Question

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. Create a 95% confidence interval for the proportion of complaints that were about the slow speed of the service. Use Excel to create the confidence interval, rounding to four decimal places.

Answer 1

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).

Answer 2

`0.1375` to
`0.3624`

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:
`\widecap{p}-z_{(\alpha)/(2)}\sqrt{\frac{\widecap{p}\widecap{q}}{n}}` to
`\widecap{p}+z_{(\alpha)/(2)}\sqrt{\frac{\widecap{p}\widecap{q}}{n}}`

`\widecap{p}=(x)/(n)`

`\widecap{p}=(57)/(227)`

`\widecap{p}=0.25`

`\widecap{q}=1-\widecap{p}`

`\widecap{q}=1-0.25`

`\widecap{q}=0.75`

z at 95% is 1.96

Substitute the values in the formula :

Confidence for proportion:
`0.25-1.96\sqrt{(0.25 * 0.75)/(57)}` to
`0.25+1.96\sqrt{(0.25 * 0.75)/(57)}`

Confidence for proportion:
`0.1375` to
`0.3624`

Hence 95% confidence interval for the proportion of complaints that were about the slow speed of the service is
`0.1375` to
`0.3624`