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

Answer:

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

Step-by-step 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  $\frac{\alpha }{2}$ from the normal distribution table , the value is  

         $Z_{\frac{\alpha }{2} } = 1.96$

Considering question a

The sample proportion is mathematically represented as

           $\r p = \frac{e}{n}$

=>        $\r p = \frac{732}{1003}$

=>        $\r p = 0.73$

Generally the margin of error is mathematically represented as

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

          $E = 1.96* \sqrt{ \frac{ 0.73 (1- 0.73)}{1003} }$

          $E = 0.01402$

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

           $\r p = \frac{q}{n}$

=>        $\r p = \frac{381}{1003}$

=>        $\r p = 0.3799$

Generally the margin of error is mathematically represented as

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

          $E = 1.96* \sqrt{ \frac{ 0.3799 (1- 0.3799)}{1003} }$

          $E = 0.0300$

The 95% confidence interval is  

        $\r p - E < p < \r p +E$

        $0.3798 - 0.0300 < p < 0.3798 + 0.0300$

        $0.3498 < p < 0.4089$