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Sagot :
Answer:
a. The minimum sample size required is of 60,025.
b. The minimum sample size required is of 20,465.
c. The minimum sample size required is 9,589.
Step-by-step explanation:
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 z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].
The margin of error is of:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
Question a:
95% confidence level
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
p and q unknown means that we use [tex]\pi = 0.5[/tex], which is when the largest sample size is needed.
Margin of error: 0.004
We have to find n for which M = 0.004. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.004 = 1.96\sqrt{\frac{0.5*0.5}{n}}[/tex]
[tex]0.004\sqrt{n} = 1.96*0.5[/tex]
[tex]\sqrt{n} = \frac{1.96*0.5}{0.004}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.96*0.5}{0.004})^2[/tex]
[tex]n = 60025[/tex]
The minimum sample size required is of 60,025.
b. Margin of error: 0.009, confidence level: 99%; p and q unknown
99% confidence level
So [tex]\alpha = 0.01[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.01}{2} = 0.995[/tex], so [tex]Z = 2.575[/tex].
Margin of error: 0.009
We have to find n for which M = 0.009. So
[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:
The minimum sample size required is of 20,465.
c. Margin of error: 0.01; confidence level: 95%; from a prior study, p is estimated by the decimal equivalent of 52%.
Here we have [tex]\pi = 0.52[/tex].
The minimum sample size 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{0.52*0.48}{n}}[/tex]
[tex]0.01\sqrt{n} = 1.96\sqrt{0.52*0.48}[/tex]
[tex]\sqrt{n} = \frac{1.96\sqrt{0.52*0.48}}{0.01}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.96\sqrt{0.52*0.48}}{0.01})^2[/tex]
[tex]n = 9588.6[/tex]
Rounding up:
The minimum sample size required is 9,589.
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