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)?

a

b

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  

The number that complained about dirty or ill-equipped bathrooms is

 The number that complained about loud or distracting diners at other tables is  

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

Next we obtain the critical value of   from the normal distribution table , the value is  

Considering question a

The sample proportion is mathematically represented as

=>        

=>        

Generally the margin of error is mathematically represented as

The 95% confidence interval is  

Considering question b

The sample proportion is mathematically represented as

=>        

=>        

Generally the margin of error is mathematically represented as

The 95% confidence interval is  

the answer is c hope it

step-by-step explanation: