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sweet [91]
3 years ago
12

In a survey, the planning value for the population proportion is p* = 0.35. How large a sample should be taken to provide a 95%

confidence interval with a margin of error of 0.05?
Mathematics
1 answer:
lions [1.4K]3 years ago
6 0

Answer:

n=\frac{0.35(1-0.35)}{(\frac{0.05}{1.96})^2}=349.59  

And rounded up we have that n=350  

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".  

p represent the real population proportion of interest

\aht p represent the estimated proportion for the sample

n is the sample size required (variable of interest)

z represent the critical value for the margin of error

Solution to the problem

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})  

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 \alpha=1-0.95=0.05 and \alpha/2 =0.025. And the critical value would be given by:  

z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96  

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}} (a)  

And on this case we have that ME =\pm 0.05 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2} (b)

And replacing into equation (b) the values from part a we got:  

n=\frac{0.35(1-0.35)}{(\frac{0.05}{1.96})^2}=349.59  

And rounded up we have that n=350  

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