The rates of **6/23** and **318/811** for the age groups gives a **confidence**

**interval **for the difference snoring rate

<h3>What method is used to calculate the confidence interval for the difference in rate?</h3>

The **number **of **younger age group**, **n₁** = 48 + 136 = 184

**Proportion **of the younger adult that snored, = 48 ÷ 184 =

The number of older age group, n₂ = 318 + 493 = **811**

Proportion of the** older adult **that snored, = 318 ÷ 811 =

The **confidence interval**, CI,** **for the **difference **in two rate (proportion) is

given as follows;

The **z-score **at **96% **confidence **level** is **2.05**, which gives;

Which gives;

Minimum value in the interval ≈ -0.206

Interval maximum value ≈ -0.056

The confidence interval for the different in the snoring rates between the younger and the older age is therefore;

CI = (-0.206, -0.056) =

Learn more about the finding the **confidence interval** for the **mean** of a sample here:

brainly.com/question/6156233