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Sagot :
Answer:
The 80% confidence interval for the proportion of companies that are planning to increase their workforce is (0.033, 0.331).
A sample size of 8852 is needed.
Step-by-step explanation:
First question:
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].
Only 2 of the 11 companies were planning to increase their workforce.
This means that [tex]n = 11, \pi = \frac{2}{11} = 0.1818[/tex]
80% confidence level
So [tex]\alpha = 0.2[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.2}{2} = 0.9[/tex], so [tex]Z = 1.28[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.1818 - 1.28\sqrt{\frac{0.1818*0.8182}{11}} = 0.033[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.1818 + 1.28\sqrt{\frac{0.1818*0.8182}{11}} = 0.331[/tex]
The 80% confidence interval for the proportion of companies that are planning to increase their workforce is (0.033, 0.331).
Second question:
The margin of error is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
A poll taken in July 2010 estimates this proportion to be 0.36.
This means that [tex]\pi = 0.36[/tex]
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].
Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.01?
This is n for which M = 0.01. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.01 = 1.96\sqrt{\frac{0.36*0.64}{n}}[/tex]
[tex]0.01\sqrt{n} = 1.96\sqrt{0.36*0.64}[/tex]
[tex]\sqrt{n} = \frac{1.96\sqrt{0.36*0.64}}{0.01}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.96\sqrt{0.36*0.64}}{0.01})^2[/tex]
[tex]n = 8851.04[/tex]
Rounding up(as a sample of 8851 will have a margin of error slightly over 0.01):
A sample size of 8852 is needed.
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