Question

Sleep apnea: Sleep apnea is a disorder in which there are pauses in breathing during sleep. People with this condition must w up frequently to breathe. In a sample of 432 people aged 65 and over, 108 of them had sleep apnea.
Construct a 99.5% confidence interval for the proportion of those aged 65 and over who have sleep apnea. Round the answer to three decimal places.

Answer 1

Answer:The required interval is (0.196, 0.030).

Step-by-step explanation:

Since we have given that

n = 432

x = 108

So, $\hat{p}=\dfrac{x}{n}=\dfrac{108}{432}=0.25$

at 99% level of confidence, z = 2.576

So, Margin of error would be

$z\times \sqrt{\dfrac{pq}{n}}\\\\=2.576\times \sqrt{\dfrac{0.25\times 0.75}{432}}\\\\=2.576\times 0.021\\\\=0.0536$

So, margin of error = 0.0536

Now, confidence interval would be

$\hat{p}\pm 0.0536\\\\=(0.25-0.0536,0.25+0.0536)\\\\=(0.1964,0.3036)\\\\\approx (0.196,0.030)$

Hence, the required interval is (0.196, 0.030).

Answer 2

Answer:

99.5% confidence interval = [0.191 , 0.309]

Step-by-step explanation:

We are given that in a sample of 432 people aged 65 and over, 108 of them had sleep apnea.

The pivotal quantity for 99.5% confidence interval for the proportion of those aged 65 and over who have sleep apnea is given by;

           P.Q. = $\frac{\hat p - p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion = 108/432 = 0.25

            p = population proportion

            n = sample size = 432

So, 99.5% confidence interval for the population proportion, p is given by;

P(-2.813 < N(0,1) < 2.813) = 0.995 {At 0.5% level of significance z table gives

                                                          critical value of 2.813}

P(-2.813 < $\frac{\hat p - p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 2.813) = 0.995

P(-2.813 * ${\sqrt{\frac{\hat p(1-\hat p)}{n} }$ < ${\hat p - p}$ < 2.813 * ${\sqrt{\frac{\hat p(1-\hat p)}{n} }$ ) = 0.995

P($\hat p$ - 2.813 * ${\sqrt{\frac{\hat p(1-\hat p)}{n} }$ < p < $\hat p$ + 2.813 * ${\sqrt{\frac{\hat p(1-\hat p)}{n} }$ ) = 0.995

99.5% confidence interval for p = [$\hat p$ - 2.813 * ${\sqrt{\frac{\hat p(1-\hat p)}{n} }$ , $\hat p$ + 2.813 * ${\sqrt{\frac{\hat p(1-\hat p)}{n} }$ ]

                               = [0.25 - 2.813 * ${\sqrt{\frac{0.25(1-0.25)}{432} }$ , 0.25 + 2.813 * ${\sqrt{\frac{0.25(1-0.25)}{432} }$ ]

                               = [0.191 , 0.309]

Therefore, 99.5% confidence interval for the proportion of those aged 65 and over who have sleep apnea is [0.191 , 0.309] .