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A researcher wishes to estimate the number of households with two cars. How large a sample is needed in order to be 99% confident that the sample proportion will not differ from the true proportion by more than 4%? A previous study indicates that the proportion of households with two cars is 21%.

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A researcher wants to estimate the proportion of households with two cars. The researcher wants to be very sure of their estimation, so they choose a high level of confidence: 99%. This means that they want to capture the true proportion 99 times out of 100. The margin of error that they are willing to accept is 4% (0.04), which means that if the true proportion is, say, 21%, their estimate can range between 17% and 25%.

The formula to calculate the required sample size for estimating a proportion with a certain level of confidence and margin of error is:

n = (Z^2 * P * (1 - P)) / E^2

Here:

- n is the sample size.

- Z is the Z-score, which is a measure of how many standard deviations an element is from the mean. For a 99% confidence level, the Z-score is approximately 2.58.

- P is the estimated proportion of households with two cars. In this case, a previous study indicates it's 21%, or 0.21.

- E is the margin of error, which is 4% or 0.04 in this case.

Plugging in the values:

n = (2.58^2 * 0.21 * (1 - 0.21)) / 0.04^2

≈ (6.67 * 0.21 * 0.79) / 0.0016

≈ (1.10) / 0.0016

≈ 689.06

Since you can't have a fraction of a household, round up to the nearest whole number, which is 690.

This means that the researcher needs to randomly select a sample of at least 690 households in order to be 99% confident that the sample proportion will not differ from the true proportion by more than 4%.

Keep in mind that this is a simplified explanation and in practice, researchers may take into account other factors, such as design effect, in determining sample size.

User Aung Si Min Htet
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