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You take a sample of size 35 from a very large population in which the true proportion of males is \( p=0.52 \). Which of the statements below best describes what is known about the sampling distribution of \( \hat{p} \), the sample proportion of males?

Ask by Fuentes Boone. in the United States
Feb 26,2025

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Answer

The sampling distribution of \( \hat{p} \) is approximately normal with a mean of 0.52 and a standard deviation of approximately 0.084.

Solution

To determine what is known about the sampling distribution of \( \hat{p} \), the sample proportion of males, we can use the properties of the sampling distribution of a proportion. Given: - Sample size \( n = 35 \) - True proportion of males in the population \( p = 0.52 \) The sampling distribution of \( \hat{p} \) is approximately normal with mean \( p \) and standard deviation \( \sqrt{\frac{p(1-p)}{n}} \). Therefore, the sampling distribution of \( \hat{p} \) is approximately normal with mean \( 0.52 \) and standard deviation \( \sqrt{\frac{0.52(1-0.52)}{35}} \). Let's calculate the standard deviation of the sampling distribution of \( \hat{p} \). Calculate the value by following steps: - step0: Calculate: \(\sqrt{\frac{0.52\left(1-0.52\right)}{35}}\) - step1: Subtract the numbers: \(\sqrt{\frac{0.52\times 0.48}{35}}\) - step2: Multiply the numbers: \(\sqrt{\frac{0.2496}{35}}\) - step3: Divide the terms: \(\sqrt{\frac{156}{21875}}\) - step4: Use the properties of radicals: \(\frac{\sqrt{156}}{\sqrt{21875}}\) - step5: Simplify the expression: \(\frac{2\sqrt{39}}{25\sqrt{35}}\) - step6: Multiply by the Conjugate: \(\frac{2\sqrt{39}\times \sqrt{35}}{25\sqrt{35}\times \sqrt{35}}\) - step7: Multiply the numbers: \(\frac{2\sqrt{1365}}{25\sqrt{35}\times \sqrt{35}}\) - step8: Multiply the numbers: \(\frac{2\sqrt{1365}}{875}\) The standard deviation of the sampling distribution of \( \hat{p} \) is approximately \( 0.084448 \). Therefore, the statement that best describes what is known about the sampling distribution of \( \hat{p} \) is that it is approximately normal with mean \( 0.52 \) and standard deviation \( 0.084448 \).

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