Question

In a survey of 1,003 adults concerning complaints about restaurants, 732 complained about dirty or ill-equipped bathrooms and 381 complained about loud or distracting diners at other tables.

a. Construct a 95% confidence interval estimate for the population proportion of adults who complained about dirty or ill-equipped bathrooms )
b. Construct a 95% confidence interval estimate for the population proportion of adults who complained about loud or distracting diners at other tables.
c. How would the manager of a chain of restaurants use the results of (a) and (b)?

Answer 1

a

`0.716 < p < 0.744`

b

`0.3498 < p < 0.4089`

c

With the result obtained from a and b the manager can be 95 % confidence that the proportion of the population that complained about dirty or ill-equipped bathrooms are within the interval obtained at a

and that

the proportion of the population that complained about loud or distracting diners at other tables are within the interval obtained at b

Explanation:

From the question we are told that

The sample size is
`n = 1003`

The number that complained about dirty or ill-equipped bathrooms is
`e = 732`

The number that complained about loud or distracting diners at other tables is
`q = 381`

Given that the the confidence level is 95% then the level of significance is mathematically represented as

`\alpha = (100- 95)\%`

`\alpha = 0.05`

Next we obtain the critical value of
`(\alpha )/(2)` from the normal distribution table , the value is

`Z_{(\alpha )/(2) } = 1.96`

Considering question a

The sample proportion is mathematically represented as

`\r p = (e)/(n)`

=>
`\r p = (732)/(1003)`

=>
`\r p = 0.73`

Generally the margin of error is mathematically represented as

`E = Z_{(\alpha )/(2) } * \sqrt{ ( \r p (1- \r p))/(n) }`

`E = 1.96* \sqrt{ ( 0.73 (1- 0.73))/(1003) }`

`E = 0.01402`

The 95% confidence interval is

`\r p - E < p < \r p +E`

`0.73 - 0.01402 < p < 0.73 + 0.01402`

`0.716 < p < 0.744`

Considering question b

The sample proportion is mathematically represented as

`\r p = (q)/(n)`

=>
`\r p = (381)/(1003)`

=>
`\r p = 0.3799`

Generally the margin of error is mathematically represented as

`E = Z_{(\alpha )/(2) } * \sqrt{ ( \r p (1- \r p))/(n) }`

`E = 1.96* \sqrt{ ( 0.3799 (1- 0.3799))/(1003) }`

`E = 0.0300`

The 95% confidence interval is

`\r p - E < p < \r p +E`

`0.3798 - 0.0300 < p < 0.3798 + 0.0300`

`0.3498 < p < 0.4089`