Respuesta :
The rates of 6/23 and 318/811 for the age groups gives a confidence
interval for the difference snoring rate [tex]\underline{-0.206 < \hat{p}_1 - \hat{p}_2 < -0.056}[/tex]
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, [tex]\mathbf{\hat{p}_1}[/tex] = 48 ÷ 184 = [tex]\dfrac{6}{23}[/tex]
The number of older age group, n₂ = 318 + 493 = 811
Proportion of the older adult that snored, [tex]\mathbf{\hat{p}_2}[/tex] = 318 ÷ 811 = [tex]\dfrac{318}{811}[/tex]
The confidence interval, CI, for the difference in two rate (proportion) is
given as follows;
[tex]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}}}}[/tex]
The z-score at 96% confidence level is 2.05, which gives;
[tex]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}}}[/tex]
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) = [tex]\underline{-0.206 < \hat{p}_1 - \hat{p}_2 < -0.056}[/tex]
Learn more about the finding the confidence interval for the mean of a sample here:
https://brainly.com/question/6156233
