Confidence in Public Schools A 2012 Gallup Poll reported that only 581 out of a total of 2004 U.S. adults said they had a “great deal of confidence” or “quite a lot of confidence” in the public school...

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Confidence in Public Schools A 2012 Gallup Poll reported that only 581 out of a total of 2004
U.S. adults said they had a “great deal of confidence” or “quite a lot of confidence” in the public
school system. This was down 5 percentage points from the previous year. Assume the
conditions for using the CLT are met.
a. Find a 99% confidence interval for the proportion that express a great deal of confidence or
quite a lot of confidence in the public schools, and interpret this interval.
b. Find a 90% confidence interval and interpret it.
c. Find the width of each interval by subtracting the lower proportion from the upper proportion,
and state which interval is wider.
d. How would a 95% interval compare with the others in width?
Answer:
Expressed confidence Total
Sample 581 2004 (n)
Proportion (p) 581/2004 = 0.29
Proportion (q) 1 – 0.29 = 0.71
Formula for Confidence Interval for proportions is given below.
CI = p ± z * √p*q / n
a). Find a 99% confidence interval for the proportion and interpret this interval.
z for 99% is 2.576
Substituting values we get,
CI = p ± z * √p*q / n = 0.29 ± 2.576 √0.29* 0.71 / 2004
= 0.29 ± 2.576 √0.29* 0.71 / 2004 = 0.29 ±...