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3 years ago

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

Advanced Placement (AP)

1 answer:

Nata [24]3 years ago

3 0

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

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

<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, $\mathbf{\hat{p}_1}$ = 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:

brainly.com/question/6156233

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Solve the following differential equation with initial conditions: y''=e^-2t+10e^4t ; y(0)=1, y'(0)=0

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Answer:

Option A. $y = \frac{1}{4} e^{-2t} + \frac{5}{8} e^{4t} - 2 t + \frac{1}{8}$

Explanation:

This is a second order DE, so we'll need to integrate twice, applying initial conditions as we go. At a couple points, we'll need to apply u-substitution.

Round 1:

To solve the differential equation, write it as differentials, move the differential, and integrate both sides:

$y''=e^{-2t}+10e^{4t}$

$\frac{dy'}{dt}=e^{-2t}+10e^{4t}$

$dy'=[e^{-2t}+10e^{4t}]dt$

$\int dy'=\int [e^{-2t}+10e^{4t}]dt$

Applying various properties of integration:

$\int dy'=\int e^{-2t} dt + \int 10e^{4t}dt\\\int dy'=\int e^{-2t} dt + 10\int e^{4t}dt$

Prepare for integration by u-substitution

$\int dy'=\int e^{u_1} dt + 10\int e^{u_2}dt$ , letting $u_1=-2t$ and $u_2=4t$

Find dt in terms of $u_1 \text{ and } u_2$

$u_1=-2t\\du_1=-2dt\\-\frac{1}{2}du_1=dt$ $u_2=4t\\du_2=4dt\\\frac{1}{4}du_2=dt$

$\int dy'=\int e^{u_1} dt + 10\int e^{u_2}dt\\\int dy'=\int e^{u_1} (-\frac{1}{2} du_1) + 10\int e^{u_2} (\frac{1}{4} du_2)\\\int dy'=-\frac{1}{2} \int e^{u_1} (du_1) + 10 *\frac{1}{4} \int e^{u_2} (du_2)$

Using the Exponential rule (don't forget your constant of integration):

$y'=-\frac{1}{2} e^{u_1} + 10 *\frac{1}{4}e^{u_2} +C_1$

Back substituting for $u_1 \text{ and } u_2$ :

$y'=-\frac{1}{2} e^{(-2t)} + 10 *\frac{1}{4}e^{(4t)} +C_1\\y'=-\frac{1}{2} e^{-2t} + \frac{5}{2}e^{4t} +C_1\\$

Finding the constant of integration

Given initial condition $y'(0)=0$

$y'(t)=-\frac{1}{2} e^{-2t} + \frac{5}{2}e^{4t} +C_1\\0=y'(0)=-\frac{1}{2} e^{-2(0)} + \frac{5}{2}e^{4(0)} +C_1\\0=-\frac{1}{2} (1) + \frac{5}{2}(1) +C_1\\-2=C_1\\$

The first derivative with the initial condition applied: $y'(t)=-\frac{1}{2} e^{-2t} + \frac{5}{2}e^{4t} -2\\$

Round 2:

Integrate again:

$y' =-\frac{1}{2} e^{-2t} + \frac{5}{2}e^{4t} -2\\\frac{dy}{dt} =-\frac{1}{2} e^{-2t} + \frac{5}{2}e^{4t} -2\\dy =[-\frac{1}{2} e^{-2t} + \frac{5}{2}e^{4t} -2]dt\\\int dy =\int [-\frac{1}{2} e^{-2t} + \frac{5}{2}e^{4t} -2]dt\\\int dy =\int -\frac{1}{2} e^{-2t} dt + \int \frac{5}{2}e^{4t} dt - \int 2 dt\\\int dy = -\frac{1}{2} \int e^{-2t} dt + \frac{5}{2} \int e^{4t} dt - 2 \int dt\\$

$y = -\frac{1}{2} * -\frac{1}{2} e^{-2t} + \frac{5}{2} * \frac{1}{4} e^{4t} - 2 t + C_2\\y(t) = \frac{1}{4} e^{-2t} + \frac{5}{8} e^{4t} - 2 t + C_2$

Finding the constant of integration :

Given initial condition $y(0)=1$

$1=y(0) = \frac{1}{4} e^{-2(0)} + \frac{5}{8} e^{4(0)} - 2 (0) + C_2\\1 = \frac{1}{4} (1) + \frac{5}{8} (1) - (0) + C_2\\1 = \frac{7}{8} + C_2\\\frac{1}{8}=C_2$

So, $y(t) = \frac{1}{4} e^{-2t} + \frac{5}{8} e^{4t} - 2 t + \frac{1}{8}$

Checking the solution

$y(t) = \frac{1}{4} e^{-2t} + \frac{5}{8} e^{4t} - 2 t + \frac{1}{8}$

This matches our initial conditions here $y(0) = \frac{1}{4} e^{-2(0)} + \frac{5}{8} e^{4(0)} - 2 (0) + \frac{1}{8} = 1$

Going back to the function, differentiate:

$y' = [\frac{1}{4} e^{-2t} + \frac{5}{8} e^{4t} - 2 t + \frac{1}{8}]'\\y' = [\frac{1}{4} e^{-2t}]' + [\frac{5}{8} e^{4t}]' - [2 t]' + [\frac{1}{8}]'\\y' = \frac{1}{4} [e^{-2t}]' + \frac{5}{8} [e^{4t}]' - 2 [t]' + [\frac{1}{8}]'$

Apply Exponential rule and chain rule, then power rule

$y' = \frac{1}{4} e^{-2t}[-2t]' + \frac{5}{8} e^{4t}[4t]' - 2 [t]' + [\frac{1}{8}]'\\y' = \frac{1}{4} e^{-2t}(-2) + \frac{5}{8} e^{4t}(4) - 2 (1) + (0)\\y' = -\frac{1}{2} e^{-2t} + \frac{5}{2} e^{4t} - 2$

This matches our first order step and the initial conditions there.

$y'(0) = -\frac{1}{2} e^{-2(0)} + \frac{5}{2} e^{4(0)} - 2=0$

Going back to the function y', differentiate:

$y' = -\frac{1}{2} e^{-2t} + \frac{5}{2} e^{4t} - 2\\y'' = [-\frac{1}{2} e^{-2t} + \frac{5}{2} e^{4t} - 2]'\\y'' = [-\frac{1}{2} e^{-2t}]' + [\frac{5}{2} e^{4t}]' - [2]'\\y'' = -\frac{1}{2} [e^{-2t}]' + \frac{5}{2} [e^{4t}]' - [2]'$

Applying the Exponential rule and chain rule, then power rule

$y'' = -\frac{1}{2} e^{-2t}[-2t]' + \frac{5}{2} e^{4t}[4t]' - [2]'\\y'' = -\frac{1}{2} e^{-2t}(-2) + \frac{5}{2} e^{4t}(4) - (0)\\y'' = e^{-2t} + 10 e^{4t}$

So our proposed solution is a solution to the differential equation, and satisfies the initial conditions given.

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