Question

A newspaper reports that the​ governor's approval rating stands at 54​%. The article adds that the poll is based on a random sample of 4134 adults and has a margin of error of 2%. What level of confidence did the pollsters​ use?

Answer 1

Answer on Question #67666 – Math – Statistics and Probability

Question

A newspaper reports that the governor's approval rating stands at 54%. The article adds that the poll is based on a random sample of 4134 adults and has a margin of error of 2%. What level of confidence did the pollsters use?

Solution

If $n\hat{p} \geq 10$ and $n(1 - \hat{p}) \geq 10$ we can use the following formula to compute the margin of error for a sample proportion:

$$ME = z^* \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}},$$

where $\hat{p}$ is a sample proportion, $n$ is the sample size, $z^*$ is the multiplier dependent on the level of confidence:

In our case $\hat{p} = 0.54$, $n = 4134$, $ME = 0.02$.

Conditions $n\hat{p} \geq 10$ and $n(1 - \hat{p}) \geq 10$ are met. Hence

$$z^* = \frac{ME}{\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}} = \frac{0.02}{\sqrt{\frac{0.54(1 - 0.54)}{4134}}} \approx 2.58,$$

which corresponds to approximately 99% level of confidence.

Answer: 99%.

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