Mathematics: Amedical study is to be designed to estimate the population proportion p having a certain disease. how many people should be examined if the researcher wishes to be 95% certain that the error of estimation is below 0.1 when a. there is no prior knowledge about the value of p? b. the proportion is known to be about 0.2?

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Amedical study is to be designed to estimate the

Ответ:

alleshia2007

13.01.2022

Mathematics

a) $n=\frac{0.5(1-0.5)}{(\frac{0.1}{1.96})^2}=96.04$

And rounded up we have that n=97

b) $n=\frac{0.2(1-0.2)}{(\frac{0.1}{1.96})^2}=61.47$

And rounded up we have that n=62

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".

The population proportion have the following distribution

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

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 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: