Mathematics, 07.04.2020 21:48 ashleyaparicio876

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?

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Answer from: garciatania837

A sample size of 2166 is needed.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error of a confidence interval is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

98% confidence level

So $\alpha = 0.02$ , z is the value of Z that has a pvalue of $1 - \frac{0.02}{2} = 0.99$ , so Z = 2.327 .

Solving

We need to find n when M = 0.025.

We dont know the proportion, so we use $\pi = 0.5$ , which is when we are going to need the largest sample size for this estimate.

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.025 = 2.327\sqrt{\frac{0.5*0.5}{n}}$

$0.025\sqrt{n} = 2.327*0.5$

$\sqrt{n} = \frac{2.327*0.5}{0.025}$

$(\sqrt{n})^{2} = (\frac{2.327*0.5}{0.025})^{2}$

n = 2166

A sample size of 2166 is needed.

Answer from: Quest

1,8,15,22,29,36,43,50

step-by-step explanation:

Answer from: Quest

step-by-step explanation:

i found the life tables for the year 2012.

life table of males at age 30. probability of dying between ages x and x+1

total population:

29–30 . . . . . . . . . . . . 0.001030

30–31 . . . . . . . . . . . . 0.001061

males:

29–30 . . . . . . . . . . . . 0.001420

30–31 . . . . . . . . . . . . 0.001448

white population

29–30 . . . . . . . . . . . . 0.001005

30–31 . . . . . . . . . . . . 0.001037

white males

29–30 . . . . . . . . . . . . 0.001372

30–31 . . . . . . . . . . . . 0.001399

non hispanic white population

29–30 . . . . . . . . . . . . 0.001083

30–31 . . . . . . . . . . . . 0.001116

non hispanic white males

29–30 . . . . . . . . . . . . 0.001477

30–31 . . . . . . . . . . . . 0.001503

jacob, non hispanic white male. 30 years old.

probability that jacob lives to be 30 years old.

since the given probabilities are probabilities of dying, i think we simply subtract it from 1 to get the probability of living. we will use the probability of non hispanic white male.

probability of living to be 30 y.o.

29 - 30 : 1 - 0.001477 = 0.998523

30 - 31 : 1 - 0.001503 = 0.998497

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