Respuesta :
Answer:
[tex]0.1375[/tex] to [tex]0.3624[/tex]
Step-by-step explanation:
Given :The owner of a restaurant is reviewing customer complaints. In a random sample of 227 complaints, 57 complaints were about the slow speed of the service.
To Find :Create a 95% confidence interval for the proportion of complaints that were about the slow speed of the service.
Solution:
n = 227
x = 57
Formula of confidence for proportion: [tex]\widecap{p}-z_{\frac{\alpha}{2}}\sqrt{\frac{\widecap{p}\widecap{q}}{n}}[/tex] to [tex]\widecap{p}+z_{\frac{\alpha}{2}}\sqrt{\frac{\widecap{p}\widecap{q}}{n}}[/tex]
[tex]\widecap{p}=\frac{x}{n}[/tex]
[tex]\widecap{p}=\frac{57}{227}[/tex]
[tex]\widecap{p}=0.25[/tex]
[tex]\widecap{q}=1-\widecap{p}[/tex]
[tex]\widecap{q}=1-0.25[/tex]
[tex]\widecap{q}=0.75[/tex]
z at 95% is 1.96
Substitute the values in the formula :
Confidence for proportion: [tex]0.25-1.96\sqrt{\frac{0.25 \times 0.75}{57}}[/tex] to [tex]0.25+1.96\sqrt{\frac{0.25 \times 0.75}{57}}[/tex]
Confidence for proportion: [tex]0.1375[/tex] to [tex]0.3624[/tex]
Hence 95% confidence interval for the proportion of complaints that were about the slow speed of the service is [tex]0.1375[/tex] to [tex]0.3624[/tex]
Answer:
0.1947-0.3075
The answer for the margin of error, rounded to four decimal places, is z⋆p^(1−p^)n−−−−−−−−√≈0.0564. The confidence interval for the population proportion has a lower limit of A4−A7=0.1947 and an upper limit of A4+A7=0.3075. Thus, the 95% confidence interval for the population proportion of complaints that were about the slow speed of the service, based on this sample, is approximately (0.1947, 0.3075).