Answer:
a) The 99% confidence interval on the proportion of aircraft that have such wiring errors is (0.0005, 0.0095).
b) A sample of 408 is required.
c) A sample of 20465 is required.
Step-by-step explanation:
Question a:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence level of
, we have the following confidence interval of proportions.
![\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=%5Cpi%20%5Cpm%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
In which
z is the zscore that has a pvalue of
.
Of 1600 randomly selected aircraft, eight were found to have wiring errors that could display incorrect information to the flight crew.
This means that ![n = 1600, \pi = \frac{8}{1600} = 0.005](https://tex.z-dn.net/?f=n%20%3D%201600%2C%20%5Cpi%20%3D%20%5Cfrac%7B8%7D%7B1600%7D%20%3D%200.005)
99% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
The lower limit of this interval is:
![\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.005 - 2.575\sqrt{\frac{0.005*0.995}{1600}} = 0.0005](https://tex.z-dn.net/?f=%5Cpi%20-%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D%20%3D%200.005%20-%202.575%5Csqrt%7B%5Cfrac%7B0.005%2A0.995%7D%7B1600%7D%7D%20%3D%200.0005)
The upper limit of this interval is:
![\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.005 + 2.575\sqrt{\frac{0.005*0.995}{1600}} = 0.0095](https://tex.z-dn.net/?f=%5Cpi%20%2B%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D%20%3D%200.005%20%2B%202.575%5Csqrt%7B%5Cfrac%7B0.005%2A0.995%7D%7B1600%7D%7D%20%3D%200.0095)
The 99% confidence interval on the proportion of aircraft that have such wiring errors is (0.0005, 0.0095).
b. Suppose we use the information in this example to provide a preliminary estimate of p. How large a sample would be required to produce an estimate of p that we are 99% confident differs from the true value by at most 0.009?
The margin of error is of:
![M = z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
A sample of n is required, and n is found for M = 0.009. So
![M = z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
![0.009 = 2.575\sqrt{\frac{0.005*0.995}{n}}](https://tex.z-dn.net/?f=0.009%20%3D%202.575%5Csqrt%7B%5Cfrac%7B0.005%2A0.995%7D%7Bn%7D%7D)
![0.009\sqrt{n} = 2.575\sqrt{0.005*0.995}](https://tex.z-dn.net/?f=0.009%5Csqrt%7Bn%7D%20%3D%202.575%5Csqrt%7B0.005%2A0.995%7D)
![\sqrt{n} = \frac{2.575\sqrt{0.005*0.995}}{0.009}](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%3D%20%5Cfrac%7B2.575%5Csqrt%7B0.005%2A0.995%7D%7D%7B0.009%7D)
![(\sqrt{n})^2 = (\frac{2.575\sqrt{0.005*0.995}}{0.009})^2](https://tex.z-dn.net/?f=%28%5Csqrt%7Bn%7D%29%5E2%20%3D%20%28%5Cfrac%7B2.575%5Csqrt%7B0.005%2A0.995%7D%7D%7B0.009%7D%29%5E2)
![n = 407.3](https://tex.z-dn.net/?f=n%20%3D%20407.3)
Rounding up:
A sample of 408 is required.
c. Suppose we did not have a preliminary estimate of p. How large a sample would be required if we wanted to be at least 99% confident that the sample proportion differs from the true proportion by at most 0.009 regardless of the true value of p?
Since we have no estimate, we use ![\pi = 0.5](https://tex.z-dn.net/?f=%5Cpi%20%3D%200.5)
![M = z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
![0.009 = 2.575\sqrt{\frac{0.5*0.5}{n}}](https://tex.z-dn.net/?f=0.009%20%3D%202.575%5Csqrt%7B%5Cfrac%7B0.5%2A0.5%7D%7Bn%7D%7D)
![0.009\sqrt{n} = 2.575*0.5](https://tex.z-dn.net/?f=0.009%5Csqrt%7Bn%7D%20%3D%202.575%2A0.5)
![\sqrt{n} = \frac{2.575*0.5}{0.009}](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%3D%20%5Cfrac%7B2.575%2A0.5%7D%7B0.009%7D)
![(\sqrt{n})^2 = (\frac{2.575*0.5}{0.009})^2](https://tex.z-dn.net/?f=%28%5Csqrt%7Bn%7D%29%5E2%20%3D%20%28%5Cfrac%7B2.575%2A0.5%7D%7B0.009%7D%29%5E2)
![n = 20464.9](https://tex.z-dn.net/?f=n%20%3D%2020464.9)
Rounding up:
A sample of 20465 is required.