Answer:
(a) The confidence interval is: 0.0304 ≤ π ≤ 0.0830.
(b) Upper confidence bound = 0.0787
Step-by-step explanation:
(a) The confidence interval for p (proportion) can be calculated as
[tex]p \pm z*\sigma_{p}[/tex]
[tex]\sigma=\sqrt{\frac{\pi*(1-\pi)}{N} }\approx\sqrt{\frac{p(1-p)}{N} }[/tex]
NOTE: π is the proportion ot the population, but it is unknown. It can be estimated as p.
[tex]p=17/300=0.0567\\\\\sigma=\sqrt{\frac{p(1-p)}{N} }=\sqrt{\frac{0.0567(1-0.0567)}{300} }=0.0134[/tex]
For a 95% two-sided confidence interval, z=±1.96, so
[tex]\\LL = p-z*\sigma=0.0567 - (1.96)(0.0134) = 0.0304\\UL =p+z*\sigma= 0.0567 + (1.96)(0.0134) = 0.0830\\\\[/tex]
The confidence interval is: 0.0304 ≤ π ≤ 0.0830.
(b) The confidence interval now has only an upper limit, so z is now 1.64.
[tex]UL =p+z*\sigma= 0.0567 + (1.64)(0.0134) = 0.0787[/tex]
The confidence interval is: -∞ ≤ π ≤ 0.0787.
