You are a pollster conducting a survey to estimate the proportion of voters who support a certain candidate in an upcoming election. You take a random sample of 500 voters and find that 280 of them support the candidate. Estimate the population proportion of the voters who support the candidate using two standard errors. Round your answer to two decimal places. (2 points) ≤p≤

1 answer

To estimate the population proportion of voters who support the candidate, we first need to calculate the sample proportion (\( \hat{p} \)) and the standard error (SE) of that sample proportion.

  1. Calculate the sample proportion (\( \hat{p} \)): \[ \hat{p} = \frac{\text{Number of supporters}}{\text{Total sample size}} = \frac{280}{500} = 0.56 \]

  2. Calculate the standard error (SE) for the proportion: The standard error for a sample proportion can be calculated using the formula: \[ SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \] where \( n \) is the sample size.

    Plugging in the values: \[ SE = \sqrt{\frac{0.56(1 - 0.56)}{500}} = \sqrt{\frac{0.56 \times 0.44}{500}} = \sqrt{\frac{0.2464}{500}} = \sqrt{0.0004928} \approx 0.0222 \]

  3. Calculate the interval using two standard errors: The interval will be \( \hat{p} \pm 2 \times SE \). \[ 2 \times SE = 2 \times 0.0222 \approx 0.0444 \]

    Therefore, the interval can be calculated as follows: \[ \text{Lower bound} = \hat{p} - 2 \times SE = 0.56 - 0.0444 \approx 0.5156 \] \[ \text{Upper bound} = \hat{p} + 2 \times SE = 0.56 + 0.0444 \approx 0.6044 \]

  4. Round the bounds to two decimal places:

    • Lower bound: \( 0.52 \)
    • Upper bound: \( 0.60 \)

Putting it all together, the estimated proportion of voters who support the candidate, ranging ± two standard errors, is:

\[ 0.52 \leq p \leq 0.60 \]