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mihalych1998 [28]
3 years ago
7

In the past, 21% of all homes with a stay-at-home parent, the father is the stay-at-home parent. An independent research firm ha

s been charged with conducting a sample survey to obtain more current information. (a) What sample size is needed if the research firm's goal is to estimate the current proportion of homes with a stay-at-home parent in which the father is the stay-at-home parent with a margin of error of 0.09? Use a 95% confidence level. (Round your answer up to the nearest whole number.)
Mathematics
1 answer:
Afina-wow [57]3 years ago
4 0

Answer:

A sample size of 79 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}.

For this problem, we have that:

\pi = 0.21

The margin of error is:

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

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.  

What sample size is needed if the research firm's goal is to estimate the current proportion of homes with a stay-at-home parent in which the father is the stay-at-home parent with a margin of error of 0.09?

A sample size of n is needed.

n is found when M = 0.09. So

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

0.09 = 1.96\sqrt{\frac{0.21*0.79}{n}}

0.09\sqrt{n} = 1.96\sqrt{0.21*0.79}

\sqrt{n} = \frac{1.96\sqrt{0.21*0.79}}{0.09}

(\sqrt{n})^{2} = (\frac{1.96\sqrt{0.21*0.79}}{0.09})^{2}

n = 78.68

Rounding up to the nearest whole number.

A sample size of 79 is needed.

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