Question

K1: I want to know whether snoring rates differ for those under and over 30 years old. The data

Answer 1

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

interval for the difference snoring rate $\underline{-0.206 < \hat{p}_1 - \hat{p}_2 < -0.056}$

What method is used to calculate the confidence interval for the difference in rate?

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

Proportion of the younger adult that snored, $\mathbf{\hat{p}_1}$ = 48 ÷ 184 = $\dfrac{6}{23}$

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

Proportion of the older adult that snored, $\mathbf{\hat{p}_2}$ = 318 ÷ 811 = $\dfrac{318}{811}$

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

given as follows;

$C.I. = \mathbf{\hat{p}_1-\hat{p}_2\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}_1\left (1-\hat{p}_1 \right )}{n_{1}}+\dfrac{\hat{p}_2\left (1-\hat{p}_2 \right )}{n_{2}}}}$

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

$CI;\ \mathbf{\left(\dfrac{6}{23}-\dfrac{318}{811} \right)\pm 2.05 \times \sqrt{\dfrac{\dfrac{6}{23} \imes \left (1-\dfrac{6}{23} \right )}{184}+\dfrac{\dfrac{318}{811} \times \left (1-\dfrac{318}{811} \right )}{811}}}$

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) = $\underline{-0.206 < \hat{p}_1 - \hat{p}_2 < -0.056}$

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

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