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From a survey of 297 McMaster students, you find that 51% take the bus to McMaster everyday. Construct an approximate 99% confidence interval for the true proportion p and report the upper limit of the interval (assume z₀.₀₀₅​=2.58 ).

User AndASM
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Final answer:

To construct an approximate 99% confidence interval for the true proportion p, calculate p' using the formula (x + 2) / (n + 4) and find the upper limit using the formula p' + z₀.₀₀₅ * sqrt(p' * (1 - p') / (n + 4)). For the given survey results, the upper limit of the confidence interval is 0.592.

Step-by-step explanation:

To construct an approximate 99% confidence interval for the true proportion p, we can use the formula:

p' = (x + 2) / (n + 4)

where x is the number of McMaster students who take the bus (297 * 0.51 = 151.47, rounded to 151) and n is the sample size (297).

Using this formula, we can calculate p' as (151 + 2) / (297 + 4) = 153 / 301 = 0.508.

The upper limit of the confidence interval can be found using the formula:

Upper limit = p' + z₀.₀₀₅ * sqrt(p' * (1 - p') / (n + 4))

Substituting the given values, the upper limit is:

Upper limit = 0.508 + 2.58 * sqrt(0.508 * (1 - 0.508) / 301) = 0.508 + 0.084 = 0.592.

Therefore, the approximate 99% confidence interval for the true proportion p is (0.508, 0.592), and the upper limit of the interval is 0.592.

User Dpfauwadel
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