Respuesta :
Answer:
[tex]n=\frac{0.005(1-0.005)}{(\frac{0.001}{1.96})^2}=19111.96[/tex] Â
And rounded up we have that n=19112
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". Â
The margin of error is the range of values below and above the sample statistic in a confidence interval. Â
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". Â
Solution to the problem
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 95% of confidence, our significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. And the critical value would be given by:
[tex]z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96[/tex]
The margin of error for the proportion interval is given by this formula: Â
[tex] ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex] Â Â (a) Â
And on this case we have that [tex]ME =\pm 0.001[/tex] and we are interested in order to find the value of n, if we solve n from equation (a) we got: Â
[tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex] Â (b) Â
And replacing into equation (b) the values from part a we got:
[tex]n=\frac{0.005(1-0.005)}{(\frac{0.001}{1.96})^2}=19111.96[/tex] Â
And rounded up we have that n=19112
Answer:
We need a mailing list of at least 191112 people.
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]
For this problem, we have that:
[tex]\pi = 0.005[/tex]
95% confidence level
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
To be within a tenth of a percentage point (0.001) of the true rate with 95% confidence, how big does the test mailing have to be?
They need at least n people
n is found when [tex]M = 0.001[/tex]. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.001 = 1.96\sqrt{\frac{0.005*0.995}{n}}[/tex]
[tex]0.001\sqrt{n} = 1.96\sqrt{0.005*0.995}[/tex]
[tex]\sqrt{n} = \frac{1.96\sqrt{0.005*0.995}}{0.001}[/tex]
[tex](\sqrt{n})^{2} = (\frac{1.96\sqrt{0.005*0.995}}{0.001})^{2}[/tex]
[tex]n = 19111.96[/tex]
We need a mailing list of at least 191112 people.
