Question

A may 2000 Gallup poll found that 38% of a random sample of 1023 adults said that they believe in ghosts. The standard deviation is 1.5%. What is the margin of error for this survey? Determine an interval that is likely to contain the percent of all adults who believe in ghosts.

Answer 1

Margin of error = 0.09

Confidence interval = (0.29, 0.41)

Explanation:

Sample size, n = 1023

p = 38%

= 0.38

Standard deviation, σ = 1.5

Consider the level of confidence to be 95%

So, corresponding z* value for 95% confidence level = 1.96

`\text{Now, Margin of error = }z^** (\sigma)/(√(n))\\\\\text{Margin of error = }1.96* (1.5)/(√(1023))\\\\\textbf{Margin of error = }\bf 0.09`

Hence, confidence interval = p ± margin of error

= 0.38 ± 0.09

= (0.29, 0.47)