Respuesta :
Answer:
First question: LCL = 522, UCL = 1000.5
Second question: A sample size no smaller than 418 is needed.
Step-by-step explanation:
First question:
Lower bound:
0.36 of 1450. So
0.36*1450 = 522
Upper bound:
0.69 of 1450. So
0.69*1450 = 1000.5
LCL = 522, UCL = 1000.5
Second 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].
The margin of error is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
The project manager believes that p will turn out to be approximately 0.11.
This means that [tex]\pi = 0.11[/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].
The project manager wants to estimate the proportion to within 0.03
This means that the sample size needed is given by n, and n is found when M = 0.03. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.03 = 1.96\sqrt{\frac{0.11*0.89}{n}}[/tex]
[tex]0.03\sqrt{n} = 1.96\sqrt{0.11*0.89}[/tex]
[tex]\sqrt{n} = \frac{1.96\sqrt{0.11*0.89}}{0.03}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.96\sqrt{0.11*0.89}}{0.03})^2[/tex]
[tex]n = 417.9[/tex]
Rounding up
A sample size no smaller than 418 is needed.
