What sample size would be required to estimate the true proportion of American female business executives who prefer the title "

Question

3 years ago

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What sample size would be required to estimate the true proportion of American female business executives who prefer the title "

Ms.," with an error of ±0.025 and 98 percent confidence? (Enter your answer as a whole number (no decimals). Use a z-value taken to three decimal places in your calculations.) Sample size

Mathematics

1 answer:

monitta · Answer 1 · 2022-05-05T16:24:29+03:00

Answer:

$n=\frac{0.5(1-0.5)}{(\frac{0.05}{2.326})^2}=2164.1104$

And rounded up we have that n=2165

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

$\hat p$ 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

The population proportion have the following distribution

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

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 98% of confidence, our significance level would be given by $\alpha=1-0.98=0.02$ and $\alpha/2 =0.01$ . And the critical value would be given by:

$z_{\alpha/2}=-2.326, z_{1-\alpha/2}=2.326$

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 $ME =\pm 0.025$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

(b)

We can assume that the estimated proportion is 0.5 since we don't have other info provided to assume a different value. And replacing into equation (b) the values from part a we got:

$n=\frac{0.5(1-0.5)}{(\frac{0.05}{2.326})^2}=2164.1104$

And rounded up we have that n=2165