Answer:
The sample size is [tex]n = 451 [/tex]
Step-by-step explanation:
From the question we are told that
The margin of error is E = 0.03
The sample proportion is [tex]\^ p = 0.12[/tex]
From the question we are told the confidence level is 95% , hence the level of significance is
[tex]\alpha = (100 - 95 ) \%[/tex]
=> [tex]\alpha = 0.05[/tex]
Generally from the normal distribution table the critical value of [tex]\frac{\alpha }{2}[/tex] is
[tex]Z_{\frac{\alpha }{2} } = 1.96[/tex]
Generally the sample size is mathematically represented as
[tex]n = [\frac{Z_{\frac{\alpha }{2} }}{E} ]^2 * \^ p (1 - \^ p ) [/tex]
=> [tex]n = [\frac{1.96}{0.03} ]^2 *0.12 (1 - 0.12 ) [/tex]
=> [tex]n = 451 [/tex]
This question is based on concept of normal distribution.Therefore, 451 individuals must survey to estimate the proportion who approve of Congress at that point to within 3% at 95% confidence.
Given:
The sample size is n=451.
According to question,
The margin of error is E = 0.03
The sample proportion is [tex]\bold{\hat{p}}[/tex] =0.12
From the question, the confidence level is 95% , hence the level of significance is,
[tex]\alpha =(100-95)\%\\\bold{\alpha =5\%=0.05}[/tex]
Normally, as per the normal distribution table the critical value of,
[tex]\dfrac{\alpha }{2}[/tex] is
[tex]\bold{Z_\frac{\alpha }{2} =1.96}[/tex]
And ,the sample size in mathematically represented as,
[tex]\begin{aligned}n&=(\frac{Z_\frac{a}{2} }{E})^{2} \times \hat p(1-\hat p)\\n&=(\frac{1.96}{0.03})^2 \times 0.12(1-0.12)\\\bold{n&=451} \end{aligned}[/tex]
Hence, there are 451 individuals must we survey to estimate the proportion who approve of Congress at that point to within 3% at 95% confidence.
For further details, please prefer this link:
https://brainly.com/question/15103234
