Mathematics, 23.08.2020 01:01 gabby5155

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)?

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Answer from: lwilhite33105

0.716 < p < 0.744

0.3498 < p < 0.4089

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

Answer from: Quest

the solution is h=10 because you add 9 to each side of the equation.

Answer from: Quest

In all honesty, there really is no “work” per say to show. mainly it’s just using the process of elimination. the question is asking which domain(s) would be the most appropriate for the function c(n). n is the number of observed vehicles so right away, 1, 2, and 3 can be eliminated.why? 1, 2 start in the negatives and therefore it is impossible to observe negative cars. so, #4 would be the best answer to the given problem.

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