In a recent poll^1 of 1000 American adults, the number saying that exercise is an important part of daily life was 753. Use strategy or other technology to find a 90% confidence interval for the proportion of American adults who think exercise is an important part of daily life. The 90% confidence interval is.

The 90% confidence interval is between a lower limit of 0.7306 and an upper limit of 0.7754.

Step-by-step explanation:

Confidence interval for a proportion is given as p +/- margin of error (E)

p is sample proportion = 753/1000 = 0.753

n is sample size = 1000

confidence level = 90%

critical value corresponding to 90% confidence level and infinity degree of freedom is 1.645

E = critical value×sqrt[p(1-p) ÷ n] = 1.645×sqrt[0.753(1-0.753) ÷ 1000] = 1.645×sqrt[1.85991×10^-4] = 1.645×0.0136 = 0.0224

Lower limit of proportion = p - E = 0.753 - 0.0224 = 0.7306

Upper limit of proportion = p + E = 0.753 + 0.0224 = 0.7754

90% confidence interval is (0.7306, 0.7754)

the answer is d

step-by-step explanation:

you can also use mathway to get you answer + steps

a+b=c is your answer