Respuesta :
Answer:
We need a sample size of at least 271
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 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]
90% confidence level
So [tex]\alpha = 0.1[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.1}{2} = 0.95[/tex], so [tex]Z = 1.645[/tex].
How large a sample is needed in order to be 90% confident that the sample proportion will not differ from the true proportion by more than 5%
We need a sample size of at least n, in which n is found when [tex]M = 0.05[/tex]
We do not know the true proportion, so we use [tex]\pi = 0.5[/tex], which is when we are going to need to use the largest sample size.
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.05 = 1.645\sqrt{\frac{0.5*0.5}{n}}[/tex]
[tex]0.05\sqrt{n} = 1.645*0.5[/tex]
[tex]\sqrt{n} = \frac{1.645*0.5}{0.05}[/tex]
[tex](\sqrt{n})^{2} = (\frac{1.645*0.5}{0.05})^{2}[/tex]
[tex]n = 270.6[/tex]
Rounding up
We need a sample size of at least 271
The size of the sample that is needed in order to be 90% confident that the sample proportion will not differ from the true proportion by more than 5% is; n = 271
We are given;
Confidence level of 90%
Margin of error; E = 5% = 0.05
We are not given the population proportion but we will assume that it is 0.5
Formula for margin of error here is;
E = z(√(p(1 - p)/n)
where z is the critical value at the confidence level
n is the sample size
Now, z at CL of 90% is 1.645
Making n the subject gives;
n = z²(p(1 - p)/E²
Plugging in the relevant values gives;
n = 1.645²(0.5(1 - 0.5))/0.05²
n ≈ 271
Read more about confidence intervals with proportion at; https://brainly.com/question/17030168
