Answer: a 95% confidence interval for the proportion of all customers who have experienced a service interruption: (0.076, 0.127).
Step-by-step explanation:
Let p be the population proportion of all customers who have experienced a service interruption.
Confidence interval for p:[tex]\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex], where [tex]\hat{p}[/tex] = sample proportion, n= sample size, z* = critical z-value.
Given: n = 540
[tex]\hat{p}=\dfrac{55}{540}=0.102[/tex]
z* = 1.96
The required confidence interval:
[tex]0.102\pm (1.96)\sqrt{\dfrac{0.102(1-0.102)}{540}}\\\\=0.102\pm (1.96)(0.013024)\\\\=0.102\pm$0.02552704\\\\=(0.102-$0.02552704, 0.102+0.02552704)\\\\\approx(0.076, 0.127)[/tex]
Hence, a 95% confidence interval for the proportion of all customers who have experienced a service interruption: (0.076, 0.127).
