Given:
n = 4000, sample size
p = 16% = 0.16, the proportionof theose who watch60 minnures.
Confidence interval (CI) = 95%
The z* parameter is 1.96 for 95% CI, so the confidence interval is
[tex]p \pm z* \sqrt{ \frac{p(1-p)}{n} } [/tex]
That is
0.16 +/- 1.96*√[(0.16*0.81)/4000]
= 0.16 +/- 0.114
= (0.1486, 0.1714)
Answer:
The 95% confidence interval is approximately (15%, 17%) or 15% ≤ p ≤ 17%.
Note:
To correct for the fact that we are using a discrete distribution to match a continuous distribution, it is customary to subtract 0.5/n from the lower limit and to add it to the upper limit.
If this correction is applied, the 95% confidence interval becomes
(0.1486-0.5/4000, 0.1714+0.5/4000)
= (0.1485, 0.1715)
or
15% ≤ p ≤ 17%
The correction does not affect the result in a significant way.
