Answer:
a) 0.24356 or 24.36%
b) [102.39 , 105.61]
c) The interpretation of this confidence interval is that in samples of 427 people aged 65 years and older, there is a 99% probability that the number of people that suffers sleep apnea is between 103 and 105.
Step-by-step explanation:
a)
This can be considered a binomial distribution (a person either has sleep apnea or not).
Based on the sample we have the probability of suffering the condition is
p = 104/427 = 0.24356
and q (the probability of not suffering the condition) is
q=1-0.24356=0.75644
So the proportion of people aged 65 years and older who have sleep apnea is 0.24356 or 24.36%
b)
To check if we can <em>approximate this binomial distribution with the Normal distribution</em> we must see that
np ≥ 5 and nq ≥ 5
where n is the sample size. Since
427*0.24356 ≥ 5 and 427*0.75644 ≥ 5
we can approximate the binomial with a Normal distribution with mean
np = 427*0.24356 = 104
and standard deviation
The 99% confidence interval (without the continuity correction factor) is given by the interval
![\large [\bar x-z^*\frac{s}{\sqrt n}, \bar x+z^*\frac{s}{\sqrt n}]](https://tex.z-dn.net/?f=%5Clarge%20%5B%5Cbar%20x-z%5E%2A%5Cfrac%7Bs%7D%7B%5Csqrt%20n%7D%2C%20%5Cbar%20x%2Bz%5E%2A%5Cfrac%7Bs%7D%7B%5Csqrt%20n%7D%5D)
where
<em>is the sample mean </em>
<em>s is the sample standard deviation </em>
<em>n is the sample size
</em>
<em>is the 0.01 (99%) upper critical value for
</em>
<em>the Normal distribution N(0;1).
</em>
The value of can be found either by using a table or a computer to find it equals to
and our 99% confidence interval <em>(without continuity correction) </em>is
![\large [104-2.576*\frac{8.867}{\sqrt{427}}, 104+2.576*\frac{8.867}{\sqrt {427}}]=[102.8946,105.1054]](https://tex.z-dn.net/?f=%5Clarge%20%5B104-2.576%2A%5Cfrac%7B8.867%7D%7B%5Csqrt%7B427%7D%7D%2C%20104%2B2.576%2A%5Cfrac%7B8.867%7D%7B%5Csqrt%20%7B427%7D%7D%5D%3D%5B102.8946%2C105.1054%5D)
We can now introduce the continuity correction factor. This should be done because <em>we are approximating a discrete distribution (Binomial) with a continuous one (Normal).
</em>
This is simply done by widening the interval in 0.5 at each end, so our final 99% confidence interval is
[102.3946 , 105.6054] = [102.39 , 105.61] rounded to 2 decimal places.
c)
The interpretation of this confidence interval is that in samples of 427 people aged 65 years and older, there is a 99% probability that the number of people that suffers sleep apnea is between 103 and 105.
If another study found a 15% of elderly people suffered sleep apnea, that would mean that in a sample of 427 only 64 would have the condition. Since that number is less than 103 by far, that would give us a reason to doubt about the conditions that framed the study (sample size, sampling method, age of people, etc.)
