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Sagot :
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 [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
Of 1600 randomly selected aircraft, eight were found to have wiring errors that could display incorrect information to the flight crew.
This means that [tex]n = 1600, \pi = \frac{8}{1600} = 0.005[/tex]
99% confidence level
So [tex]\alpha = 0.01[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.01}{2} = 0.995[/tex], so [tex]Z = 2.575[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.005 - 2.575\sqrt{\frac{0.005*0.995}{1600}} = 0.0005[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.005 + 2.575\sqrt{\frac{0.005*0.995}{1600}} = 0.0095[/tex]
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:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
A sample of n is required, and n is found for M = 0.009. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.009 = 2.575\sqrt{\frac{0.005*0.995}{n}}[/tex]
[tex]0.009\sqrt{n} = 2.575\sqrt{0.005*0.995}[/tex]
[tex]\sqrt{n} = \frac{2.575\sqrt{0.005*0.995}}{0.009}[/tex]
[tex](\sqrt{n})^2 = (\frac{2.575\sqrt{0.005*0.995}}{0.009})^2[/tex]
[tex]n = 407.3[/tex]
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 [tex]\pi = 0.5[/tex]
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.009 = 2.575\sqrt{\frac{0.5*0.5}{n}}[/tex]
[tex]0.009\sqrt{n} = 2.575*0.5[/tex]
[tex]\sqrt{n} = \frac{2.575*0.5}{0.009}[/tex]
[tex](\sqrt{n})^2 = (\frac{2.575*0.5}{0.009})^2[/tex]
[tex]n = 20464.9[/tex]
Rounding up:
A sample of 20465 is required.
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