Find the margin of error of this poll, at the 90% confidence level.

Question

In a recent poll, 330 people were asked if they liked dogs, and 57% said they did. Find the margin of error of this poll, at the 90% confidence level.

Give your answer to three decimals

Answer 1

Hi Kaitlyn,

First, we need to see if we can assume normality. For this, we need to know if:

np>=10

n(1-p)>=10

n=sample size

p=probability we were given

n=330

p=0.58

0.58*330=191.4

.42*330=138.6

We can assume normality. Formula for confidence interval for population proportion:

p +/- z* sqrt(p(1-p)/n)

where:

p=probability we were given, sometimes denoted p-hat

z*=z-critical value

n=sample size

In this problem:

p=0.58

1-p=0.42

z*=1.645, worth memorizing for 90% confidence intervals

n=330

CI= 0.58 +/- 1.645*sqrt((0.58*0.42)/330)

CI=0.580 +/- 0.045

CI=(0.535, 0.625)

I hope this helps.

Sorry--margin of error is second part of confidence interval--z*sqrt(p(1-p)/n) MoE=0.045