Respuesta :
Answer:
a) point estimate for the population proportion of all customers who experienced an interruption
E = 0.029
b)
90% confidence interval for the proportion of all customers who experienced an interruption
(0.11441 , 0.16319)
C)
Size of the sample 'n' = 509
Step-by-step explanation:
Step(i):-
a)
An Internet service provider sampled 540 customers and found that 75 of them experienced an interruption in high-speed service during the previous month.
Sample proportion
[tex]p = \frac{x}{n} = \frac{75}{540} = 0.1388[/tex]
The point estimate of all customers who experienced an interruption
[tex]E = \frac{Z_{\alpha }\sqrt{p(1-p)} }{\sqrt{n} }[/tex]
Level of significance
Z₀.₀₅ = 1.96
[tex]E = \frac{1.96\sqrt{0.138(1-0.1388)} }{\sqrt{540} }[/tex]
E = 0.029
b)
90% confidence interval for the proportion of all customers who experienced an interruption
[tex]( p - Z_{0.10} \sqrt{\frac{p(1-p)}{n} } ,p + Z_{0.10} \sqrt{\frac{p(1-p)}{n} })[/tex]
[tex]( 0.1388 -1.645 \sqrt{\frac{0.1388(1-0.1388)}{540} } ,0.1388 + 1.645 \sqrt{\frac{0.1388(1-0.1388)}{540} })[/tex]
( 0.1388 - 0.02439 , 0.1388 + 0.02439)
(0.11441 , 0.16319)
c)
Margin of error
[tex]M.E = \frac{Z_{0.05} \sqrt{p(1-p)} }{\sqrt{n} }[/tex]
[tex]0.03 = \frac{1.96 \sqrt{0.1388(1-0.1388)} }{\sqrt{n} }[/tex]
√n = 22.58
Squaring on both sides, we get
n = 509
