Verify that the requirements for constructing a confidence interval about p are satisfied. The sample ▼ a simple random​ sample, the value of ▼ n ModifyingAbove p with caret ModifyingAbove p with caret (1 minus ModifyingAbove p with caret )n ModifyingAbove p with caret (1 minus ModifyingAbove p with caret )ModifyingAbove p with caret n is nothing​, which is ▼ greater than or equal to less than ​10, and the ▼ population proportion population size sample proportion sample size ▼ can be assumed to be is stated to be is stated to not be cannot be assumed to be less than or equal to​ 5% of the ▼ population proportion. population size. sample proportion. sample size. ​(Round to three decimal places as​ needed.)

Respuesta :

Answer:

a) ^p= 0.52

b)

Check explanation.

c) A) We are 95% confident the proportion of adults in the country who believe that televisions are a luxury they could do without is between  [0.49; 0.55]

d)

Check explanation.

e) [0.46; 0.52]

Step-by-step explanation:

Hello!

Original text

A random sample of 1001 adults in a certain large country was asked​ "Do you pretty much think televisions are a necessity or a luxury you could do​ without?" Of the 1001 adults​ surveyed, 521 indicated that televisions are a luxury they could do without. Complete parts​ (a) through​ (e) below.

The variable of interest is:

X: Number of people surveyed that think television is a luxury they can do without, out of  1001.

X~Bi(n;p)

a)

The estimate of the population proportion is the sample proportion (p-hat) you can calculate it as the ratio between the number of successes "x" and the sample size "n":

^p= x/n= 521/1001= 0.52

b)

To  estimate the population proportion using a Confidence interval, the following conditions should be met:

- The sampling method must be randomized.

- The sample proportion must have an approximate normal distribution (E(^p)=p and V(^p)=(pq)/n) and there must be at least 10 expected successes and 10 expected failures.

- The observations need to be independent.

c)

95% CI

[tex]Z_{1-\alpha /2}= Z_{0.975}= 1.96[/tex]

^p ± [tex]Z_{1-\alpha /2}[/tex]*[tex]\sqrt{\frac{p'(1-p')}{n} }[/tex]

0.52 ± 1.96 * [tex]\sqrt{\frac{0.52(1-0.52)}{1001} }[/tex]

[0.4890; 0.5509]≅ [0.49; 0.55]

With a 95% confidence level you'd expect that the population proportion of the people that think that television is a luxury they can do without is contained by the interval  [0.49; 0.55].

d)

The question asks if there is a super majority of people who think television is a luxury they can do without (p>0.60)

The interval of item c) has limits 0.49 and 0.55, meaning that with a 95% confidence you'd expect the true value of the population proportion to be between these two values.

If it is so then it is not likely that the proportion of people who think television is a luxury they can do without to be a super majority.

e)

Using the same sample, you have to calculate a confidence interval for the proportion of people that think television is a necessity.

The number of people who believe television is a necessity is: 1001-512=  489

And the sample proportion: ^p= 489/1001= 0.49

95% CI

[tex]Z_{1-\alpha /2}= Z_{0.975}= 1.96[/tex]

^p ± [tex]Z_{1-\alpha /2}[/tex]*[tex]\sqrt{\frac{p'(1-p')}{n} }[/tex]

0.49 ± 1.96 * [tex]\sqrt{\frac{0.49(1-0.49)}{1001} }[/tex]

[0.459; 0.521] ≅ [0.46; 0.52]

Hope you have a nice day!

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