In a poll of 175 randomly selected U.S. adults, 94 said they favored a new propo

Question

In a poll of 175 randomly selected U.S. adults, 94 said they favored a new proposition. Based on this poll, compute a 999% confidence interval for the proportion of all U.S. adults in favor of the proposition (at the time of the poll). Then complete the table below. Carry your intermediate computations to at least three decimal places. Round your answers to two decimal places. (If necessary, consult a list of formulas.) what is the lower limit of the 99% confidence interval? what is the upper limit of the 99% confidence interval? ClearUndo Hel I don't know Next>>

Explanation / Answer

The statistical software output for this problem is:

One sample proportion summary confidence interval:
p : Proportion of successes
Method: Standard-Wald

99% confidence interval results:

Hence,

Lower limit = 0.44

Upper limit = 0.63

Proportion Count Total Sample Prop. Std. Err. L. Limit U. Limit p 94 175 0.53714286 0.037692016 0.44005466 0.63423106

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