Respuesta :
Answer:
a) [tex]n=\frac{0.52(1-0.52)}{(\frac{0.02}{2.58})^2}=4153.59[/tex]
And rounded up we have that n=4154
b) [tex]n=\frac{0.5(1-0.5)}{(\frac{0.02}{2.58})^2}=4160.25[/tex]
And rounded up we have that n=4161
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
Part a
In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by [tex]\alpha=1-0.99=0.01[/tex] and [tex]\alpha/2 =0.005[/tex]. And the critical value would be given by:
[tex]z_{\alpha/2}=-2.58, z_{1-\alpha/2}=2.58[/tex]
The margin of error for the proportion interval is given by this formula:
[tex] ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex] (a)
And on this case we have that [tex]ME =\pm 0.02[/tex] and we are interested in order to find the value of n, if we solve n from equation (a) we got:
[tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex] (b)
And replacing into equation (b) the values from part a we got:
[tex]n=\frac{0.52(1-0.52)}{(\frac{0.02}{2.58})^2}=4153.59[/tex]
And rounded up we have that n=4154
Part b
Since we don't have prior estimation for p we can use [tex]\hat p =0.5[/tex]
[tex]n=\frac{0.5(1-0.5)}{(\frac{0.02}{2.58})^2}=4160.25[/tex]
And rounded up we have that n=4161
Answer:
(a)If she takes previous estimate then sample size n=4154
(b)If she does not take previous estimate then sample size n=4154
Step-by-step explanation:
Given information :
Confidence level=99%=0.99
Marginal error=[tex]\pm0.02[/tex]
Significance level,
[tex]\alpha=1-0.99=0.01[/tex]
[tex]\frac{\alpha}{2}=0.05[/tex]
Critical value,[tex]z_{\frac{\alpha}{2}=-2.58[/tex]
By use of formula,
[tex]ME=z_\frac{\alpha}{2}\sqrt\frac{\widehat{p}(1-\widehat{p})}{n}[/tex]
Where [tex]ME[/tex]= marginal error,
[tex]n=\frac{\widehat{p}(1-\widehat{p})}{(\frac{ME}{z_\frac{\alpha}{2}})^2}[/tex]
(a)Previous estimate =0.52
[tex]n=\frac{\widehat{p}(1-\widehat{p})}{(\frac{ME}{z_\frac{\alpha}{2}})^2}=\frac{{0.52}(1-{0.52})}{(\frac{0.02}{2.58})^2}=4153.5\approx4154[/tex]
(b)Not use any prior estimates,the we take [tex]\widehat{p}[/tex]=0.5
[tex]n=\frac{{0.5}(1-{0.5})}{(\frac{0.02}{2.58})^2}=4160.25\approx4161[/tex]
Hence,
(a)If she takes previous estimate then sample size n=4154
(b)If she does not take previous estimate then sample size n=4161
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