Question

In a recent poll, 230 people were asked if they liked dogs, and 90% said they did. Find the margin of error of this poll, at the 99% confidence level.

Answer 1

The margin of error of this poll is

Given that 230 people were asked if they liked dogs, and 90% said they did.

In a random sample, a margin of error is a statistical measure that takes into account the discrepancy between actual and expected results.

Sample size (N)=230

Sample proportion  $(\hat{p})$ = 90% = 0.90  and

confidence interval = 99%

The critical value at 99% confidence interval with α=0.01 is

$\begin{aligned}Z_{c}&=Z_{1-\frac{\alpha}{2}}\\ &=Z_{1-\frac{0.01}{2}}\\ &=Z_{0.995}\\ &=2.576\end$

This value is taken from the Standard normal distribution table.

Now, we will calculate the margin of error (E) by using the formula

$E=Z_{c}\sqrt{\frac{\hat{p}(1-\hat{p})}{N}}$

Substituting the values in the formula, we get

$\begin{aligned}E &=2.576\times \sqrt{\frac{0.90(1-0.90)}{230}}\\ &=2.576\times \sqrt{\frac{0.09}{230}}\\ &=2.576\times\sqrt{0.00039}\\ &=2.576\times 0.0197\\ &=0.0507\end$

Hence, the margin of error when 230 people were asked if they liked dogs, and 90% said they did is 0.0507.

Learn more about margin of error from here brainly.com/question/25779324

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