**Confidence Intervals: 1 Sample Mean Z**

**Introduction**

One Sample Mean Z confidence intervals are used when...

- You are dealing with a single sample mean (\(\bar{x}\))
- The POPULATION standard deviation (\(\sigma\)) is known

Because the population standard deviation is known, the critical value used will be z* (as opposed to t*).

Example

*You want to determine the mean score for students taking a national language test. You randomly select 35 students and find the sample mean to equal 40.714, and you know from prior studies that the population standard deviation is equal to 9. Construct a 70% cofidence interval for the mean language score. *

__Step 1__: Name the Confidence Interval: One Sample Mean Z

__Step 2__: Check the Conditions

1. Data is drawn from a random sample.

2. The sampling distribution of \(\bar{x}\) is approximately normal.

* This can be checked using the Central Limit Theorem.

3. N ≥ 10n

__Step 3__: Construct the Interval (Apply the Formula)

**1 Sample Mean Z Confidence Interval Formula: \(\bar{x}\) \(^+_-\) \((z^*)({\sigma \over {\sqrt {n}}}) \)**

\(\bar{x}\) \(^+_-\) \((z^*)({\sigma \over {\sqrt {n}}}) \)

→ 40.714 \(^+_-\) \((1.04)({9\over {\sqrt {35}}}) \)

→ 40.714 \(^+_-\) 1.582

40.714 - 1.582 = 39.13

40.714 + 1.582 = 42.3

Interval: (39.13, 42.3)

Note: The critical value was found using a z-table. A portion of the z-table is listed below with the part needed for our problem highlighted:

Confidence Level | z* Value |
---|---|

70% | 1.04 |

80% | 1.28 |

90% | 1.64 |

__Step 4__: State the Conclusion

Based on the data, I am 70% confident that the mean language test score is between 39.13 and 42.3.