Question

A survey was conducted at Acadia on attitudes toward the campus alcohol policy against drinking games in residence with 262 randomly selected residence students. In response to the question, “Are you in favour of the drinking games policy?” 25% were in favour. What is the 99% confidence interval to estimate the percentage of all Acadia residence students regarding the drinking game policy? Remember to report your confidence interval in a complete sentence and describe what it means.

Answer 1

Solution:

The 99% confidence interval for population proportion (percentage) is given as follows:

Where, p̂ is sample proportion, q̂ = 1 - p̂, n is sample size and Z_(0.01/2) is critical z-value to construct 99% confidence interval.

Sample proportion of Acadia residence who are in favor of the drinking games policy is, p̂ = 25/100 = 0.25

q̂ = 1 - 0.25 = 0.75 and n = 262

Using Z-table we get, Z_(0.01/2) = 2.576

Hence, 99% confidence interval for proportion of Acadia residence students who are in favor of drinking game policy is,

On converting the values into percentage we get,

(18.11%, 31.89%)

Hence, 99% confidence interval to estimate the percentage of all Acadia residence students who are in favor of drinking game policy is, (18.11%, 31.89%).

Interpretation : We are 99% confident that true percentage of all Acadia residence students who are in favor of drinking game policy lies within the calculated 99% confidence interval.

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