions for performing a significance test are met.
(a) Explain why the sample result gives some evidence for the alternative hypothesis.
(b) Calculate the standardized test statistic and P-value.
(c) What conclusion would you make?
Andersen Ross/Exactostock-1555/Superstock

25. Performing a Significance Test About p
Problem:
H0: p = 0.25
Ha: p < 0.25
The administrator selects a random sample of 200 students from the school and finds that 39 of them have a part-time job.
(a) Explain why the sample result gives some evidence for the alternative hypothesis.
Andersen Ross/Exactostock-1555/Superstock

26. Performing a Significance Test About p
Problem:
H0: p = 0.25
Ha: p < 0.25
The administrator selects a random sample of 200 students from the school and finds that 39 of them have a part-time job.
(a) Explain why the sample result gives some evidence for the alternative hypothesis.
Andersen Ross/Exactostock-1555/Superstock
(a) The sample proportion of students at this school with a part-time job is 𝑝 = 39/200 = 0.195, which is less than the national proportion of p = 0.25 (as suggested by Ha).

27. Performing a Significance Test About p
Problem:
H0: p = 0.25
Ha: p < 0.25
The administrator selects a random sample of 200 students from the school and finds that 39 of them have a part-time job.
(b) Calculate the standardized test statistic and P-value.
Andersen Ross/Exactostock-1555/Superstock
(b)
𝑧= 0.195− =−1.80

28. Performing a Significance Test About p
Problem:
H0: p = 0.25
Ha: p < 0.25
The administrator selects a random sample of 200 students from the school and finds that 39 of them have a part-time job.
(b) Calculate the standardized test statistic and P-value.
Andersen Ross/Exactostock-1555/Superstock
(b) Using Table A:
P(z ≤ 21.80) =

29. Performing a Significance Test About p
Problem:
H0: p = 0.25
Ha: p < 0.25
The administrator selects a random sample of 200 students from the school and finds that 39 of them have a part-time job.
(b) Calculate the standardized test statistic and P-value.
Andersen Ross/Exactostock-1555/Superstock
(b) Using Technology:
normalcdf(lower: 21000, upper: 21.80, mean: 0, SD: 1) =

30. Performing a Significance Test About p
Problem:
H0: p = 0.25
Ha: p < 0.25
The administrator selects a random sample of 200 students from the school and finds that 39 of them have a part-time job.
(c) What conclusion would you make?
Andersen Ross/Exactostock-1555/Superstock
(c) Because the P-value of < α = 0.05, we reject H0. We have convincing evidence that the true proportion of all students at this large high school who have a part-time job is less than 0.25.

31. Putting It All Together: One-Sample z Test for p
Significance Tests: A Four-Step Process
State: State the hypotheses you want to test and the significance level, and define any parameters you use.

32. Putting It All Together: One-Sample z Test for p
Significance Tests: A Four-Step Process
State: State the hypotheses you want to test and the significance level, and define any parameters you use.
Plan: Identify the appropriate inference method and check 3 conditions.

33. Putting It All Together: One-Sample z Test for p
Significance Tests: A Four-Step Process
State: State the hypotheses you want to test and the significance level, and define any parameters you use.
Plan: Identify the appropriate inference method and check 3 conditions.
Do: If the conditions are met, perform calculations.
• Give the sample statistic(s).
• Calculate the standardized test statistic.
• Find the P-value.

34. Putting It All Together: One-Sample z Test for p
Significance Tests: A Four-Step Process
State: State the hypotheses you want to test and the significance level, and define any parameters you use.
Plan: Identify the appropriate inference method and check 3 conditions.
Do: If the conditions are met, perform calculations.
• Give the sample statistic(s).
• Calculate the standardized test statistic.
• Find the P-value.
Conclude: Make a conclusion about the hypotheses in the context of the problem.

35. Putting It All Together: One-Sample z Test for p
There are three conditions that must be met for the formula for the standardized test statistic to be valid—one for each of the three components in the formula.

36. Putting It All Together: One-Sample z Test for p
There are three conditions that must be met for the formula for the standardized test statistic to be valid—one for each of the three components in the formula.
1. The Random condition This condition helps ensure that 𝑝 – p0 is a good estimate for the difference between the true value of p and the null value of p0.

37. Putting It All Together: One-Sample z Test for p
There are three conditions that must be met for the formula for the standardized test statistic to be valid—one for each of the three components in the formula.
2. The Large Counts condition / Normal This condition allows us to use a Normal distribution to model the sampling distribution of 𝑝 . When this condition is met and H0 is true, the standardized test statistic z has approximately the standard Normal distribution.

38. Putting It All Together: One-Sample z Test for p
There are three conditions that must be met for the formula for the standardized test statistic to be valid—one for each of the three components in the formula.
3. The 10% condition / Independent This condition allows us to use the familiar formula for the standard deviation of the sampling distribution of 𝑝 (with p0 replacing p) when we are sampling without replacement from a finite population.

39. Putting It All Together: One-Sample z Test for p
One-Sample z Test for a Proportion
Suppose the conditions are met. To perform a test of H0: p = p0, compute
the standardized test statistic
𝑧= 𝑝 − 𝑝 𝑝 0 1− 𝑝 0 𝑛
Find the P-value by calculating the probability of getting a z statistic this large or larger in the direction specified by the alternative hypothesis Ha using a standard Normal distribution.

40. Putting It All Together: One-Sample z Test for p
Problem: Recall that the potato-chip producer we met in Section 9.1 and its main supplier agree that each shipment of potatoes must meet certain quality standards. If the producer finds convincing evidence that more than 8% of the potatoes in the shipment have “blemishes,” the truck will be sent away to get another load of potatoes from the supplier. Otherwise, the entire truckload will be used to make potato chips. The potato-chip producer has just received a truckload of potatoes from the supplier. A supervisor selects a random sample of 500 potatoes from the truck. An inspection reveals that 47 of the potatoes have blemishes. Is there convincing evidence at the α = 0.10 level that more than 8% of the potatoes in the shipment have blemishes?
imagestock/Getty Images

41. Putting It All Together: One-Sample z Test for p
Problem: A supervisor selects a random sample of 500 potatoes from the truck. An inspection reveals that 47 of the potatoes have blemishes. Is there convincing evidence at the α = 0.10 level that more than 8% of the potatoes in the shipment have blemishes?
imagestock/Getty Images

42. Putting It All Together: One-Sample z Test for p
Problem: A supervisor selects a random sample of 500 potatoes from the truck. An inspection reveals that 47 of the potatoes have blemishes. Is there convincing evidence at the α = 0.10 level that more than 8% of the potatoes in the shipment have blemishes?
imagestock/Getty Images
STATE:
We want to test
H0: p = 0.08
Ha: p > 0.08
where p = the true proportion of potatoes in this shipment with blemishes, using α = 0.10.

43. Putting It All Together: One-Sample z Test for p
Problem: A supervisor selects a random sample of 500 potatoes from the truck. An inspection reveals that 47 of the potatoes have blemishes. Is there convincing evidence at the α = 0.10 level that more than 8% of the potatoes in the shipment have blemishes?
imagestock/Getty Images
PLAN:
One-sample z test for p
• Random: Random sample of 500 potatoes from the shipment. ✓
º Independent/10%: It’s reasonable to assume that 500 * 10% of all potatoes in the shipment. ✓
• Normal/Large Counts:
500(0.08) = 40 ≥ 10 and 500(0.92) = 460 ≥ 10 ✓

44. Putting It All Together: One-Sample z Test for p
Problem: A supervisor selects a random sample of 500 potatoes from the truck. An inspection reveals that 47 of the potatoes have blemishes. Is there convincing evidence at the α = 0.10 level that more than 8% of the potatoes in the shipment have blemishes?
imagestock/Getty Images
DO:
𝑝 = =0.094
𝑧= 0.094− =1.15
P-value:
Using Table A: P(z ≥ 1 .15) = 1 – =
Using technology: normalcdf(lower:1.15, upper:1000, mean:0, SD:1) =

45. Putting It All Together: One-Sample z Test for p
Problem: A supervisor selects a random sample of 500 potatoes from the truck. An inspection reveals that 47 of the potatoes have blemishes. Is there convincing evidence at the α = 0.10 level that more than 8% of the potatoes in the shipment have blemishes?
imagestock/Getty Images
CONCLUDE:
Because our P-value of > α = 0.10, we fail to reject H0. There is not convincing evidence that the true proportion of blemished potatoes in the shipment is greater than 0.08.

46. Putting It All Together: One-Sample z Test for p
AP® Exam Tip
When a significance test leads to a fail to reject H0 decision, as in the preceding example, be sure to interpret the results as “We don’t have convincing evidence for Ha.” Saying anything that sounds like you believe H0 is (or might be) true will lead to a loss of credit. For instance,
it would be wrong to conclude, “There is convincing evidence that the true proportion of blemished potatoes is 0.08.” And don’t write responses as text messages, like “FTR the H0.”

47. Putting It All Together: One-Sample z Test for p
AP® Exam Tip
You can use your calculator to carry out the mechanics of a significance test on the AP® Statistics exam. But there’s a risk involved. If you give just the calculator answer with no work, and one or more of your values are incorrect, you will probably get no credit for the “Do” step. If you opt for the calculator-only method, be sure to name the procedure (one-sample z test for a proportion) and to report the standardized test statistic (z = 1.15) and P-value (0.1243).

48. Two-Sided Tests
The P-value in a one-sided test about a proportion is the area in one tail of a standard Normal distribution—the tail specified by Ha.

49. Two-Sided Tests
The P-value in a one-sided test about a proportion is the area in one tail of a standard Normal distribution—the tail specified by Ha.

50. Two-Sided Tests
The P-value in a one-sided test about a proportion is the area in one tail of a standard Normal distribution—the tail specified by Ha.
In a two-sided test, the alternative hypothesis has the form Ha: p ≠ p0. The P-value in such a test is the probability of getting a sample proportion as far as or farther from p0 in either direction than the observed value of 𝑝 .

51. Two-Sided Tests
The P-value in a one-sided test about a proportion is the area in one tail of a standard Normal distribution—the tail specified by Ha.
In a two-sided test, the alternative hypothesis has the form Ha: p ≠ p0. The P-value in such a test is the probability of getting a sample proportion as far as or farther from p0 in either direction than the observed value of 𝑝 .

52. Two-Sided Tests
Problem: According to the Centers for Disease Control and Prevention (CDC) website, 68% of high school students have never smoked a cigarette. Yanhong wonders whether this national result holds true in her large, urban high school. For her AP® Statistics class project, Yanhong surveys a simple
random sample of 150 students from her school. She gets responses from all 150 students, and 90 say that they have never smoked a cigarette. Is there convincing evidence that the CDC’s claim does not hold true at Yanhong’s school?
snowmeo/Getty Images

53. Two-Sided Tests
Problem: According to the Centers for Disease Control and Prevention (CDC) website, 68% of high school students have never smoked a cigarette. Yanhong wonders whether this national result holds true in her large, urban high school. For her AP® Statistics class project, Yanhong surveys a simple
random sample of 150 students from her school. She gets responses from all 150 students, and 90 say that they have never smoked a cigarette. Is there convincing evidence that the CDC’s claim does not hold true at Yanhong’s school?
snowmeo/Getty Images
STATE:
We want to test
H0: p = 0.68
Ha: p ≠ 0.68
where p = the proportion of all students in Yanhong’s school who would say they have never smoked a cigarette. We’ll use α = 0.05.

54. Two-Sided Tests
Problem: According to the Centers for Disease Control and Prevention (CDC) website, 68% of high school students have never smoked a cigarette. Yanhong wonders whether this national result holds true in her large, urban high school. For her AP® Statistics class project, Yanhong surveys a simple
random sample of 150 students from her school. She gets responses from all 150 students, and 90 say that they have never smoked a cigarette. Is there convincing evidence that the CDC’s claim does not hold true at Yanhong’s school?
snowmeo/Getty Images
PLAN:
One-sample z test for p.
• Random: Yanhong surveyed an SRS of 150 students. ✓
º Independent/10%: 150 students is less than 10% of all students at her large urban school. ✓
• Normal/Large Counts:
150(0.68) = 102 ≥ 10 and 150(0.32) = 48 ≥ 10 ✓

55. Two-Sided Tests
Problem: According to the Centers for Disease Control and Prevention (CDC) website, 68% of high school students have never smoked a cigarette. Yanhong wonders whether this national result holds true in her large, urban high school. For her AP® Statistics class project, Yanhong surveys a simple
random sample of 150 students from her school. She gets responses from all 150 students, and 90 say that they have never smoked a cigarette. Is there convincing evidence that the CDC’s claim does not hold true at Yanhong’s school?
snowmeo/Getty Images
DO:
𝑝 = =0.60
𝑧= 0.60− =−2.10

56. Two-Sided Tests
Problem: According to the Centers for Disease Control and Prevention (CDC) website, 68% of high school students have never smoked a cigarette. Yanhong wonders whether this national result holds true in her large, urban high school. For her AP® Statistics class project, Yanhong surveys a simple
random sample of 150 students from her school. She gets responses from all 150 students, and 90 say that they have never smoked a cigarette. Is there convincing evidence that the CDC’s claim does not hold true at Yanhong’s school?
snowmeo/Getty Images
DO:
Using Table A:
P(z ≤ –2.10 or z ≥ 2.10) = 2(0.0179) =

57. Two-Sided Tests
Problem: According to the Centers for Disease Control and Prevention (CDC) website, 68% of high school students have never smoked a cigarette. Yanhong wonders whether this national result holds true in her large, urban high school. For her AP® Statistics class project, Yanhong surveys a simple
random sample of 150 students from her school. She gets responses from all 150 students, and 90 say that they have never smoked a cigarette. Is there convincing evidence that the CDC’s claim does not hold true at Yanhong’s school?
snowmeo/Getty Images
DO:
Using technology: normalcdf(lower: –1000, upper: –2.10, mean:0, SD:1) × 2 =

58. Two-Sided Tests
Problem: According to the Centers for Disease Control and Prevention (CDC) website, 68% of high school students have never smoked a cigarette. Yanhong wonders whether this national result holds true in her large, urban high school. For her AP® Statistics class project, Yanhong surveys a simple
random sample of 150 students from her school. She gets responses from all 150 students, and 90 say that they have never smoked a cigarette. Is there convincing evidence that the CDC’s claim does not hold true at Yanhong’s school?
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CONCLUDE:
Because our P-value of < α = 0.05, we reject H0. We have convincing evidence that the proportion of all students at Yanhong’s school who would say they have never smoked a cigarette differs
from the CDC’s claim of 0.68.

59. Two-Sided Tests AP® Exam Tip
When making a conclusion in a significance test, be sure that you are describing the parameter and not the statistic. In the preceding example, it’s wrong to say that we have convincing evidence that the proportion of students at Yanhong’s school who said they have never smoked differs from the CDC’s claim of The “proportion who said they have never smoked” is the sample proportion, which is known to be The test gives convincing evidence that the proportion of all students at Yanhong’s school who would say they have
never smoked a cigarette differs from 0.68.

60. Section Summary
STATE and CHECK the Random, 10%, and Large Counts conditions for performing a significance test about a population proportion.
CALCULATE the standardized test statistic and P-value for a test about a population proportion.
PERFORM a significance test about a population proportion.

61. Assignment
9.2 p
#36, 38, 40, 46, 50, 56, 58, all
If you are stuck on any of these, look at the odd before or after and the answer in the back of your book. If you are still not sure text a friend or me for help (before 8pm).
Tomorrow we will check homework and review for 9.2 Quiz.