Respuesta :
Answer:
The sample size is = 814(approximately)
Step-by-step explanation:
Given -
Population proportion [tex]{\widehat{(p)}[/tex] = 0.4
Confidence interval = 98[tex]\%[/tex]
[tex]\alpha =[/tex] 1 - Confidence interval = 1 - .98
= .02
[tex]\frac{\alpha}{2}[/tex] = .001
[tex]Margin\; of\; error[/tex] = .04
Let sample size = n
[tex]Margin\; of\; error = z_{\frac{\alpha }{2}}\sqrt{\frac{{\widehat{(p)}}{(1 - \widehat{p})}}{n}}[/tex]
.04 = [tex]z_{.01}\sqrt{\frac{(.4){(1 - (0.4)}}{n}}[/tex]
For 98[tex]\%[/tex] level of confidence, z = 2.33
.04 = [tex]2.33\times\sqrt{\frac{(.4){(.6)}}{n}}[/tex]
(Squaring both side)
[tex](.04)^2[/tex] = [tex](2.33)^2\times\frac{(.4)(.6)}{n}[/tex]
n = [tex]\frac{(2.33)^2}{(.04)^2}\times{(.4)(.6)}[/tex]
n = 814.33
Approximately n =814
The required large sample will be "814". To understand the calculation, check below.
Confidence level and Margin of error
According to the question,
Confidence interval (CI) = 98% or,
= 0.98
Margin of error = 0.04
α - 1 - CI
= 1 - 0.98
= 0.02
[tex]\frac{\alpha}{2}[/tex] = [tex]\frac{0.02}{2}[/tex]
= 0.01
Let, the sample size be "n".
We know the formula,
→ Margin or error = [tex]z_{\frac{\alpha}{2} }[/tex] [tex]\sqrt{\frac{\hat p (1 - \hat p)}{n} }[/tex]
By substituting the values,
0.04 = [tex]z_{0.01}\sqrt{\frac{0.04 (1-0.4)}{n} }[/tex]
0.04 = 2.33 × [tex]\sqrt{\frac{0.4\times 0.6}{n} }[/tex]
By squaring both the sides,
n = [tex]\frac{(2.33)^2}{(0.04)^2}[/tex] × 0.4 × 0.6
= 814.33 or,
= 814
Thus the approach above is correct.
Find out more information about margin of error here;
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