Question

In a survey of 3,803 adults concerning complaints about​restaurants, 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.

Answer 1

The confidence that the population proportion of adults who complained about the dirty or ill-equipped bathrooms is from 0.374 to 0.400.

Here given as,

X = 1470, n = 3803

`\bar p` (ill-equipped bathrooms) =
`(x)/(a)` =
`(1470)/(3803)` = 0.387

For 90% confidence level,
`z_{(n)/(2) }` = 1.645

The formula as the confidence interval for p =
`\bar p` ±
`z_{(n)/(2) } * {(√(\bar p (1-\bar p)) )/(n) }`

Substituting the values as,

= 0.387+ 1.645 x
`(0.387(1-0,387))/(3.803)`

Solving this get that,

p = 0.387±0.013

Then the,

Lower Bound = 0.387 -0.013 0.374

Upper Bound = 0.387 +0.013 = 0.400

Here the 90% confidence that the population proportion of adults who complained about the dirty or ill-equipped bathrooms is from 0.374 to 0.400.

Answer 2

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)

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{(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=(1470)/(3803)=0.387` estimated proportion of people who complained about dirty or​ ill-equipped bathrooms

`p` 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{(\hat p (1-\hat p))/(n)}`

If we replace the values obtained we got:

`0.387 - 1.64\sqrt{(0.387(1-0.387))/(3803)}=0.374`

`0.387 + 1.64\sqrt{(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=(1217)/(3803)=0.320` estimated proportion of people who complained about loud or distracting diners at other tables

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

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{(\hat p (1-\hat p))/(n)}`

If we replace the values obtained we got:

`0.320 - 1.64\sqrt{(0.320(1-0.320))/(3803)}=0.308`

`0.320 + 1.64\sqrt{(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)