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Sagot :
Answer:
a) A sample of 385 must be taken.
b) The 95% confidence interval for the proportion of workers who have changed jobs within the past year is (0.1216, 0.2784).
c) A sample of 246 is needed.
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].
The margin of error is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
Absence of preliminary data:
This means that we use [tex]\pi = 0.5[/tex], which is when the largest sample will be needed.
95% confidence level
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
How large a sample must be taken to ensure that a 95% confidence interval will specify the proportion to within ±0.05?
A sample of n must be taken.
n is found when [tex]M = 0.05[/tex]. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.05 = 1.96\sqrt{\frac{0.5*0.5}{n}}[/tex]
[tex]0.05\sqrt{n} = 1.96*0.5[/tex]
Dividing both sides by 0.05
[tex]\sqrt{n} = 1.96*10[/tex]
[tex](\sqrt{n})^2 = (19.6)^2[/tex]
[tex]n = 384.16[/tex]
Rounding up,
A sample of 385 must be taken.
Question b:
In a sample of 100 workers, 20 of them had changed jobs within the past year.
This means that [tex]n = 100, \pi = \frac{20}{100} = 0.2[/tex]
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2 - 1.96\sqrt{\frac{0.2*0.8}{100}} = 0.1216[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2 + 1.96\sqrt{\frac{0.2*0.8}{100}} = 0.2784[/tex]
The 95% confidence interval for the proportion of workers who have changed jobs within the past year is (0.1216, 0.2784).
Question c:
Same procedure as question a, just with [tex]\pi = 0.2[/tex]
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.05 = 1.96\sqrt{\frac{0.2*0.8}{n}}[/tex]
[tex]0.05\sqrt{n} = 1.96\sqrt{0.2*0.8}[/tex]
[tex]\sqrt{n} = \frac{1.96\sqrt{0.2*0.8}}{0.05}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.96\sqrt{0.2*0.8}}{0.05})^2[/tex]
[tex]n = 245.8[/tex]
Rounding up
A sample of 246 is needed.
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