Question

Suppose a poll of 700 voters says that 45% of the voters will vote for Candidate A for president. Compute the 99% confidence interval

Answer 1

Answer: (0.4016, 0.4984).

Step-by-step explanation:

Let p be the proportion of voters will vote for Candidate A for president.

Formula for confidence interval for proportion:

$\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

, here $\hat{p}$= sample proportion

n= sample size.

z* =two-tailed critical z- value.

As per given, we have

n= 700

$\hat{p}$ = 0.45

Critical two-tailed z-value for 99% confidence interval = 2.576

Then, the required 99% confidence interval for p would be:

$0.45\pm (2.576)\sqrt{\dfrac{0.45(1-0.45)}{700}}\\\\=0.45\pm(2.576)\sqrt{0.000353571428}\\\\=0.45\pm (2.576)(0.0188035)\\\\=0.45\pm0.04844\\\\=(0.45-0.048437,\ 0.45+0.048437)\\\\=(0.401563,\ 0.498437)\approx(0.4016,\ 0.4984)$

Hence, the required confidence interval (0.4016, 0.4984).