Question

In a poll of 125 randomly selected U.S. adults, 67 said they favored a new proposition. Based on this poll, compute a 99% 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. What is the lower limit of the 99% confidence interval? What is the upper limit of the 99% confidence interval?

Answer 1

Sol:

p^=sample proportion=x/n=67/125=0.536

z crit for 99%=2.576

99% confidence interval for p

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

0.536-2.576*sqrt(0.536*(1-0.536)/125),0.536+2.576*sqrt(0.536*(1-0.536)/125)

0.4210968, 0.6509032

99% lower limit=0.42

99% upper limit=0.65

TEXAS INSTRUMENTS TI-83 Plus 1-PropZint. 4.4211..6509)

ANSWER:

99% lower limit=0.42

99% upper limit=0.65