| Number of complaints due to identity theft (out of 500) | Value of , the sample proportion | P-value | Do you think we have convincing evidence to suggest that Florida is exceeding the national trend? Why? | |
| 148 | 0.296 | 0.098 | 0.461 | No, because a sample proportion of 0.296 is not that unlikely given the national trend of 0.294. |
| 150 | ||||
| 155 | ||||
| 160 | ||||
| 165 | ||||
| 170 |
| P-value | Reject or Do not reject the null hypothesis | |
| 7) | 0.01 | |
| 8) | 0.11 | |
| 9) | 0.50 |
| When the alternative is: | The P-value is: | The P-value equals: | |
| 4) | Lower-tailed
( null value) |
How “unlikely” it is that we observed the sample data that resulted in a test statistic of or lower.
The area on the left of the test statistic under the standard normal curve. |
|
| 5) | Upper-tailed
( null value) |
How “unlikely” it is that we observed the sample data that resulted in a test statistic of or higher.
The area on the right side of the test statistic under the standard normal curve. |
|
| 6) | Two-tailed
( null value) |
How “unlikely” it is that we observed the sample data that resulted in a test statistic of or lower OR or higher.
The area on the right of the absolute value of the test statistic and the area on the left of the negative absolute value of the test statistic (i.e., more extreme). |
| Number of complaints due to identity theft (out of 500) | Value of , the sample proportion | P-value | Do you think we have convincing evidence to suggest that Florida is exceeding the national trend? Why? | |
| 148 | 0.296 | 0.461 | No, because a sample proportion of 0.296 is not that unlikely given the national trend of 0.294. | |
| 150 | 0.3 | 0.29 | 0.3859 | No, because a sample proportion of 0.30 is not that unlikely given the national trend of 0.294. |
| Skill or Concept: I can . . . | Questions to check your understanding | Rating from 1 to 5 |
| Describe what a P–value measures. | 2–6 | |
| Identify how a P-value is represented in a statistical distribution. | 4–6 |

| Company A | P-value = 0.021 | “At the 5% significance level, the data provide convincing evidence that more than half of single people believe that online dating is right for them.” |
| Company B | P-value = 0.063 | “At the 5% significance level, the data do not provide convincing evidence that more than half of single people believe that online dating is right for them.” |
| Decision | Conclusion |
| If P-value , there is enough evidence to reject the null hypothesis. | At the 100% significance level, the data provide convincing evidence in support of the alternative hypothesis. |
| If P-value, there is not enough evidence to reject the null hypothesis. | At the 100% significance level, the data do not provide convincing evidence in support of the alternative hypothesis. |
| Skill or Concept: I can . . . | Questions to check your understanding | Rating from 1 to 5 |
| Write the conclusion of a hypothesis test in context. | 1, Parts B and D
3, Part G |
|
| Conduct a complete hypothesis test for a proportion. | 3 | |
| Understand that there are limitations on P-values.
|
4 | |
| Verify that the conditions of the one-sample z-test for proportions have been met. | 2
3, Part C |

Glossary 11C
- percent
- out of one hundred.
- significance level
- the cut-off for P-values at which we have enough evidence to reject the null hypothesis.
- P-value
- the probability of obtaining a test statistic at least as extreme (in the direction of the alternative hypothesis) as the one that is actually seen if the null hypothesis is true.