Respuesta :
Answer:
[tex]n=\frac{0.5(1-0.5)}{(\frac{0.1}{1.96})^2}=96.04[/tex]
And rounded up we have that n=97
Based on this with a sample of 100 we have a size large enough to satisfy the condition that the margin of error would be 0.1
Step-by-step explanation:
Notation and definitions
[tex]X[/tex] number of ratepayers in favor of the proposal
[tex]n[/tex] random sample taken (variable of interest)
[tex]\hat p[/tex] estimated proportion of ratepayers in favor of the proposal
[tex]p[/tex] true population proportion of ratepayers in favor of the proposal
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".
The population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]
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:
[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 0.1[/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)
We can use as estimate [tex]\hat p =0.5[/tex] since we don't have prior info for this. And replacing into equation (b) the values from part a we got:
[tex]n=\frac{0.5(1-0.5)}{(\frac{0.1}{1.96})^2}=96.04[/tex]
And rounded up we have that n=97
Based on this with a sample of 100 we have a size large enough to satisfy the condition that the margin of error would be 0.1
The considered sample size n = 100 is enough, if we continue assuming that [tex]\hat{p} = 1/2 = 0.5[/tex]
How to find the margin of error of sample proportion?
For large enough sample, let the population proportion of a quantity be denoted by random variable [tex]p[/tex]
Then, we get:
[tex]p \sim N(\hat{p}, \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}})[/tex]
where [tex]\hat{p}[/tex] = estimated(mean value) proportion of that quantity
and n = size of sample drawn.
It is visible that as we increase the value of n, the standard deviation decreases, therefore, forcing the values of population proportion to be closer to the estimated proportion.
Margin of error is the distance between the mean and one of the end point of the confidence interval(assuming its equal on both the sides of the mean).
The margin of error with level of significance [tex]\alpha[/tex] is calculated as:
[tex]MOE = Z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]
where [tex]Z_{\alpha/2}[/tex] is the critical value of the test statistic for level of significance [tex]\alpha[/tex]
For the considered case, we are given:
n = 100
Confidence interval of 95%, thus, level of significance = 100% - 95% = 5% = 0.05
At this level of significance, the critical value of the test statistic is Z = 1.96
No information for proportion is given, so we can take it half of total which is 1/2, thus getting [tex]\hat{p} = 1/2 = 0.5[/tex]
Putting these values, we get:
[tex]MOE = 1.96 \times \sqrt{\dfrac{0.5 \times 0.5}{100}} \approx 0.098[/tex]
That means, the values are within 0.1 distance from the mean (as 0.1 > 0.098)
Therefore, the considered sample size n = 100 is enough, if we continue assuming that [tex]\hat{p} = 1/2 = 0.5[/tex]
Learn more about margin of error here:
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