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

"it's not the heat, it's the humidity". that's a partly valid phrase you may have heard in the summer, but it's actually both. the heat index, also known as the apparent temperature, is what the temperature feels like to the human body when relative humidity is combined with the air temperature. this has important considerations for the human body's comfort. when the body gets too hot, it begins to perspire or sweat to cool itself off. if the perspiration is not able to evaporate, the body cannot regulate its temperature. evaporation is a cooling process. when perspiration is evaporated off the body, it effectively reduces the body's temperature. when the atmospheric moisture content (i.e. relative humidity) is high, the rate of perspiration from the body decreases. in other words, the human body feels warmer in humid conditions. the opposite is true when the relative humidity decreases because the rate of perspiration increases. the body actually feels cooler in arid conditions. there is direct relationship between the air temperature and relative humidity and the heat index, meaning as the air temperature and relative humidity increase (decrease), the heat index increases (decreases).

step-by-step explanation:

75% of the temperatures are below what value? how do you know? 75% of the temperatures are above w

Answer from: Quest

step-by-step explanation:

a triangle's full angle is 180 degrees so 180 - 156 = 24. 24/2=12

Plz asap! ! i will award find the value of x in each case. give reasons to justify your solution

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In a survey of 1,003 adults concerning complaints about restaurants, 732 complained about dirty or i...