Mathematics, 24.12.2019 20:31 JS28boss

In a survey of 3,803 adults concerning complaints aboutrestaurants, 1,470 complained about dirty or ill-equipped bathrooms and 1,217 complained about loud or distracting diners at other tables. complete parts (a) through (c) below. a. construct a 90% confidence interval estimate of the population proportion of adults who complained about dirty or ill-equipped bathrooms.

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Answer from: bnbjj

a) The 90% confidence interval would be given by (0.374;0.400)

b) The 90% confidence interval would be given by (0.308;0.332)

The manager can be 90% confident that the population proportion of all adults who complained about dirty or II equipped bathrooms lies within in the interval(a)

The manager can be 90% confident that the population proportion of all adults who complained about loud or distracting diners at other tables lies within in the interval (b)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Part a

X=1470 number of people who complained about dirty or ill-equipped bathrooms

n=3803 random sample taken

$\hat p=\frac{1470}{3803}=0.387$ estimated proportion of people who complained about dirty or ill-equipped bathrooms

true population proportion of peple who complained about dirty or ill-equipped bathrooms

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by $\alpha=1-0.9=0.1$ and $\alpha/2 =0.05$ . And the critical value would be given by:

$z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64$

The confidence interval for the mean is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.387 - 1.64\sqrt{\frac{0.387(1-0.387)}{3803}}=0.374$

$0.387 + 1.64\sqrt{\frac{0.387(1-0.387)}{3803}}=0.400$

The 90% confidence interval would be given by (0.374;0.400)

Part b

X=1217 number of people who complained about loud or distracting diners at other tables

n=3803 random sample taken

$\hat p=\frac{1217}{3803}=0.320$ estimated proportion of people who complained about loud or distracting diners at other tables

true population proportion of peple who complained about loud or distracting diners at other tables

$z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64$

The confidence interval for the mean is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.320 - 1.64\sqrt{\frac{0.320(1-0.320)}{3803}}=0.308$

$0.320 + 1.64\sqrt{\frac{0.320(1-0.320)}{3803}}=0.332$

The 90% confidence interval would be given by (0.308;0.332)

Part c

The manager can be 90% confident that the population proportion of all adults who complained about dirty or II equipped bathrooms lies within in the interval (a)

The manager can be 90% confident that the population proportion of all adults who complained about loud or distracting diners at other tables lies within in the interval (b)

Answer from: Quest

the range is 19

step-by-step explanation:

the range is biggest number minus the smallest, so in this case it would be

41 - 22 = 19

Answer from: Quest

answer: approximately 206 bones

i hope this u out a branlist and a would be greatly appreciated :

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