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Current Attempt in Progress Consider random samples of size 1200 from a population with proportion 0.70 . Find the standard error of the distribution of sample proportions. Round your answer for the standard error to three decimal places. standard error \( = \) i

Ask by Ingram Mckenzie. in the United States
Mar 18,2025

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The standard error is approximately 0.013.

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To find the standard error of the distribution of sample proportions, we can use the formula: \[ \text{Standard Error} = \sqrt{\frac{p(1-p)}{n}} \] where: - \( p \) is the population proportion (0.70 in this case), - \( n \) is the sample size (1200 in this case). Substitute the given values into the formula to calculate the standard error. Calculate the value by following steps: - step0: Calculate: \(\sqrt{\frac{0.7\left(1-0.7\right)}{1200}}\) - step1: Subtract the numbers: \(\sqrt{\frac{0.7\times 0.3}{1200}}\) - step2: Multiply the numbers: \(\sqrt{\frac{0.21}{1200}}\) - step3: Divide the terms: \(\sqrt{\frac{7}{40000}}\) - step4: Use the properties of radicals: \(\frac{\sqrt{7}}{\sqrt{40000}}\) - step5: Simplify the expression: \(\frac{\sqrt{7}}{200}\) The standard error of the distribution of sample proportions is approximately 0.0132 when the sample size is 1200 and the population proportion is 0.70.

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To find the standard error of the distribution of sample proportions, you can use the formula: \[ \text{Standard Error (SE)} = \sqrt{\frac{p(1 - p)}{n}} \] where \( p \) is the population proportion and \( n \) is the sample size. Here, \( p = 0.70 \) and \( n = 1200 \). Plugging in the values: \[ SE = \sqrt{\frac{0.70(1 - 0.70)}{1200}} \] \[ SE = \sqrt{\frac{0.70 \times 0.30}{1200}} \] \[ SE = \sqrt{\frac{0.21}{1200}} \] \[ SE = \sqrt{0.000175} \] \[ SE \approx 0.013228 \] Rounded to three decimal places, the standard error \( \approx 0.013 \).

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